🧠:[<think>Okay, let me try to figure this problem out. So, we have a function y = f(x) that's increasing on all real numbers. That means if I take two inputs, say a and b, where a < b, then f(a) ≤ f(b). Wait, actually, since it's an increasing function, maybe it's strictly less? Hmm, the problem just says "increasing," not "strictly increasing," so I think it could be non-decreasing. So, maybe f(a) ≤ f(b) whenever a < b. Got it.Then we have f(m + 3) ≤ f(5). The question is asking for the range of the real number m that satisfies this inequality. Since the function is increasing, if I have f(m + 3) ≤ f(5), that should relate the inputs m + 3 and 5 in some way. Let me recall that for an increasing function, if f(x1) ≤ f(x2), then x1 ≤ x2. Wait, is that right? Let me think.Suppose f is increasing. So, if x1 ≤ x2, then f(x1) ≤ f(x2). Conversely, if f(x1) ≤ f(x2), does that imply x1 ≤ x2? Hmm. If the function is strictly increasing, then yes, because if x1 > x2, then f(x1) > f(x2). But if the function is non-decreasing (so just "increasing"), there might be intervals where it's constant. So, if f(x1) ≤ f(x2), then x1 ≤ x2 or x1 could be in a region where the function is constant. Wait, no. Let's see.Suppose f is increasing. Then f(x1) ≤ f(x2) implies x1 ≤ x2. Is that always true? Let's test with an example. Let's say f(x) is a constant function. Wait, but a constant function isn't increasing. Wait, according to some definitions, a constant function is considered both non-decreasing and non-increasing. But the problem states that f is increasing. So maybe here "increasing" is meant in the non-strict sense? Hmm.Wait, there's some ambiguity in terminology. In some contexts, "increasing" means strictly increasing, and "non-decreasing" means allowing flat regions. But sometimes, especially in older texts, "increasing" can mean non-decreasing. The problem says "increasing function on ℝ". So maybe we have to assume it's non-decreasing? Because if it's strictly increasing, then the solution is straightforward. Let me check.If the function is strictly increasing, then f(m + 3) ≤ f(5) would imply m + 3 ≤ 5, so m ≤ 2. But if the function is non-decreasing, then f(m + 3) ≤ f(5) could imply m + 3 ≤ 5, but we have to account for the possibility that if m + 3 is in a region where the function is constant. Wait, but since the function is defined on all real numbers, and it's increasing (non-decreasing), if there's a point where it's constant over an interval, then that interval would have to be a flat region.But the problem doesn't specify any particular form of the function, just that it's increasing. So, perhaps the safest assumption is that the function is non-decreasing. Therefore, f(m + 3) ≤ f(5) implies m + 3 ≤ 5. Because even if there are flat regions, the inequality f(a) ≤ f(b) would require that a ≤ b. Wait, but if the function is non-decreasing, then f(a) ≤ f(b) does imply a ≤ b only if the function is strictly increasing. Otherwise, if the function is constant over some interval, then even if a > b, you can have f(a) = f(b). Wait, no. Wait, if the function is non-decreasing, then a ≤ b implies f(a) ≤ f(b). But the converse isn't necessarily true. For example, if the function is constant over some interval, then even if a > b in that interval, f(a) = f(b). So, f(a) ≤ f(b) doesn't necessarily imply a ≤ b. So, actually, in that case, you can have f(a) = f(b) even if a > b.Therefore, the problem is a bit trickier. If the function is non-decreasing (which is what "increasing" might mean here), then f(m + 3) ≤ f(5) would imply that either m + 3 ≤ 5 or m + 3 is in a region where the function is constant. But since the function is defined on all real numbers, unless specified otherwise, we can't assume any particular structure. Therefore, the safest approach is to consider that, for a non-decreasing function, if f(x) ≤ f(y), then x ≤ y or x is in a region where f is constant. But without knowing where the function is constant, we can't make any specific conclusions.Wait, but the problem is asking for the range of m such that f(m + 3) ≤ f(5). To guarantee that this inequality holds, regardless of where the function is constant or increasing, the most we can say is that m + 3 must be less than or equal to 5. Because if m + 3 were greater than 5, then since the function is non-decreasing, f(m + 3) would be greater than or equal to f(5). Therefore, to ensure f(m + 3) ≤ f(5), we must have m + 3 ≤ 5, which leads to m ≤ 2.But wait, suppose there is a region around 5 where the function is constant. For example, suppose the function is constant for all x ≥ 4. Then even if m + 3 is greater than 5, say m + 3 = 6, then f(6) = f(5), so the inequality f(6) ≤ f(5) would hold. Therefore, in that case, m could be greater than 2. Hmm, this complicates things.Therefore, the problem is that depending on the function's behavior, the range of m could be different. However, the problem states that the function is increasing on ℝ, but without additional information, we need to find the most restrictive possible range that works for all such functions. Alternatively, maybe since it's given that the function is increasing, and we need to find m such that f(m + 3) ≤ f(5), the answer should be m ≤ 2, because for some functions, if m + 3 > 5, then f(m + 3) would be greater than f(5), so to satisfy the inequality for all possible increasing functions, m must be ≤ 2.Alternatively, if we consider that there might exist some functions where f(m + 3) ≤ f(5) even when m + 3 > 5, but since the problem doesn't specify a particular function, we need to give the range of m that works for all possible increasing functions. Therefore, the only way to ensure f(m + 3) ≤ f(5) for any increasing function is to have m + 3 ≤ 5, hence m ≤ 2.Alternatively, if the function is allowed to be non-decreasing, then the inequality f(m + 3) ≤ f(5) could hold even if m + 3 > 5, provided that the function is constant in that interval. However, since the problem states that the function is increasing, perhaps in this context, increasing is meant to be non-decreasing. But, in some sources, an increasing function is one where x < y implies f(x) < f(y), which is strictly increasing. Wait, this is conflicting.Wait, maybe I should check the exact definition. In mathematics, a function f is called increasing on an interval I if for any x1, x2 in I with x1 < x2, we have f(x1) ≤ f(x2). If the inequality is strict, then it's strictly increasing. So, in many contexts, "increasing" can mean non-decreasing. Therefore, given that, the problem says the function is increasing on ℝ, so non-decreasing. Therefore, the answer would require that m + 3 ≤ 5, so m ≤ 2. Because if the function is non-decreasing, and if m + 3 > 5, then f(m + 3) ≥ f(5), so to have f(m + 3) ≤ f(5), we must have m + 3 ≤ 5.Wait, but if the function is non-decreasing, then if m + 3 > 5, f(m + 3) ≥ f(5). Therefore, f(m + 3) ≤ f(5) would imply that f(m + 3) = f(5). So, even if m + 3 > 5, but f(m + 3) = f(5), then the inequality holds. However, the problem doesn't specify any particular properties about the function other than it's increasing. Therefore, to determine the range of m that must satisfy the inequality regardless of the specific increasing function, we have to consider the worst-case scenario. That is, unless m + 3 ≤ 5, there exists some increasing function where f(m + 3) > f(5). Therefore, to ensure that f(m + 3) ≤ f(5) holds for any increasing function, m must satisfy m + 3 ≤ 5, i.e., m ≤ 2.Alternatively, if the problem is only asking for the range of m such that there exists an increasing function where f(m + 3) ≤ f(5), then the answer could be different. But since the problem states "the function y = f(x) is an increasing function on ℝ, and f(m + 3) ≤ f(5)", it's likely that they want the range of m that must satisfy the inequality for all such functions. Therefore, m must be ≤ 2.Wait, but the problem is phrased as: "Given that the function y = f(x) is an increasing function on ℝ, and f(m + 3) ≤ f(5), then the range of the real number m is." So, it's given that the function is increasing, and given that f(m + 3) ≤ f(5), find m. So, in this case, we don't have to consider all possible increasing functions, but rather, for a particular increasing function, given that f(m + 3) ≤ f(5), find m. Wait, but the problem is a bit ambiguous. Is the function given, and we need to find m such that f(m + 3) ≤ f(5)? But since the function is not specified, we need to find m such that for any increasing function, if f(m + 3) ≤ f(5), then m is in some range. Wait, maybe not.Wait, perhaps the problem is simply saying that if f is an increasing function, then solve the inequality f(m + 3) ≤ f(5). So, in that case, since f is increasing, then f(a) ≤ f(b) if and only if a ≤ b. Wait, but earlier we saw that if f is non-decreasing, then f(a) ≤ f(b) implies a ≤ b or a is in a region where f is constant. But since we don't know where f is constant, the safest conclusion is that a ≤ b. Therefore, maybe in this problem, since f is increasing (non-decreasing), then f(m + 3) ≤ f(5) implies m + 3 ≤ 5, so m ≤ 2.Alternatively, if the function is strictly increasing, then yes, f(m + 3) ≤ f(5) implies m + 3 ≤ 5. But the problem didn't specify strictly increasing, so it could be non-decreasing. But even in the non-decreasing case, if we have f(m + 3) ≤ f(5), that could mean that m + 3 is either less than or equal to 5 or in a flat region. But since the function is arbitrary, the only way to guarantee that f(m + 3) ≤ f(5) is to have m + 3 ≤ 5. Because if m + 3 > 5, then depending on the function, f(m + 3) could be greater than f(5). So, to ensure the inequality holds, m must satisfy m + 3 ≤ 5. Therefore, m ≤ 2.Therefore, the answer is m ≤ 2, so the range is (-∞, 2], which in boxed form is boxed{(-infty, 2]}.Final AnswerThe range of the real number ( m ) is boxed{(-infty, 2]}.</think>Given that the function ( y = f(x) ) is an increasing function on (mathbb{R}), we need to determine the range of the real number ( m ) such that ( f(m + 3) leq f(5) ).1. Understanding the function's behavior: An increasing function (non-decreasing) means that if ( a leq b ), then ( f(a) leq f(b) ). Conversely, if ( f(a) leq f(b) ), it implies ( a leq b ) or ( a ) is in a region where the function is constant. However, since we do not have specific information about the function's constancy, we consider the general case.2. Implications of the inequality: For the inequality ( f(m + 3) leq f(5) ) to hold, we need to ensure that ( m + 3 leq 5 ). This is because if ( m + 3 > 5 ), then for an increasing function, ( f(m + 3) geq f(5) ). To satisfy ( f(m + 3) leq f(5) ), it must be the case that ( m + 3 leq 5 ).3. Solving for ( m ): [ m + 3 leq 5 implies m leq 2 ]Thus, the range of the real number ( m ) is (boxed{(-infty, 2]}).,<think>Okay, so I need to figure out the range of the real number m for which f(m + 3) ≤ f(5), given that the function y = f(x) is increasing on all real numbers. Hmm, let's start by recalling what it means for a function to be increasing. If a function is increasing, then whenever the input x increases, the output f(x) also increases. In other words, if a ≤ b, then f(a) ≤ f(b). Conversely, if f(a) ≤ f(b), then it must be that a ≤ b because the function is increasing. Is that right? Wait, let me think again. If the function is strictly increasing, then yes, f(a) ≤ f(b) implies a ≤ b. But the problem just says "increasing," which might allow for some flat regions where the function doesn't increase. Wait, no, in some contexts, "increasing" can mean non-decreasing, but in others, it might mean strictly increasing. Hmm, the problem says "an increasing function on ℝ," so I need to confirm the definition here. Typically, in calculus, an increasing function is one where if x1 < x2, then f(x1) ≤ f(x2), so non-decreasing. Similarly, a strictly increasing function would have f(x1) < f(x2) whenever x1 < x2. Since the problem just says "increasing," I think it's safe to assume non-decreasing. That is, if m + 3 ≤ 5, then f(m + 3) ≤ f(5). Conversely, if f(m + 3) ≤ f(5), then since the function is increasing, we must have m + 3 ≤ 5. Wait, but if the function is non-decreasing, then the implication is one-way. That is, if a ≤ b, then f(a) ≤ f(b). However, if f(a) ≤ f(b), can we say anything about a and b? If the function is strictly increasing, then yes, f(a) ≤ f(b) implies a ≤ b. But if it's non-decreasing, then f(a) ≤ f(b) could mean that a ≤ b or that a > b but the function is constant between a and b. So in that case, we can't necessarily conclude that a ≤ b. Wait, but the problem says the function is increasing on ℝ. Hmm, maybe in this problem, they consider "increasing" to mean strictly increasing? The problem is in Chinese, but the translation says "increasing function on ℝ," which can sometimes be ambiguous. But in many math problems, unless specified otherwise, "increasing" can mean non-decreasing. Hmm. Let's see. Let's suppose that the function is non-decreasing. Then, if f(m + 3) ≤ f(5), it could be that m + 3 ≤ 5, or that m + 3 > 5 but the function is constant there. But since the function is defined on all real numbers and is increasing, if there is any interval where it's constant, then for those x-values, it's flat. However, the problem doesn't specify anything else about the function. Since we need to find the range of m for which f(m + 3) ≤ f(5), and the function is non-decreasing, the safest approach is to consider that m + 3 must be less than or equal to 5. Because if m + 3 > 5, then since the function is non-decreasing, f(m + 3) ≥ f(5). Therefore, if we have f(m + 3) ≤ f(5), then m + 3 ≤ 5. So m ≤ 2. But wait, but if the function is non-decreasing, then if m + 3 is in a region where the function is constant, even if m + 3 > 5, the function could still have f(m + 3) = f(5). So in that case, m + 3 could be greater than 5 as long as the function is constant from 5 up to m + 3. However, since the function is defined on all real numbers and is increasing (non-decreasing), we don't have any information about where it might be constant. But the problem is asking for the range of m such that f(m + 3) ≤ f(5). To cover all possibilities, m could be any real number such that either m + 3 ≤ 5, or m + 3 > 5 but the function is constant from 5 to m + 3. However, since we don't know where the function is constant, the only guaranteed values of m are those where m + 3 ≤ 5. Because if the function isn't constant beyond 5, then m + 3 > 5 would lead to f(m + 3) ≥ f(5). Therefore, to ensure that f(m + 3) ≤ f(5) holds regardless of the function's behavior (as long as it's non-decreasing), we must have m + 3 ≤ 5. Otherwise, there could exist some functions where f(m + 3) > f(5) if m + 3 > 5. Therefore, the maximum m can be is 2. So the range of m is (-∞, 2]. Wait, but let's check again. Suppose the function is strictly increasing. Then definitely m + 3 ≤ 5, so m ≤ 2. If the function is non-decreasing, but maybe it's constant after 5. For example, suppose f(x) is constant for all x ≥ 5. Then for any m + 3 ≥ 5, f(m + 3) = f(5), so m can be any real number such that m + 3 ≥ 5, which would be m ≥ 2. But in that case, the inequality would still hold. However, the problem is that the function is given to be increasing on all of ℝ. If the function is constant over some interval, it's still considered non-decreasing. But the problem is that we don't know the specific behavior of the function. Therefore, the question is asking for the range of m such that f(m + 3) ≤ f(5) must hold for any such increasing function. So in order for the inequality to hold for all possible increasing functions, we must have m + 3 ≤ 5. Because if m + 3 > 5, there exists some increasing function where f(m + 3) > f(5). For example, take f(x) = x, which is strictly increasing. Then f(m + 3) = m + 3, and f(5) = 5. So the inequality m + 3 ≤ 5 would require m ≤ 2. If m > 2, then m + 3 > 5, so f(m + 3) > f(5), hence the inequality fails. Therefore, to satisfy the inequality for all possible increasing functions, m must satisfy m + 3 ≤ 5, so m ≤ 2. However, if the problem is interpreted as the function being non-decreasing, and we just need the range of m such that there exists an increasing function where f(m + 3) ≤ f(5), then the answer could be different. But in typical math problems like this, unless specified otherwise, when given a condition involving an inequality of function values and stating the function is increasing, they usually want you to use the property that if f is increasing, then f(a) ≤ f(b) implies a ≤ b. But wait, actually, for non-decreasing functions, f(a) ≤ f(b) does not necessarily imply a ≤ b. For example, if f is constant, then f(a) = f(b) for all a and b. So in that case, the inequality would hold for all m, regardless of m + 3. But since the function is increasing (non-decreasing), if we have f(m + 3) ≤ f(5), it could be that m + 3 ≤ 5, or that m + 3 > 5 but f is constant on [5, m + 3]. However, since the problem doesn't specify any particular properties of f other than being increasing, we can't assume that f is constant on any interval. Therefore, to guarantee that f(m + 3) ≤ f(5) must hold for all such functions, m must satisfy m + 3 ≤ 5. Otherwise, there exists a function (like f(x) = x) where f(m + 3) > f(5). Therefore, the correct range for m is m ≤ 2. So I think the answer is m ≤ 2, or in interval notation, (-∞, 2]. Let me check again with an example. Suppose f(x) = x. Then f(m + 3) = m + 3 ≤ f(5) = 5 ⇒ m + 3 ≤ 5 ⇒ m ≤ 2. If f(x) is a constant function, say f(x) = c for all x, then f(m + 3) = c ≤ c = f(5) for any m. So in that case, m can be any real number. But the problem states that the function is increasing. A constant function is considered non-decreasing, but it's not strictly increasing. However, since the problem allows for non-strictly increasing functions, the answer could technically be all real numbers if we consider constant functions. But in the context of such problems, usually, when they say "increasing," they might mean non-decreasing, but the answer expected here is m ≤ 2 because if the function isn't constant, then m has to be ≤ 2. But the problem says "the range of the real number m," so unless there's a restriction on the function being strictly increasing or not, we have to consider both possibilities. Wait, no. The key is that the problem states "the function y = f(x) is an increasing function on ℝ." If "increasing" allows for non-decreasing, then m could be any real number if f is constant. But if f is allowed to be any increasing function (non-decreasing), then the most restrictive condition is when the function is strictly increasing. Therefore, the range of m is all real numbers m such that m ≤ 2. Because if m > 2, there exists an increasing function (like f(x) = x) for which the inequality fails, but if m ≤ 2, then for all increasing functions, the inequality holds. Therefore, the answer should be m ≤ 2. So the range is (-∞, 2]. Therefore, the final answer is m ≤ 2, which in box form is boxed{(-infty, 2]}.Final AnswerThe range of the real number ( m ) is boxed{(-infty, 2]}.</think>Given that the function ( y = f(x) ) is an increasing function on (mathbb{R}), we need to determine the range of the real number ( m ) such that ( f(m + 3) leq f(5) ).1. Understanding the Function Behavior: - An increasing function (non-decreasing) means that if ( a leq b ), then ( f(a) leq f(b) ). - For the inequality ( f(m + 3) leq f(5) ) to hold, we need to consider the implications of the function's increasing nature.2. Implications of the Inequality: - If ( f(m + 3) leq f(5) ), then since the function is non-decreasing, it must be that ( m + 3 leq 5 ). This is because if ( m + 3 > 5 ), the non-decreasing property would imply ( f(m + 3) geq f(5) ), which contradicts the given inequality unless the function is constant in that interval.3. Considering Constant Functions: - If the function is constant for all ( x geq 5 ), then ( f(m + 3) = f(5) ) for any ( m geq 2 ). However, since we need the inequality to hold for all possible increasing functions, we must consider the strictest case where the function is not constant beyond 5.4. Conclusion: - To ensure ( f(m + 3) leq f(5) ) holds for any increasing function, ( m + 3 ) must be less than or equal to 5. This gives ( m leq 2 ).Thus, the range of the real number ( m ) is (boxed{(-infty, 2]}).]

🔑:Since y=f(x) is an increasing function on mathbb{R}, and f(m+3) leq f(5), it follows that m+3 leq 5. Solving this inequality, we get: m leq 2.Therefore, the answer is (-infty, 2].Analysis: According to the property of increasing functions: the larger the function value, the larger the independent variable. By removing the symbol "f", we can find the range of m.Thus, the range of the real number m is boxed{(-infty, 2]}.

🧠:[<think>Okay, so I need to figure out whether proposition p (sin A > sin B) is a sufficient condition, necessary condition, both, or neither for proposition q (A > B) in triangle ABC. Let me start by recalling some basic trigonometry and triangle properties.First, in any triangle, the sum of the angles is 180 degrees. So all angles A, B, and C are between 0 and 180 degrees. Also, the sine function has some properties here. The sine of an angle between 0 and 180 degrees is always positive, which makes sense because all angles in a triangle are in that range. Also, the sine function increases from 0 to 90 degrees and then decreases from 90 to 180 degrees. So, sin θ is increasing in (0°, 90°] and decreasing in [90°, 180°). Therefore, two angles A and B, if they are both acute (less than 90°), then sin A > sin B implies A > B. But if one angle is acute and the other is obtuse (greater than 90°), then the sine of the obtuse angle is actually equal to the sine of its supplementary angle. For example, sin(150°) = sin(30°). So, if A is obtuse and B is acute, even if A > B (since 150° > 30°, for example), sin A would equal sin(180° - A), which is sin(30°) in this case, so sin A would be less than sin B if B is greater than 30°.Wait, let me think through that again. Suppose angle A is 150°, so sin A = sin 150° = 0.5. If angle B is 30°, then sin B = 0.5 as well, so sin A = sin B even though A > B. But in a triangle, if angle A is 150°, then angles B and C must add up to 30°, so maybe angle B can't be 30° in that case. Wait, let's take a concrete example.Suppose triangle ABC has angles A = 150°, B = 20°, and C = 10°. Then sin A = sin 150° = 0.5, sin B = sin 20° ≈ 0.3420. Here, sin A > sin B (0.5 > 0.3420) and A > B (150° > 20°). So in this case, p and q both hold. But what if angle A is 100°, angle B is 80°, then sin A = sin 100° ≈ 0.9848, sin B = sin 80° ≈ 0.9848 as well. Wait, sin 100° is equal to sin 80°, so in this case, even though A > B (100° > 80°), sin A = sin B. So here, q holds but p does not. Therefore, p is not a necessary condition for q because there are cases where q is true but p is not.Alternatively, if we have angle A = 80°, angle B = 100°, then sin A = sin 80° ≈ 0.9848, sin B = sin 100° ≈ 0.9848, so sin A = sin B. But here, A < B (80° < 100°), so in this case, even though A < B, sin A = sin B, so p does not hold here either. Wait, this seems like angles A and B on either side of 90° can have their sines equal. So sin A = sin B if A = B or A + B = 180°, but in a triangle, A + B + C = 180°, so A + B = 180° - C. Since C > 0°, then A + B < 180°, so the case where A + B = 180° is impossible. Therefore, in a triangle, two angles can only have equal sines if the angles themselves are equal or if one is supplementary to the other, but the latter can't happen because their sum is less than 180°. Wait, no, for example, angle A = 100°, angle B = 80°, their sum is 180°, but in a triangle, the sum must be less than 180°, so that can't happen. Wait, in a triangle, if angle A is 100°, then angles B and C must sum to 80°, so angle B can't be 80°. Wait, no, if angle A is 100°, then angles B and C are 80° total. So angle B could be, say, 70°, and angle C 10°, then sin B = sin 70° ≈ 0.9397, and sin A = sin 100° ≈ 0.9848. So here, sin A > sin B even though A > B. Wait, but if angle A is 100°, angle B is 80°, but that would make angle C = 0°, which is impossible. So in a valid triangle, if angle A is 100°, then angle B must be less than 80°, right? So angle B can't be 80°, since angle C has to be positive. Therefore, in a valid triangle, if angle A is obtuse (greater than 90°), then angles B and C must be acute, each less than 90°, and their sum less than 90°. So, in such a case, let's say angle A is 100°, angle B is 80°, but angle C would be 0°, which is impossible. So, angle B can't be 80°, so in reality, angle B has to be less than 80°. Let's take angle A = 100°, angle B = 70°, angle C = 10°. Then sin A = sin 100° ≈ 0.9848, sin B = sin 70° ≈ 0.9397. So sin A > sin B here, and A > B. So in this case, both p and q hold.But what if angle A is 100°, angle B is 30°, angle C = 50°. Wait, sin A = sin 100° ≈ 0.9848, sin B = sin 30° = 0.5. Then sin A > sin B, and A > B. Still holds. So maybe when A is obtuse, sin A is still greater than sin B if A > B. Wait, let's test a case where A is obtuse, but B is acute, and A > B, but sin A < sin B. Let's see. Let angle A = 100°, so sin A ≈ 0.9848. Suppose angle B is 80°, but as we saw, that's not possible. But angle B can be 80° only if angle C is 0°, which isn't allowed. So angle B must be less than 80°. Let's try angle B = 85°, which would make angle C = 180 - 100 -85 = -5°, which is invalid. Wait, so if angle A is 100°, the maximum angle B can be is 80°, but angle C would be 0°, which is invalid. Therefore, angle B must be less than 80°. So angle B can be, say, 70°, angle C = 10°, then sin B = sin 70° ≈ 0.9397, sin A ≈ 0.9848. Still sin A > sin B. Hmm. What if angle A is 150°, angle B = 20°, angle C = 10°. Then sin A = sin 150° = 0.5, sin B = sin 20° ≈ 0.3420. So sin A > sin B, and A > B. Alternatively, angle A = 150°, angle B = 25°, angle C = 5°, sin A = 0.5, sin B ≈ 0.4226. Still sin A > sin B. Wait, but sin A here is only 0.5. So even though angle A is 150°, sin A is 0.5. If angle B is, say, 60°, but angle C would be 180 - 150 -60 = -30°, which is invalid. So angle B can't be 60°. Wait, in any triangle where angle A is obtuse, angles B and C must be acute and their sum less than 90°, so angle B can't be too large. So, in such cases, sin A is equal to sin(180° - A). Since A is obtuse, 180° - A is acute. So sin A = sin(180° - A). So if angle A is 150°, sin A = sin 30° = 0.5. If angle B is 20°, sin B ≈ 0.3420. So 0.5 > 0.3420. If angle B is, say, 25°, sin B ≈ 0.4226, still less than 0.5. Wait, sin 45° ≈ 0.7071, but angle B can't be 45° if angle A is 150°, since angle C would be 180 - 150 -45 = -15°, which is invalid. So angle B can only be up to 30°, if angle A is 150°, angle C must be 30° - angle B. Wait, no, 180 - 150 = 30°, so angles B + C = 30°, so if angle B is 25°, angle C is 5°, etc. So angle B is at most 30°, but even if angle B is 30°, sin B = 0.5, which is equal to sin A (since sin 150° = 0.5). Wait, so if angle A is 150°, and angle B is 30°, angle C is 0°, which is invalid. Therefore, in a valid triangle, angle B must be less than 30° if angle A is 150°, so sin B would be less than 0.5. So sin A = 0.5 > sin B. Therefore, in all valid triangles where angle A is obtuse and angle B is acute, sin A > sin B if and only if angle A > angle B. Wait, but angle A is always greater than angle B in this case because angle A is obtuse and angle B is acute, so A > B is always true. Therefore, in such cases, sin A = sin(180 - A) which is sin(some acute angle). If angle B is acute, then whether sin A > sin B depends on whether 180 - A > B. Wait, for example, if angle A is 150°, then 180 - A = 30°, so sin A = sin 30° = 0.5. Then if angle B is 25°, sin B ≈ 0.4226, so sin A > sin B. If angle B is 15°, sin B ≈ 0.2588, so sin A still > sin B. But if angle B is, say, 31°, but wait, in reality, angle B can't be 31°, since angle A is 150°, angle B + angle C = 30°, so angle B can't exceed 30°. So in all cases where angle A is obtuse and angle B is acute, angle B is less than 180° - angle A. Therefore, 180° - angle A = angle A' (acute angle whose sine is equal to sin A). Then angle B is less than angle A', so sin B < sin A' = sin A. Therefore, in such cases, sin A > sin B. So when angle A is obtuse and angle B is acute, sin A > sin B, and since angle A is obtuse, angle A > angle B. Therefore, in that case, p and q are equivalent.But wait, what if angle A is acute and angle B is acute. For example, angle A = 50°, angle B = 40°, angle C = 90°. Then sin A ≈ 0.7660, sin B ≈ 0.6428, so sin A > sin B, and A > B. If angle A is 50°, angle B = 60°, angle C = 70°, then sin A ≈ 0.7660, sin B ≈ 0.8660, so sin A < sin B, even though A < B. So in this case, if A < B, then sin A < sin B. So in the acute angles, the sine function preserves the inequality. Therefore, if angles A and B are both acute, then sin A > sin B if and only if A > B. If angles A and B are both obtuse, but in a triangle, only one angle can be obtuse. Therefore, in a triangle, only one angle can be obtuse. So angles A and B can't both be obtuse.Therefore, possible cases:1. Both angles A and B are acute: Then sin A > sin B iff A > B. So p and q are equivalent here.2. One angle is obtuse (say A), the other is acute (B): Then as above, since A is obtuse, angle B is acute, so A > B. And since sin A = sin(180 - A), which is an acute angle. Since 180 - A is less than 90°, and angle B is also acute. So, sin A > sin B iff (180 - A) > B. Wait, but (180 - A) = B + C. Since angles B and C are both positive, so (180 - A) = B + C > B. Therefore, (180 - A) > B, which implies sin A = sin(180 - A) > sin B? Wait, no, if (180 - A) > B, then since both (180 - A) and B are acute angles, then sin(180 - A) > sin B. Therefore, sin A = sin(180 - A) > sin B. But (180 - A) is equal to B + C, which is definitely greater than B because C > 0. Therefore, sin(180 - A) = sin A > sin B. Therefore, if angle A is obtuse, then sin A > sin B, and angle A > angle B. Therefore, in this case, p and q are equivalent as well. Wait, but this contradicts the earlier thought where if angle A is 100°, angle B is 80°, but we saw that in reality, angle B can't be 80°, since angle C would be 0°. So in reality, angle B must be less than 80°, so (180 - A) = 80°, so angle B must be less than 80°, so (180 - A) > B. Therefore, sin A = sin(80°) > sin B. So in all cases where angle A is obtuse and angle B is acute, sin A > sin B, and angle A > angle B.Therefore, combining both cases:- If angles A and B are both acute, then sin A > sin B iff A > B.- If angle A is obtuse and angle B is acute, then sin A > sin B and A > B.Therefore, in all cases, sin A > sin B if and only if A > B. Wait, but earlier, I thought of a case where angle A is 100°, angle B is 80°, but realized that angle B can't be 80° because angle C would be 0°, which is invalid. So in a valid triangle, angle B must be less than 80°, so sin A = sin(80°) > sin B (since angle B is less than 80°, so sin B < sin 80°). So in that case, sin A > sin B and A > B.But wait, what if angle A is acute and angle B is obtuse? Wait, but in a triangle, only one angle can be obtuse. So if angle A is acute, angle B can't be obtuse. So only one angle can be obtuse. Therefore, the possible cases are:1. All angles acute: then sin A > sin B iff A > B.2. One angle obtuse (say angle A), others acute: then A > B and sin A > sin B.Therefore, in all cases, p (sin A > sin B) is equivalent to q (A > B). Therefore, p is a necessary and sufficient condition for q. Therefore, the answer is C.Wait, but wait, let me check again. Suppose angle A is 30°, angle B is 150°, but in a triangle, angle B can't be 150° if angle A is 30°, because angle C would be 0°, which is invalid. So angle B can't be 150°. So actually, in a valid triangle, if angle A is 30°, the maximum angle B can be is less than 150°, but since angles sum to 180°, angle B can be up to 150° only if angle C is 0°, which isn't allowed. Therefore, angle B must be acute if angle A is 30°, so in such a case, angle B would be acute, so sin B < sin(90°) = 1. Then if angle A is 30°, sin A = 0.5. So if angle B is, say, 100°, but that's not possible. Wait, only one angle can be obtuse. So if angle A is 30°, then angles B and C sum to 150°, so if angle B is obtuse, say 100°, angle C is 50°, which is valid. Wait, so angle B = 100°, angle A = 30°, angle C = 50°. Then sin A = sin 30° = 0.5, sin B = sin 100° ≈ 0.9848. So here, sin A < sin B even though A < B. Wait, but in this case, angle B is obtuse (100°), angle A is acute (30°). So A < B (30° < 100°), and sin A < sin B (0.5 < 0.9848). Therefore, in this case, q (A > B) is false, and p (sin A > sin B) is also false, so the equivalence holds. Wait, but hold on, if angle A is 30°, angle B is 100°, angle C is 50°, then A < B and sin A < sin B, which is consistent with the equivalence. But what if angle A is 30°, angle B is 20°, angle C is 130°. Then sin A = 0.5, sin B ≈ 0.3420, sin C = sin 130° ≈ 0.7660. Here, angle C is the largest angle (130°), so q would be A > B? Well, in this triangle, angle C is the largest angle. So if we're comparing angles A and B, A is 30°, B is 20°, so A > B (30° > 20°), sin A = 0.5 > sin B ≈ 0.3420, so here p and q both hold. So that's consistent. If angle C is the obtuse angle, and angles A and B are both acute, then comparing A and B, if A > B, then sin A > sin B.But wait, let's take another case where angle C is the obtuse one. Suppose angle A = 40°, angle B = 50°, angle C = 90°. Wait, angle C isn't obtuse here. Let me try angle A = 40°, angle B = 30°, angle C = 110°. Then sin A ≈ 0.6428, sin B = 0.5, sin C ≈ sin 70° ≈ 0.9397. So here, angle C is obtuse. Comparing angles A and B, A > B (40° > 30°), sin A > sin B (0.6428 > 0.5). So p and q hold. Another example: angle A = 50°, angle B = 60°, angle C = 70°. All acute. Then A < B (50° < 60°), sin A ≈ 0.7660 < sin B ≈ 0.8660. So again, p and q align.Wait, but let's see if there's a case where p and q don't align. Suppose we have angle A = 45°, angle B = 135°, but that's impossible because in a triangle, only one angle can be obtuse. So angle B can't be 135° if angle A is 45°, because angle C would have to be 0°, which isn't allowed. So angle B can't be obtuse if angle A is 45°. Therefore, all other angles must be acute if one angle is obtuse. Wait, no, only one angle can be obtuse. So if angle B is obtuse, angles A and C must be acute. So if we have angle B = 100°, angle A = 50°, angle C = 30°, then sin B = sin 100° ≈ 0.9848, sin A = sin 50° ≈ 0.7660, sin C = 0.5. Comparing angles A and C: A > C (50° > 30°), sin A > sin C (0.7660 > 0.5). Comparing angles B and A: B > A (100° > 50°), sin B > sin A (0.9848 > 0.7660). So again, p and q align.Wait, but let's try to find a case where A > B but sin A ≤ sin B. Is that possible? Suppose angle A is 170°, angle B is 5°, angle C is 5°. Then sin A = sin 170° = sin 10° ≈ 0.1736, sin B = sin 5° ≈ 0.0872. Here, A > B (170° > 5°), sin A > sin B (0.1736 > 0.0872). So still holds. If angle A is 170°, angle B is 9°, angle C = 1°, then sin A ≈ sin 10° ≈ 0.1736, sin B ≈ sin 9° ≈ 0.1564. So sin A > sin B. Still holds. If angle A is 91°, angle B is 89°, angle C = 0°, but angle C can't be 0°. So angle B has to be less than 89°. Let's take angle A = 91°, angle B = 88°, angle C = 1°. Then sin A = sin 91° ≈ 0.9998, sin B = sin 88° ≈ 0.9994. Here, A > B (91° > 88°), sin A > sin B (0.9998 > 0.9994). Still holds. So in this case, even though angle A is just barely obtuse and angle B is barely acute, sin A is still slightly larger.Wait, so is there any case where A > B but sin A ≤ sin B? From all these examples, it seems not. Because when A is acute and B is acute, sin increases with angle. When A is obtuse, sin A is sin(180 - A), which is less than sin 90°, but since B is acute and less than 180 - A (because angles sum to 180°, so B < 180 - A), so sin B < sin(180 - A) = sin A. Therefore, in all cases, A > B implies sin A > sin B. Conversely, if sin A > sin B, does that imply A > B? Suppose sin A > sin B.If both angles are acute, then A > B. If one angle is obtuse (say A), then sin A = sin(180 - A). Since 180 - A is acute, and sin A > sin B implies sin(180 - A) > sin B. Since B is acute, and 180 - A is acute, so 180 - A > B, which implies A < 180 - B. But since A is obtuse, 180 - B is greater than 90°, so A < 180 - B. Wait, how does this relate to A > B? Let's see. If sin A > sin B, and A is obtuse, then sin(180 - A) > sin B. Since both 180 - A and B are acute, this implies 180 - A > B, so A < 180 - B. But since A is obtuse, B must be acute. Let me rearrange 180 - A > B: A + B < 180°. But in a triangle, A + B + C = 180°, so C = 180° - A - B. Since C > 0°, then A + B < 180°, so 180 - A > B is always true. Wait, so sin A > sin B is equivalent to 180 - A > B, but since in a triangle, 180 - A = B + C, so B + C > B, which is always true because C > 0°. Wait, this seems confusing. Let's take specific values.Suppose angle A is 100°, angle B is 80°, but angle C would be 0°, which is invalid. So angle B must be less than 80°, say 70°. Then 180 - A = 80°, so sin A = sin 80° ≈ 0.9848. Angle B is 70°, sin B ≈ 0.9397. So sin A > sin B. And since A is 100°, B is 70°, A > B. So sin A > sin B implies A > B. Conversely, if A > B, then in both cases (both acute or A obtuse), sin A > sin B.Therefore, p and q are equivalent. Therefore, the correct answer is C: Necessary and sufficient condition.Wait, but I thought earlier there might be a case where A > B but sin A = sin B. For example, if angle A is 150°, angle B is 30°, but in a triangle, angle C would be 0°, which is invalid. So such a case isn't possible. Similarly, angle A = 135°, angle B = 45°, angle C = 0°, invalid. So in a valid triangle, if angle A is obtuse, angles B and C must sum to 180° - A, which is less than 90°, so each of B and C is less than 90°, and their sines are less than sin(90°) = 1. But sin A = sin(180° - A), which is sin of an acute angle (since 180° - A < 90°). So sin A = sin(acute angle) which is less than 1. But since angle B is acute and less than 180° - A (since angle B < 180° - A, because angle B + angle C = 180° - A, and angle C > 0°), then sin B < sin(180° - A) = sin A. Therefore, sin A > sin B.Therefore, in all valid triangles, A > B if and only if sin A > sin B. Therefore, p is a necessary and sufficient condition for q. So the answer is C.Final Answerboxed{C}</think>Given a triangle ( triangle ABC ), we need to determine the relationship between the propositions ( p ) ( (sin A > sin B) ) and ( q ) ( (A > B) ).1. Properties of the Sine Function: - The sine function is increasing from (0^circ) to (90^circ) and decreasing from (90^circ) to (180^circ). - For angles (A) and (B) in a triangle, both are between (0^circ) and (180^circ).2. Case Analysis: - Both Angles Acute: If (A) and (B) are both acute (less than (90^circ)), (sin A > sin B) if and only if (A > B). This is because the sine function is increasing in this interval. - One Angle Obtuse: If one angle is obtuse (greater than (90^circ)) and the other is acute, the obtuse angle (A) will have (sin A = sin(180^circ - A)), where (180^circ - A) is acute. Since the sum of angles in a triangle is (180^circ), the acute angle (B) must be less than (180^circ - A). Therefore, (sin(180^circ - A) > sin B), implying (sin A > sin B) and (A > B).3. Verification with Examples: - For an obtuse angle (A = 100^circ) and acute angle (B = 70^circ), (sin 100^circ = sin 80^circ approx 0.9848) and (sin 70^circ approx 0.9397), showing (sin A > sin B) and (A > B). - For acute angles (A = 50^circ) and (B = 40^circ), (sin 50^circ approx 0.7660) and (sin 40^circ approx 0.6428), showing (sin A > sin B) and (A > B).4. Conclusion: - In all valid triangles, (sin A > sin B) if and only if (A > B). This is because the sine function's behavior in both acute and obtuse angles, combined with the triangle angle sum property, ensures the equivalence.Thus, proposition (p) is a necessary and sufficient condition for proposition (q).[boxed{C}],<think>Okay, so I need to figure out whether proposition p (sin A > sin B) is a sufficient condition, necessary condition, both, or neither for proposition q (A > B) in triangle ABC. The options are A, B, C, D as listed. Let me start by recalling some trigonometric properties related to triangles and angles.First, in any triangle, the sum of the angles is 180 degrees. So all angles A, B, and C are between 0 and 180 degrees. Also, the sine function is positive in this range since all angles are between 0 and π radians (0 and 180 degrees). Now, the sine function on the interval from 0 to π radians has some specific behavior. It increases from 0 to π/2 (0 to 90 degrees) and then decreases from π/2 to π (90 to 180 degrees). That means that for angles between 0 and 90 degrees, a larger angle will have a larger sine value. However, for angles between 90 and 180 degrees, a larger angle will actually have a smaller sine value. For example, sin(60°) is about 0.866, sin(90°) is 1, and sin(120°) is also about 0.866. So, sin(120°) = sin(60°), even though 120° is larger than 60°.But in a triangle, any angle greater than 90° must mean that the other two angles are less than 90°, because all three angles add up to 180°. So if angle A is greater than 90°, then angles B and C must each be less than 90°, right? Similarly, if angle A is 90°, the other two angles add up to 90°, so they are both acute. So, let's consider the relationship between angles A and B. If A > B, does that necessarily mean sin A > sin B? Well, let's see. If both A and B are acute angles (less than 90°), then since sine is increasing in [0, π/2], if A > B, then sin A > sin B. But if one of them is obtuse (greater than 90°), then since sine is decreasing in [π/2, π], a larger angle would have a smaller sine. Wait, but in a triangle, only one angle can be obtuse. So if angle A is obtuse, then angles B and C must be acute. Similarly, if angle B is obtuse, then A and C are acute. So, if we have A > B, there are two possibilities: either both A and B are acute, in which case A > B implies sin A > sin B, or A is obtuse and B is acute. However, if A is obtuse, then since A > B, and B is acute, how does sin A compare to sin B? Let's take an example.Suppose A is 120°, so sin A = sin(120°) = √3/2 ≈ 0.866. If B is, say, 30°, then sin B = 0.5. So in this case, sin A > sin B. Wait, even though A is obtuse and greater than B, which is acute, sin A is still greater than sin B. Hmm. Wait, but if A is 150°, sin A = 0.5. If B is 15°, sin B ≈ 0.2588. Then sin A (0.5) is still greater than sin B (0.2588). Wait, but what if A is 100°, sin A ≈ 0.9848, and B is 80°, sin B ≈ 0.9848 as well. Wait, no, sin(100°) is sin(80°), which is the same. Wait, hold on, sin(100°) = sin(80°) because sin(π - θ) = sin θ. So if A is 100° and B is 80°, then A > B, but sin A = sin B. So in this case, even though A > B, sin A is equal to sin B. That contradicts the idea that A > B implies sin A > sin B. Wait, so that's a problem. So if angle A is 100° and angle B is 80°, then A > B, but sin A = sin B. Therefore, proposition q (A > B) is true, but proposition p (sin A > sin B) is false. Therefore, q does not necessarily imply p. Therefore, p is not a necessary condition for q? Wait, hold on. Wait, but the question is about p being a condition for q, not the other way around. Wait, let me make sure.Wait, the question is: "p is the (　　) of q." So in other words, what is the relationship between p and q? So, if p is a sufficient condition for q, then p implies q. If p is a necessary condition for q, then q implies p. If it's a necessary and sufficient condition, then p and q imply each other. If neither, then no implication either way.But in the problem statement, p is "sin A > sin B" and q is "A > B". So the question is: is p a sufficient condition for q? That is, does sin A > sin B imply that A > B? Is p a necessary condition for q? That is, does A > B imply that sin A > sin B?So let's analyze both directions.First, suppose that sin A > sin B. Does this imply that A > B?If both angles are acute (i.e., less than 90°), then since sin is increasing in [0, π/2], sin A > sin B would indeed imply A > B. However, if one angle is acute and the other is obtuse, then sin of the obtuse angle is equal to sin(180° - θ), where θ is the acute angle. For example, sin(120°) = sin(60°). So, if angle A is obtuse (say 120°) and angle B is acute (say 60°), then sin A = sin B, but A > B. Wait, but in this case, sin A is equal to sin B, not greater. So, if sin A > sin B, could that happen when one angle is obtuse and the other is acute?Suppose angle A is obtuse, say 100°, and angle B is acute, say 50°. Then sin(100°) ≈ 0.9848, sin(50°) ≈ 0.7660. So here, sin A > sin B, and A > B. However, suppose angle A is 150°, angle B is 30°. Then sin(150°) = 0.5, sin(30°) = 0.5. Wait, so sin A = sin B here, even though A > B. But if A is 160°, sin(160°) = sin(20°) ≈ 0.3420. If B is 10°, sin(10°) ≈ 0.1736. So sin A > sin B here as well, and A > B. Wait, but in the case where A is 150°, B is 30°, sin A = sin B. Hmm, so in that case, even though A > B, sin A is equal to sin B. So in that case, q is true but p is false. Therefore, q does not necessarily imply p. However, when we have sin A > sin B, does that imply A > B?Suppose sin A > sin B. Let's consider different cases.Case 1: Both A and B are acute (less than 90°). Then since sin is increasing in [0, π/2], sin A > sin B implies A > B.Case 2: One angle is obtuse (say A), the other is acute (B). Then sin A = sin(180° - A). Since A > 90°, 180° - A is less than 90°. So sin A = sin(180° - A). So if angle A is obtuse, sin A = sin(acute angle), where the acute angle is 180° - A. So if sin A > sin B, then sin(180° - A) > sin B. Since both 180° - A and B are acute, this implies that 180° - A > B. Therefore, 180° - A > B => 180° > A + B. But since in a triangle, A + B + C = 180°, so A + B = 180° - C. Therefore, 180° > (180° - C) => C > 0°, which is always true. Wait, maybe this approach is getting too convoluted.Alternatively, if A is obtuse, then sin A = sin(180° - A). So sin A > sin B would imply that sin(180° - A) > sin B. Since both 180° - A and B are acute angles, sin is increasing here, so 180° - A > B. Therefore, 180° - A > B => A + B < 180°. But since in a triangle, A + B + C = 180°, so A + B < 180° implies C > 0°, which is always true. Therefore, sin A > sin B when A is obtuse and B is acute would mean that 180° - A > B. But how does this relate to A > B?Wait, if A is obtuse, then A > 90°, and B is acute, so B < 90°, so certainly A > B. However, the condition sin A > sin B translates to sin(180° - A) > sin B, so 180° - A > B. Therefore, sin A > sin B (when A is obtuse) is equivalent to 180° - A > B. But since A > 90°, 180° - A is 90° - (A - 90°), which is less than 90°, so 180° - A is an acute angle. Therefore, if sin A > sin B when A is obtuse, then 180° - A > B. But since A + B + C = 180°, 180° - A = B + C. Therefore, 180° - A > B => B + C > B => C > 0°, which is again always true. Hmm, so maybe my previous reasoning was flawed.Wait, perhaps I need to approach this differently. Let's consider that in a triangle, if we have angles A and B, and we know that sin A > sin B, what can we say about the relationship between A and B?If both angles are acute, then as before, sin A > sin B implies A > B. If one angle is obtuse and the other is acute, then since sin(180° - θ) = sin θ, if the obtuse angle is A, then sin A = sin(180° - A). So sin A > sin B implies sin(180° - A) > sin B. Since 180° - A is acute, and B is acute, this implies 180° - A > B. Therefore, 180° - A > B => 180° > A + B. But since A + B + C = 180°, this implies that C > 0°, which is always true. Therefore, sin A > sin B when A is obtuse and B is acute just tells us that 180° - A > B, but since A is obtuse and B is acute, A is already greater than B. Therefore, sin A > sin B in this case also leads to A > B. Wait, but in the case where A is 150°, B is 30°, sin A = sin(30°) = 0.5, and sin B = 0.5 as well. So here, sin A = sin B, even though A > B. So in that case, even though A is greater than B, their sines are equal. Therefore, if sin A > sin B, then perhaps A > B, but the converse is not true. Because A can be greater than B without sin A being greater than sin B.Wait, so let's check. Suppose sin A > sin B. Then, if both angles are acute, then A > B. If one is obtuse and the other is acute, then since sin A = sin(180° - A), which is less than 90°, and sin B is sin of an acute angle. So if sin A > sin B, then sin(180° - A) > sin B, which implies 180° - A > B (since both angles are acute). Then, since A is obtuse, 180° - A is acute. So 180° - A > B. But since A + B + C = 180°, 180° - A = B + C. Therefore, B + C > B => C > 0°, which is always true. Wait, so perhaps this doesn't really give us a restriction? Hmm.Wait, let's take an example. Let A be 100°, so 180° - A = 80°, and suppose B is 70°. Then sin A = sin(100°) ≈ 0.9848, sin B = sin(70°) ≈ 0.9397. So sin A > sin B. Here, A is 100°, B is 70°, so A > B. Alternatively, take A as 120°, so 180° - A = 60°, and B as 50°. Then sin(120°) = sin(60°) ≈ 0.866, sin(50°) ≈ 0.766. So sin A > sin B, and A > B. But if A is 150°, B is 20°, sin A = sin(30°) = 0.5, sin B = sin(20°) ≈ 0.3420. So here, sin A > sin B (0.5 > 0.3420) and A > B. Wait, even when A is 150°, sin A is 0.5, which is still greater than sin B if B is small enough. Wait, so maybe sin A > sin B in a triangle does imply A > B? Because even if A is obtuse, sin A is sin(180° - A), which is a decreasing function as A increases beyond 90°, but as long as B is smaller than 180° - A, then sin A would be greater than sin B. But since in a triangle, A + B + C = 180°, and if A is obtuse, then B and C are acute. Let me see. Suppose A is 100°, so 180° - A = 80°, and since A + B + C = 180°, then B + C = 80°, so both B and C are less than 80°, right? So if A is 100°, B is, say, 50°, then C is 30°. Then sin A ≈ 0.9848, sin B ≈ 0.7660, so sin A > sin B, and A > B. If A is 100°, B is 70°, then C is 10°, sin B ≈ 0.9397, sin A ≈ 0.9848, so still sin A > sin B. Wait, but if A is 100°, and B is 80°, that would make C = 0°, which is impossible. So B cannot be 80° if A is 100°, because then C would be 0°. So the maximum possible B can be is 80°, but even then, in reality, B has to be less than 80°, since C must be greater than 0°. So in reality, if A is 100°, then B must be less than 80°, so 180° - A is 80°, so sin A = sin(80°), and B is less than 80°, so sin B < sin(80°). Therefore, sin A > sin B. Similarly, if A is 150°, then 180° - A = 30°, so sin A = sin(30°). Then, since A is 150°, B + C = 30°, so B is less than 30°, hence sin B < sin(30°). Therefore, sin A > sin B. Therefore, in any case where A is obtuse and B is acute, sin A > sin B. Wait, but in this case, even if A is obtuse and B is acute, sin A is equal to sin(180° - A), which is an acute angle. Since B is also acute and less than 180° - A (because A is obtuse, so 180° - A is less than 90°, and since A + B + C = 180°, B must be less than 180° - A, because C has to be positive). Therefore, since B < 180° - A, and both angles are acute, sin B < sin(180° - A) = sin A. Therefore, sin A > sin B.Therefore, if A is obtuse and B is acute, then sin A > sin B and A > B. So in that case, sin A > sin B still implies A > B. Similarly, if A and B are both acute, sin A > sin B implies A > B. Therefore, in all cases, sin A > sin B implies A > B. So p implies q. Therefore, p is a sufficient condition for q. But is it a necessary condition?Earlier, I had a case where A is 150°, B is 30°, sin A = sin(30°) = 0.5, and sin B = 0.5. Wait, so in that case, A > B (150° > 30°), but sin A = sin B. Therefore, q is true but p is false. Therefore, q does not imply p. Therefore, p is not a necessary condition for q. Therefore, p is a sufficient condition but not a necessary condition. Therefore, the answer should be option A: Sufficient but not necessary condition.Wait, but let me check another example to be sure. Let's take A = 100°, B = 40°, C = 40°. Then sin A ≈ 0.9848, sin B ≈ 0.6428. So sin A > sin B, and A > B. Another example: A = 45°, B = 30°, sin A ≈ 0.7071, sin B = 0.5. So sin A > sin B and A > B. If A is 135°, B is 15°, sin A ≈ 0.7071, sin B ≈ 0.2588, so sin A > sin B, and A > B. So in all these cases, sin A > sin B implies A > B. However, there are cases where A > B but sin A = sin B. For example, A = 100°, B = 80°, but wait, in that case, C would have to be 0°, which is impossible. So in reality, in a valid triangle, if A is obtuse, then B must be less than 180° - A. So if A is 100°, B must be less than 80°. So in reality, when A is obtuse, B is less than 180° - A, so sin A = sin(180° - A) > sin B, since B is less than 180° - A and both are acute. Therefore, in any valid triangle where A > B, sin A >= sin B? Wait, but earlier I thought of A = 150°, B = 30°, but in that case, sin A = sin(30°) = 0.5, sin B = 0.5. So sin A = sin B. Wait, but in this triangle, A = 150°, B = 30°, C = 0°, which is not possible. Wait, no, C can't be zero. Wait, so in reality, if A = 150°, then B + C = 30°, so B must be less than 30°, so let's take A = 150°, B = 20°, C = 10°. Then sin A = sin(150°) = 0.5, sin B = sin(20°) ≈ 0.3420. So sin A > sin B. If A is 150°, and B is 15°, sin A = 0.5, sin B ≈ 0.2588, so sin A > sin B. Wait, so in any valid triangle where A is obtuse and greater than B, sin A is greater than sin B. Wait, but how about A = 120°, B = 60°, C = 0°, which is invalid. So in a valid triangle, if A is 120°, B + C = 60°, so B must be less than 60°, say B = 50°, C = 10°. Then sin A = sin(120°) ≈ 0.866, sin B ≈ 0.766. So sin A > sin B. Therefore, in any valid triangle, if A > B, then sin A > sin B. Wait, but earlier I thought of a case where A = 100°, B = 80°, but that would require C = 0°, which is impossible. So in actual valid triangles, A > B does imply sin A > sin B? That contradicts my previous thought where A = 150°, B = 30°, but that case isn't possible because C would be 0°. So maybe in any valid triangle, if A > B, then sin A > sin B? Wait, but if A is 100°, and B is 80°, that's not possible because then C = 0°, which is invalid. So perhaps in all valid triangles, if A > B, then sin A > sin B?Wait, let's check. Suppose A and B are both acute. Then since sin is increasing, A > B implies sin A > sin B. If A is obtuse, then B must be acute. Then, since A > B (because A is obtuse and B is acute), and sin A = sin(180° - A). Since A is obtuse, 180° - A is acute. Also, since in the triangle, A + B + C = 180°, so B + C = 180° - A. Since C > 0°, B < 180° - A. Therefore, B is less than 180° - A. Since both B and 180° - A are acute angles, and B < 180° - A, sin B < sin(180° - A) = sin A. Therefore, sin A > sin B. Therefore, in all valid triangles, if A > B, then sin A > sin B. Wait, then that would mean that q implies p. So if A > B, then sin A > sin B, so q implies p. Therefore, p is a necessary condition for q.But earlier, I thought that if A = 150°, B = 30°, then sin A = sin B, but in reality, that would require C = 0°, which is impossible. So in reality, in any valid triangle, if A > B, then sin A > sin B. Therefore, q implies p. Therefore, p is a necessary condition for q. Additionally, if sin A > sin B, then does that imply A > B?Yes, because if sin A > sin B, then either:1. Both angles are acute, so A > B.2. One is obtuse (A) and the other is acute (B). Then, sin A = sin(180° - A) > sin B. Since 180° - A and B are both acute, 180° - A > B. But since A is obtuse, 180° - A = 90° - (A - 90°). But how does 180° - A > B relate to A > B?Wait, since A + B + C = 180°, then 180° - A = B + C. Therefore, B + C > B (since C > 0°), so this is always true. Wait, maybe this isn't the right approach.Alternatively, if A is obtuse and sin A > sin B, then since sin A = sin(180° - A) > sin B, and both 180° - A and B are acute, this implies 180° - A > B. But since A is obtuse, 180° - A < 90°, and B is acute, so B < 90°. Therefore, 180° - A > B implies that 180° - B > A. But since A is obtuse, this would mean 180° - B > A > 90°, so 180° - B > 90°, which implies B < 90°, which is true. However, how does this relate to A > B?Since A is obtuse (>90°), and B is acute (<90°), A is automatically greater than B. Therefore, if sin A > sin B, then regardless of whether A is acute or obtuse, sin A > sin B implies A > B. Therefore, p implies q. Therefore, p is a sufficient condition for q. But earlier, we saw that q also implies p. So if q implies p and p implies q, then p and q are equivalent. Therefore, p is a necessary and sufficient condition for q. But wait, earlier I thought there was a case where q is true but p is false, but that case isn't possible in a valid triangle.Wait, let me check this again. Suppose A = 100°, B = 80°, then C = 0°, which is invalid. Therefore, such a triangle cannot exist. Therefore, in any valid triangle, if A > B, then sin A > sin B. Similarly, if sin A > sin B, then A > B. Therefore, p and q are equivalent. Therefore, p is a necessary and sufficient condition for q. Therefore, the answer is option C. But this contradicts my earlier conclusion. So what's the mistake here?Wait, in the case where A and B are both acute and A > B, sin A > sin B. If A is obtuse and B is acute, A > B and sin A = sin(180° - A). Since 180° - A < 90°, and B is acute, but how does sin(180° - A) compare to sin B? Since A is obtuse, B must be less than 180° - A (because A + B + C = 180°, so B + C = 180° - A, and since C > 0°, B < 180° - A). Therefore, B < 180° - A, and both angles B and 180° - A are acute. Therefore, since B < 180° - A, sin B < sin(180° - A) = sin A. Therefore, sin A > sin B. Therefore, in any valid triangle, A > B implies sin A > sin B. Therefore, q implies p.Similarly, if sin A > sin B, then in the case where both angles are acute, A > B. If one angle is obtuse (A) and the other is acute (B), then sin A > sin B implies sin(180° - A) > sin B. Since 180° - A and B are both acute, this implies 180° - A > B. But since A is obtuse, A > 90°, so 180° - A < 90°, so B < 180° - A < 90°, so A > 90° > B. Therefore, A > B. Therefore, in either case, sin A > sin B implies A > B. So p implies q.Therefore, p and q imply each other. Therefore, p is a necessary and sufficient condition for q. Therefore, the answer is C.Wait, but this contradicts my initial example where I thought of A = 150°, B = 30°, but in reality, that would require C = 0°, which is impossible. Therefore, in any valid triangle, A > B if and only if sin A > sin B. Therefore, p is equivalent to q, so the answer is C. Therefore, the answer should be option C: Necessary and sufficient condition.But wait, let's check an example where A and B are both obtuse. Wait, in a triangle, only one angle can be obtuse. Therefore, if A is obtuse, B must be acute, and vice versa. So in all cases, A > B would either have both acute, with A > B leading to sin A > sin B, or A obtuse and B acute, leading to A > B and sin A > sin B. Therefore, indeed, q implies p. And p implies q as well. Therefore, they are equivalent. Therefore, the correct answer is C.But why was I initially thinking that there might be a case where A > B but sin A = sin B? Because if A and B are supplementary angles, then sin A = sin B. But in a triangle, two angles cannot be supplementary unless the third angle is zero, which is impossible. Therefore, in any valid triangle, A and B cannot be supplementary. Therefore, sin A will never equal sin B unless A = B. But if A > B, then sin A > sin B. Therefore, in a valid triangle, A > B if and only if sin A > sin B. Therefore, the answer is C.Final Answerboxed{C}</think>Given in (triangle ABC), angles (A), (B), and (C) are the three interior angles. We need to determine the relationship between the propositions (p): (sin A > sin B) and (q): (A > B).1. Behavior of the Sine Function: - The sine function is increasing from (0) to (90^circ) and decreasing from (90^circ) to (180^circ). - In a triangle, the sum of angles is (180^circ), so each angle is between (0^circ) and (180^circ).2. Analysis of Propositions: - Proposition (p): (sin A > sin B) - Proposition (q): (A > B)3. Case Analysis: - Both angles are acute (less than (90^circ)): - If (A > B), then (sin A > sin B) because sine is increasing in ([0^circ, 90^circ]). - One angle is obtuse (greater than (90^circ)) and the other is acute (less than (90^circ)): - If (A) is obtuse and (B) is acute, (sin A = sin(180^circ - A)). Since (180^circ - A) is acute and (B < 180^circ - A) (as (A + B + C = 180^circ)), (sin(180^circ - A) > sin B). Thus, (sin A > sin B) implies (A > B).4. Implications: - If (A > B): - In all valid triangles, (A > B) implies (sin A > sin B) because if (A) is obtuse, (B) must be acute, and (sin A = sin(180^circ - A)) which is greater than (sin B). - If (sin A > sin B): - If both angles are acute, (sin A > sin B) implies (A > B). If one angle is obtuse, (sin A > sin B) also implies (A > B) because the obtuse angle is greater than the acute angle.5. Conclusion: - (A > B) if and only if (sin A > sin B). Therefore, (p) is a necessary and sufficient condition for (q).(boxed{C})]

🔑:In triangle ABC, according to the Law of Sines, we have frac{a}{sin A} = frac{b}{sin B} = k > 0, where a > b Leftrightarrow ksin A > ksin B Leftrightarrow sin A > sin B. And a > b Leftrightarrow A > B. Therefore, in triangle ABC, sin A > sin B Leftrightarrow A > B, which means p Leftrightarrow q. Thus, p is a necessary and sufficient condition for q. Hence, the correct choice is: boxed{C}. In triangle ABC, by the Law of Sines frac{a}{sin A} = frac{b}{sin B}, a > b Leftrightarrow sin A > sin B. And a > b Leftrightarrow A > B. This allows us to conclude. This question tests knowledge of the Law of Sines, the relationship between the sides of a triangle, simple logic determination methods, and examines reasoning and computational skills. It is considered a medium-level question.

🧠:[<think>Okay, let me try to work through this problem. So, we have triangle ABC with AB = 2, BC = 3, and there's a point D inside the triangle such that CD = 2 and angle ADC = 180° minus angle B. We need to prove that 1/4 < cos B < 3/4. Hmm, okay. Let me start by drawing a rough sketch to visualize the problem.First, triangle ABC. Let me note the sides: AB is 2 units, BC is 3 units. The third side, AC, isn't given, so maybe we'll need to relate it using the Law of Cosines or something later. Point D is inside the triangle, with CD = 2. Also, angle ADC is equal to 180° minus angle B. That seems important. So angle ADC and angle B are supplementary. Let me recall that in triangle ABC, angle B is at vertex B, between sides AB and BC. So angle B is angle ABC. Then angle ADC is 180° minus that. So if angle B is θ, then angle ADC is 180° - θ. I need to find the range for cos B. Since the problem gives inequalities, I might need to set up some equations using the Law of Cosines or Law of Sines, and then manipulate them to find bounds on cos B.Let me start by considering triangle ABC. Let's denote angle B as θ. Then, by the Law of Cosines in triangle ABC:AC² = AB² + BC² - 2 * AB * BC * cos θAC² = 2² + 3² - 2 * 2 * 3 * cos θAC² = 4 + 9 - 12 cos θAC² = 13 - 12 cos θSo AC = sqrt(13 - 12 cos θ). That's the length of side AC.Now, point D is inside triangle ABC such that CD = 2 and angle ADC = 180° - θ. Let me consider triangle ADC. In triangle ADC, we have CD = 2, angle at D is 180° - θ, and point D is inside the triangle. Let me try to apply the Law of Sines or Law of Cosines here.Wait, in triangle ADC, we can use the Law of Sines. Let's denote:In triangle ADC:- angle at D: angle ADC = 180° - θ- sides: CD = 2, AD = ?, AC = sqrt(13 - 12 cos θ)So by Law of Sines:AD / sin(angle at C) = CD / sin(angle at A) = AC / sin(angle ADC)But I'm not sure about the angles at A and C in triangle ADC. Maybe I need to denote some variables.Alternatively, maybe coordinates would help here. Let me try coordinate geometry. Let's place point B at the origin (0, 0), point C at (3, 0) since BC = 3. Then point A is somewhere such that AB = 2. Let me find coordinates of point A.Let me place coordinate system with B at (0, 0). Then, since BC = 3, let's place C at (3, 0). Then, point A is somewhere in the plane such that AB = 2. Let me assume coordinates of A are (x, y). Then distance from A to B is sqrt(x² + y²) = 2, so x² + y² = 4.Also, the coordinates of D are such that D is inside triangle ABC, CD = 2, and angle ADC = 180° - θ, where θ is angle at B.Wait, angle ADC is 180° - angle B. So angle ADC is supplementary to angle B. Hmm. Maybe we can use some trigonometric identities here.Alternatively, since angle ADC = 180° - θ, then in quadrilateral ABDC (if that's a quadrilateral), but D is inside the triangle, so maybe not. Alternatively, maybe we can relate triangles ABC and ADC.Alternatively, since angle ADC is supplementary to angle B, perhaps there's a cyclic quadrilateral involved? Wait, in cyclic quadrilaterals, opposite angles are supplementary, but here angle ADC and angle B are supplementary. Hmm, but angle B is not part of the quadrilateral unless D is connected to B. Maybe not directly. Hmm.Alternatively, consider triangle ADC. If angle ADC = 180° - θ, then the triangle ADC has an angle of 180° - θ. Let me consider the Law of Cosines in triangle ADC.In triangle ADC:AC² = AD² + CD² - 2 * AD * CD * cos(angle ADC)But angle ADC = 180° - θ, so cos(angle ADC) = cos(180° - θ) = -cos θ.Therefore,AC² = AD² + CD² - 2 * AD * CD * (-cos θ)AC² = AD² + 4 + 4 AD cos θBut earlier, we found AC² = 13 - 12 cos θ. So,AD² + 4 + 4 AD cos θ = 13 - 12 cos θRearranging:AD² + 4 AD cos θ + 4 = 13 - 12 cos θThen,AD² + 4 AD cos θ = 9 - 12 cos θHmm. So AD² + 4 AD cos θ + 12 cos θ - 9 = 0Hmm, this is a quadratic in terms of AD, but AD is a variable here. Maybe we can relate AD to other parts of the triangle.Alternatively, perhaps considering triangle ABD or triangle BDC. Wait, but we don't have much information about BD or AD. Maybe another approach.Wait, let's think about coordinates again. Let me place point B at (0, 0), point C at (3, 0). Then point A is somewhere with AB = 2, so coordinates (x, y) with x² + y² = 4. Then, point D is inside triangle ABC with CD = 2, so coordinates of D must satisfy distance from C (3,0) is 2. So D lies on a circle centered at (3,0) with radius 2. Also, angle ADC = 180° - θ. But angle ADC is the angle at D between points A, D, C. So in coordinates, if we can express the angle at D in terms of coordinates of A and D, maybe we can set up some equations.This might get complicated, but let's try.Let me denote coordinates:- B: (0, 0)- C: (3, 0)- A: (x, y) with x² + y² = 4- D: (a, b) inside triangle ABC, so (a, b) is such that it's inside the triangle formed by (0,0), (3,0), and (x,y). Also, distance from D to C is 2: sqrt((a - 3)^2 + b^2) = 2 => (a - 3)^2 + b^2 = 4.Angle ADC is 180° - θ. Let's express angle ADC in terms of coordinates. The angle at D between points A, D, C. To compute this angle, we can use the vectors DA and DC.Vector DA is (a - x, b - y)Vector DC is (a - 3, b - 0) = (a - 3, b)The angle between DA and DC is 180° - θ. So, the cosine of angle ADC is equal to cos(180° - θ) = -cos θ.Using the dot product formula:cos(angle ADC) = (DA · DC) / (|DA| |DC|)But |DC| is 2, since CD = 2. So,- cos θ = [ (a - x)(a - 3) + (b - y)b ] / ( |DA| * 2 )Therefore,- cos θ = [ (a - x)(a - 3) + b(b - y) ] / ( 2 |DA| )But |DA| is the distance from D to A: sqrt( (a - x)^2 + (b - y)^2 )This seems complicated. Maybe there's a better way. Alternatively, since we have coordinates, perhaps we can set up some equations.But perhaps coordinate geometry is too involved here. Let me think again.Earlier, we had from triangle ABC: AC² = 13 - 12 cos θFrom triangle ADC: AC² = AD² + 4 + 4 AD cos θTherefore, setting equal:AD² + 4 + 4 AD cos θ = 13 - 12 cos θRearranged:AD² + 4 AD cos θ + 12 cos θ - 9 = 0This is a quadratic equation in AD:AD² + (4 cos θ) AD + (12 cos θ - 9) = 0Let me denote AD as t. Then,t² + (4 cos θ) t + (12 cos θ - 9) = 0For real solutions, the discriminant must be non-negative:(4 cos θ)^2 - 4 * 1 * (12 cos θ - 9) ≥ 0Calculating discriminant:16 cos² θ - 48 cos θ + 36 ≥ 0Divide both sides by 4:4 cos² θ - 12 cos θ + 9 ≥ 0This factors as (2 cos θ - 3)^2 ≥ 0, which is always true because a square is non-negative. So the quadratic equation has real solutions for any θ. Hmm, that doesn't help in restricting θ. But maybe there's another condition?Wait, but AD is a length inside triangle ABC. So AD must be positive and also less than some maximum value. Since D is inside the triangle, AD must be less than AB + BD, but I'm not sure. Alternatively, perhaps AD must satisfy triangle inequalities with other sides.Alternatively, since D is inside triangle ABC, the position of D is constrained. Maybe we can use the fact that D is inside the triangle to get inequalities on AD.Alternatively, maybe there's another relation we can use. Let me think.In triangle ABC, point D is inside the triangle, so maybe applying Stewart's theorem or something like that. Hmm, Stewart's theorem relates the lengths of a side of a triangle to the length of a cevian. But we don't know where D is exactly, other than CD = 2.Alternatively, since CD = 2, and BC = 3, then the distance from C to D is 2, which is less than BC, so point D lies somewhere along the circle with radius 2 centered at C, inside triangle ABC.Alternatively, perhaps use coordinates again. Let me proceed with coordinates.Since I placed B at (0,0), C at (3,0), and A at (x,y) with x² + y² = 4. Let me assume coordinates for point A. Wait, but AB = 2, so coordinates of A satisfy x² + y² = 4, but we also need to relate AC. Wait, AC = sqrt(13 - 12 cos θ). Hmm, but θ is angle at B, which is the angle between BA and BC. So θ is the angle at the origin between points A (x,y) and C (3,0). Therefore, we can compute cos θ using the dot product.Indeed, vectors BA and BC are vectors from B to A and from B to C. So BA is (x, y), BC is (3, 0). The angle between BA and BC is θ, so:cos θ = (BA · BC) / (|BA| |BC|) = (3x + 0*y) / (2 * 3) = (3x)/6 = x/2Therefore, cos θ = x/2. So x = 2 cos θ. Since x² + y² = 4, then (2 cos θ)^2 + y^2 = 4 => 4 cos² θ + y² = 4 => y² = 4 - 4 cos² θ => y = 2 sin θ (assuming y is positive, since the triangle is drawn above the x-axis). So coordinates of A are (2 cos θ, 2 sin θ). Okay, that's helpful.So point A is (2 cos θ, 2 sin θ). Now, point D is inside triangle ABC, with CD = 2 and angle ADC = 180° - θ. Let me write coordinates for D.Since CD = 2, and point C is at (3,0), coordinates of D must satisfy (a - 3)^2 + b^2 = 4. So D lies on the circle centered at (3,0) with radius 2. Also, D is inside triangle ABC. Since triangle ABC has vertices at (0,0), (3,0), and (2 cos θ, 2 sin θ). So the triangle is formed by these three points. We need to find D inside this triangle on the circle (a - 3)^2 + b^2 = 4 such that angle ADC = 180° - θ.Let me compute angle ADC. Angle at D between points A, D, C. Let's use vectors again. Vector DA = (2 cos θ - a, 2 sin θ - b), vector DC = (3 - a, 0 - b) = (3 - a, -b). The angle between DA and DC is 180° - θ. So the cosine of this angle is equal to cos(180° - θ) = -cos θ.Using the dot product formula:cos(angle ADC) = (DA · DC) / (|DA| |DC|) = -cos θCompute DA · DC:(2 cos θ - a)(3 - a) + (2 sin θ - b)(-b)= (2 cos θ * 3 - 2 cos θ * a - 3a + a²) + (-2 sin θ * b + b²)= 6 cos θ - 2a cos θ - 3a + a² - 2b sin θ + b²|DA| is sqrt( (2 cos θ - a)^2 + (2 sin θ - b)^2 )|DC| is sqrt( (3 - a)^2 + (-b)^2 ) = sqrt( (3 - a)^2 + b^2 ) = 2, since CD = 2.Therefore,[6 cos θ - 2a cos θ - 3a + a² - 2b sin θ + b²] / [ |DA| * 2 ] = -cos θMultiply both sides by 2 |DA|:6 cos θ - 2a cos θ - 3a + a² - 2b sin θ + b² = -2 |DA| cos θBring all terms to left side:6 cos θ - 2a cos θ - 3a + a² - 2b sin θ + b² + 2 |DA| cos θ = 0Hmm, this seems complicated. But maybe we can substitute |DA|. Let's compute |DA|:|DA| = sqrt( (2 cos θ - a)^2 + (2 sin θ - b)^2 )Square both sides:|DA|² = (2 cos θ - a)^2 + (2 sin θ - b)^2 = 4 cos² θ - 4a cos θ + a² + 4 sin² θ - 4b sin θ + b²= 4 (cos² θ + sin² θ) - 4a cos θ - 4b sin θ + a² + b²= 4(1) - 4a cos θ - 4b sin θ + a² + b²= 4 - 4a cos θ - 4b sin θ + a² + b²But since D is on the circle centered at (3,0) with radius 2, we have (a - 3)^2 + b² = 4, which expands to a² - 6a + 9 + b² = 4 => a² + b² = 6a - 5Substitute a² + b² = 6a -5 into |DA|²:|DA|² = 4 - 4a cos θ - 4b sin θ + (6a -5)= 4 + 6a -5 -4a cos θ -4b sin θ= ( -1 + 6a ) -4a cos θ -4b sin θHmm, not sure if that helps. Let's go back to the equation:6 cos θ - 2a cos θ - 3a + a² - 2b sin θ + b² + 2 |DA| cos θ = 0We can use a² + b² = 6a -5 here as well. Substitute a² + b² =6a -5:6 cos θ -2a cos θ -3a + (6a -5) -2b sin θ + 2 |DA| cos θ = 0Simplify:6 cos θ -2a cos θ -3a +6a -5 -2b sin θ + 2 |DA| cos θ = 0Combine like terms:6 cos θ -2a cos θ +3a -5 -2b sin θ + 2 |DA| cos θ = 0Hmm. Let me group terms with cos θ:(6 cos θ -2a cos θ + 2 |DA| cos θ) + (3a -5 -2b sin θ) = 0Factor cos θ:cos θ (6 -2a + 2 |DA| ) + (3a -5 -2b sin θ) = 0This still looks complicated. Maybe another approach.Wait, let's recall that point D is inside triangle ABC. So the coordinates of D must satisfy certain inequalities. For example, being inside the triangle, the barycentric coordinates should be positive, but maybe that's too vague.Alternatively, since we know coordinates of A, B, C, we can express the condition that D is inside the triangle ABC by checking if D is on the same side of AB as C, on the same side of BC as A, and on the same side of AC as B. But this might get too involved.Alternatively, perhaps using parametric coordinates. Let me parametrize point D on the circle centered at C (3,0) with radius 2. So we can write D as (3 + 2 cos φ, 0 + 2 sin φ), where φ is some angle. Wait, but since D is inside the triangle, φ is constrained. For example, φ would be between 90° and 180°, maybe? Let me check.If D is inside the triangle ABC, which has vertices at B(0,0), C(3,0), and A(2 cos θ, 2 sin θ). So point A is somewhere in the plane with distance 2 from B. Depending on θ, the triangle could be acute or obtuse. Hmm, but parametrizing D as (3 + 2 cos φ, 2 sin φ) would place it on the circle, but need to ensure it's inside the triangle. Maybe that's a good approach.Let me try this. Let me set D as (3 + 2 cos φ, 2 sin φ). Wait, actually, the standard parametrization of a circle centered at (3,0) with radius 2 is (3 + 2 cos φ, 0 + 2 sin φ). So D = (3 + 2 cos φ, 2 sin φ). Now, φ is the angle parameter. Now, since D is inside triangle ABC, the coordinates must satisfy certain conditions. For example, the y-coordinate must be positive but less than the height of the triangle.But the height of triangle ABC from A is y-coordinate of A, which is 2 sin θ. So 2 sin φ < 2 sin θ, so sin φ < sin θ. Hmm, but this might not necessarily hold because D is inside the triangle. Alternatively, the position of D is constrained by the edges of the triangle.Alternatively, maybe it's better to proceed with this parametrization and see where it leads.So D = (3 + 2 cos φ, 2 sin φ). Now, we need to compute angle ADC and set it equal to 180° - θ.First, compute vectors DA and DC.Point A is (2 cos θ, 2 sin θ), point D is (3 + 2 cos φ, 2 sin φ), point C is (3,0).Vector DA = A - D = (2 cos θ - 3 - 2 cos φ, 2 sin θ - 2 sin φ)Vector DC = C - D = (3 - (3 + 2 cos φ), 0 - 2 sin φ) = (-2 cos φ, -2 sin φ)The angle between vectors DA and DC is 180° - θ.So, using the dot product:DA · DC = |DA| |DC| cos(180° - θ) = - |DA| |DC| cos θCompute DA · DC:(2 cos θ - 3 - 2 cos φ)(-2 cos φ) + (2 sin θ - 2 sin φ)(-2 sin φ)= (-2 cos φ)(2 cos θ - 3 - 2 cos φ) + (-2 sin φ)(2 sin θ - 2 sin φ)= -4 cos φ cos θ + 6 cos φ + 4 cos² φ - 4 sin φ sin θ + 4 sin² φ= -4 cos φ cos θ + 6 cos φ + 4 (cos² φ + sin² φ) - 4 sin φ sin θSince cos² φ + sin² φ = 1:= -4 cos φ cos θ + 6 cos φ + 4 * 1 - 4 sin φ sin θ= -4 cos φ cos θ + 6 cos φ + 4 - 4 sin φ sin θCombine the terms with cos φ and sin φ:= -4 (cos φ cos θ + sin φ sin θ) + 6 cos φ + 4Notice that cos φ cos θ + sin φ sin θ = cos(θ - φ). So:= -4 cos(θ - φ) + 6 cos φ + 4On the other hand, |DA| is the magnitude of vector DA:|DA| = sqrt( (2 cos θ - 3 - 2 cos φ)^2 + (2 sin θ - 2 sin φ)^2 )Similarly, |DC| is the magnitude of DC, which is sqrt( (-2 cos φ)^2 + (-2 sin φ)^2 ) = sqrt(4 cos² φ + 4 sin² φ) = sqrt(4) = 2Therefore, the dot product equation becomes:-4 cos(θ - φ) + 6 cos φ + 4 = - |DA| * 2 * cos θBut we need to compute |DA|. Let's compute |DA| squared:|DA|² = (2 cos θ - 3 - 2 cos φ)^2 + (2 sin θ - 2 sin φ)^2Expand the first term:= (2 cos θ - 3 - 2 cos φ)^2 + (2 sin θ - 2 sin φ)^2= [4 cos² θ + 9 + 4 cos² φ - 12 cos θ - 8 cos θ cos φ + 12 cos φ] + [4 sin² θ - 8 sin θ sin φ + 4 sin² φ]Combine terms:= 4 cos² θ + 9 + 4 cos² φ - 12 cos θ - 8 cos θ cos φ + 12 cos φ + 4 sin² θ - 8 sin θ sin φ + 4 sin² φGroup like terms:= 4 (cos² θ + sin² θ) + 4 (cos² φ + sin² φ) + 9 - 12 cos θ + 12 cos φ - 8 (cos θ cos φ + sin θ sin φ )= 4(1) + 4(1) + 9 -12 cos θ +12 cos φ -8 cos(θ - φ )= 4 + 4 + 9 -12 cos θ +12 cos φ -8 cos(θ - φ )= 17 -12 cos θ +12 cos φ -8 cos(θ - φ )Therefore, |DA| = sqrt(17 -12 cos θ +12 cos φ -8 cos(θ - φ ) )This is getting quite involved. Let's recap the equation we had from the dot product:-4 cos(θ - φ) + 6 cos φ + 4 = -2 |DA| cos θSubstituting |DA|:-4 cos(θ - φ) + 6 cos φ + 4 = -2 sqrt(17 -12 cos θ +12 cos φ -8 cos(θ - φ )) cos θThis equation relates θ and φ. However, solving this equation seems extremely complicated. Maybe there's a smarter approach.Wait, let's recall that angle ADC = 180° - θ. Let me consider triangle ADC again. In triangle ADC, angle at D is 180° - θ, sides CD = 2, and AC = sqrt(13 - 12 cos θ). If we can express AD in terms of θ, maybe we can use the Law of Sines.Law of Sines in triangle ADC:AD / sin(angle at C) = CD / sin(angle at A) = AC / sin(angle ADC)So,AD / sin(angle at C) = 2 / sin(angle at A) = sqrt(13 - 12 cos θ) / sin(180° - θ)But sin(180° - θ) = sin θ, so:sqrt(13 - 12 cos θ) / sin θ = 2 / sin(angle at A)Therefore,sin(angle at A) = 2 sin θ / sqrt(13 - 12 cos θ)Similarly,AD = [sqrt(13 - 12 cos θ) / sin θ] * sin(angle at C)But angle at C in triangle ADC is angle ACD. Let me denote angle at C as γ. Similarly, angle at A as α.In triangle ADC, angles sum to 180°, so:angle ADC (180° - θ) + angle at C (γ) + angle at A (α) = 180°Therefore,γ + α = θSo γ = θ - αFrom the Law of Sines:AD / sin γ = 2 / sin αSo,AD = 2 sin γ / sin α = 2 sin (θ - α) / sin αBut from earlier, sin α = 2 sin θ / sqrt(13 - 12 cos θ)Therefore,AD = 2 sin (θ - α) / (2 sin θ / sqrt(13 - 12 cos θ)) )= sqrt(13 - 12 cos θ) * sin (θ - α) / sin θBut α is angle at A in triangle ADC, which is different from angle at A in triangle ABC. Wait, maybe this isn't helping. Let me think differently.Alternatively, using the Law of Cosines in triangle ADC again. We have:AC² = AD² + CD² - 2 * AD * CD * cos(angle ADC)Which gives:13 - 12 cos θ = AD² + 4 - 2 * AD * 2 * (-cos θ)So:13 - 12 cos θ = AD² + 4 + 4 AD cos θWhich simplifies to:AD² + 4 AD cos θ - 9 + 12 cos θ = 0As before. So this quadratic in AD: AD² + (4 cos θ) AD + (12 cos θ - 9) = 0The solutions for AD are:AD = [ -4 cos θ ± sqrt(16 cos² θ - 4(12 cos θ -9)) ] / 2Compute discriminant:sqrt(16 cos² θ - 48 cos θ + 36) = sqrt( (4 cos θ -6)^2 ) = |4 cos θ -6|Since 4 cos θ -6 is likely negative (as we need to prove cos θ < 3/4), so sqrt becomes 6 -4 cos θ.Therefore,AD = [ -4 cos θ ± (6 -4 cos θ) ] / 2We have two solutions:1. [ -4 cos θ + 6 -4 cos θ ] / 2 = [6 -8 cos θ]/2 = 3 -4 cos θ2. [ -4 cos θ -6 +4 cos θ ] /2 = (-6)/2 = -3Since AD is a length, it can't be negative. So AD = 3 -4 cos θTherefore, AD = 3 -4 cos θ must be positive, since D is inside the triangle. So:3 -4 cos θ > 0 => 4 cos θ <3 => cos θ < 3/4Which is the upper bound we need to prove. So that's one part: cos θ < 3/4.Now, for the lower bound: 1/4 < cos θ.We need to find why cos θ must be greater than 1/4. Since D is inside the triangle, there must be some other condition that imposes this lower bound.Let me think. Since AD = 3 -4 cos θ must be positive, and also, AD must be less than AC + CD? Wait, no, since D is inside the triangle, AD must be less than AC + CD? Wait, triangle inequality in triangle ADC: AD < AC + CD, but AC = sqrt(13 -12 cos θ), CD = 2. So,3 -4 cos θ < sqrt(13 -12 cos θ) + 2Subtract 2:1 -4 cos θ < sqrt(13 -12 cos θ)But sqrt(13 -12 cos θ) is always positive. Let me square both sides (noting that left side must be positive for the inequality to hold as sqrt is positive):If 1 -4 cos θ < sqrt(13 -12 cos θ), then first, 1 -4 cos θ must be positive or negative.If 1 -4 cos θ is negative, then the inequality holds automatically since the right side is positive. If 1 -4 cos θ is positive, then we can square both sides:(1 -4 cos θ)^2 < 13 -12 cos θExpand left side:1 -8 cos θ +16 cos² θ <13 -12 cos θBring all terms to left:1 -8 cos θ +16 cos² θ -13 +12 cos θ <0Simplify:-12 +4 cos θ +16 cos² θ <0Divide by 4:-3 + cos θ +4 cos² θ <0Rearranged:4 cos² θ + cos θ -3 <0Quadratic inequality: 4 cos² θ + cos θ -3 <0Solve this quadratic inequality for cos θ.Let me compute the roots:4x² +x -3 =0Discriminant: 1 + 48 =49Roots: (-1 ±7)/8Thus, x=( -1 +7 )/8=6/8=3/4 and x=( -1 -7 )/8= -8/8= -1So the quadratic is positive outside the roots (-∞, -1) and (3/4, ∞), and negative between -1 and 3/4.But since cos θ is between -1 and 1, but in our case, since θ is an angle in a triangle, 0 < θ < 180°, so cos θ is between -1 and 1. However, since AB=2 and BC=3, the triangle inequality holds: AB + BC > AC, so AC <5; AB + AC > BC => AC >1; BC + AC > AB => AC >1. So AC is between 1 and 5. From AC²=13 -12 cos θ, 1 < sqrt(13 -12 cos θ) <5, so 1 <13 -12 cos θ <25 => subtract 13: -12 < -12 cos θ <12 => divide by -12 (reverse inequalities): 1 > cos θ > -1. So cos θ is between -1 and 1, which is obvious.But from the quadratic inequality 4 cos² θ + cos θ -3 <0, the solution is -1 < cos θ <3/4. However, our previous condition from AD >0 is cos θ <3/4, so combining these, the inequality 4 cos² θ + cos θ -3 <0 holds for -1 < cos θ <3/4. But we need to find when 1 -4 cos θ < sqrt(13 -12 cos θ). So if 1 -4 cos θ is positive, we must have cos θ <3/4 (already known), and additionally, 4 cos² θ + cos θ -3 <0, which is always true for cos θ <3/4 (since from the quadratic, it's true for -1 < cos θ <3/4). But if 1 -4 cos θ is negative, then the inequality holds automatically. So the condition from AD < AC + CD doesn't give a new restriction.Therefore, we need another condition to get the lower bound of 1/4. Let me think. Maybe the position of D inside the triangle ABC. Since D is inside the triangle, certain inequalities must hold. For example, in triangle ABC, point D must lie within the boundaries, so the barycentric coordinates are all positive. But how to translate this into a condition on cos θ.Alternatively, since D is inside triangle ABC, the distance from D to the sides must be positive. But this might be complicated.Alternatively, recall that in triangle ADC, AD =3 -4 cos θ. Since D is inside triangle ABC, AD must be less than AB + BD, but BD is another length we don't know. Alternatively, perhaps in triangle ABD or CBD.Alternatively, consider that in triangle ABC, point D is such that CD =2. So the distance from C to D is 2, which is less than BC =3, so D lies somewhere on the circle with radius 2 from C, inside the triangle. The minimal value of cos θ could be related to the position where D approaches one of the edges of the triangle.Alternatively, perhaps using the Law of Cosines in triangle BDC or another triangle.Wait, point D is inside triangle ABC. Let's consider triangle BDC. We know BC =3, CD=2, but we don't know BD. Similarly, in triangle ABD, we know AB=2, AD=3 -4 cos θ, but don't know BD.Alternatively, using the Law of Cosines in triangle ABD and triangle CBD.Alternatively, think about coordinates again. Recall that point A is (2 cos θ, 2 sin θ), point D is (3 +2 cos φ, 2 sin φ). Wait, earlier I tried parametrizing D as (3 + 2 cos φ, 2 sin φ), but if D is inside the triangle ABC, then coordinates must satisfy certain conditions. For example, the y-coordinate of D, which is 2 sin φ, must be less than the y-coordinate of A, which is 2 sin θ. Therefore, sin φ < sin θ. Since both φ and θ are between 0 and π (as angles in a triangle), this implies φ < θ. So φ is less than θ. Therefore, θ must be greater than φ.But how does that help? Maybe not directly.Alternatively, think about the position of D. Since D is inside triangle ABC, the line AD must intersect BC at some point. But perhaps not helpful.Alternatively, since we have AD =3 -4 cos θ, and D is inside the triangle, so AD must be less than AB + BD, but BD is less than BC + CD? Wait, not sure.Alternatively, since in triangle ABC, the length from A to D must be less than the length from A to C plus CD. Wait, that's the triangle inequality in triangle ADC: AD < AC + CD. Which we already considered earlier. That gives us that AD < sqrt(13 -12 cos θ) +2. But we already have AD =3 -4 cos θ, so 3 -4 cos θ < sqrt(13 -12 cos θ) +2. But as we saw, this leads to the quadratic inequality which didn't give a lower bound.Alternatively, maybe consider that point D is inside triangle ABC, so the area of triangle ADC must be positive and less than the area of triangle ABC.Wait, the area of triangle ABC can be computed as (1/2)*AB*BC*sin θ = (1/2)*2*3*sin θ = 3 sin θ.The area of triangle ADC can be computed using coordinates. Point A (2 cos θ, 2 sin θ), point D (3 +2 cos φ, 2 sin φ), point C (3,0). The area is |(2 cos θ*(2 sin φ -0) + (3 +2 cos φ)*(0 -2 sin θ) +3*(2 sin θ -2 sin φ))/2|But this seems complicated. Alternatively, using the formula 1/2 * base * height. If we take CD as the base, which is 2, then the height is the distance from A to line CD.But line CD connects points C (3,0) and D (3 +2 cos φ, 2 sin φ). The equation of line CD can be parametrized as x =3 +2 t cos φ, y =0 +2 t sin φ, where t ranges from 0 to1.The distance from A (2 cos θ, 2 sin θ) to line CD is given by:| ( (3 -2 cos θ)*2 sin φ - (0 -2 sin θ)*2 cos φ ) | / sqrt( (2 cos φ)^2 + (2 sin φ)^2 )Wait, the formula for the distance from a point (x0, y0) to the line ax + by +c =0 is |ax0 + by0 +c| / sqrt(a² +b²). Let me find the equation of line CD.Points C (3,0) and D (3 +2 cos φ, 2 sin φ). The slope of CD is (2 sin φ -0)/(2 cos φ) = tan φ. So the equation is y = tan φ (x -3). Rearranged: tan φ x - y -3 tan φ =0.Therefore, distance from A (2 cos θ, 2 sin θ) to line CD is | tan φ *2 cos θ -2 sin θ -3 tan φ | / sqrt( tan² φ +1 )Simplify denominator: sqrt( tan² φ +1 ) = sec φ = 1/cos φNumerator: |2 tan φ cos θ -2 sin θ -3 tan φ|= |2 (sin φ / cos φ) cos θ -2 sin θ -3 (sin φ / cos φ)|= |2 sin φ cos θ / cos φ -2 sin θ -3 sin φ / cos φ|Factor sin φ / cos φ:= | [2 sin φ cos θ -3 sin φ]/cos φ -2 sin θ |= | sin φ (2 cos θ -3)/cos φ -2 sin θ |= | sin φ (2 cos θ -3) / cos φ -2 sin θ |This is complicated. Maybe not the right approach.Alternatively, since the area of triangle ADC is 1/2 * CD * height from A, which must be positive. But since D is inside ABC, the area of ADC is less than area of ABC, which is 3 sin θ. So:1/2 *2 * height < 3 sin θ => height <3 sin θBut height is the distance from A to line CD. Not sure.Alternatively, maybe use coordinates again. Given that D is (3 +2 cos φ, 2 sin φ) inside triangle ABC. The condition for D being inside ABC is that it lies below the line AB and below the line AC.Wait, line AB connects points B(0,0) and A(2 cos θ, 2 sin θ). The equation of line AB can be parametrized as x =2 t cos θ, y=2 t sin θ, t ∈ [0,1].Line AC connects A(2 cos θ, 2 sin θ) and C(3,0). The equation for line AC can be parametrized as x =2 cos θ + t(3 -2 cos θ), y=2 sin θ -2 t sin θ, t ∈ [0,1].Point D must lie below both lines AB and AC.Alternatively, the barycentric coordinates must be positive. But this is getting too vague.Alternatively, think about the maximum possible value of AD. Since D is inside the triangle ABC, AD can't be too large. But we have AD =3 -4 cos θ. So if cos θ decreases, AD increases. So the minimal value of cos θ (which would be the lower bound we need) corresponds to the maximal AD that keeps D inside the triangle.Therefore, the maximum possible AD is when D is on the edge of the triangle. For example, when D is on side AB or AC or BC. But since CD=2, D can't be on BC because BC=3, and CD=2 would mean D is somewhere between B and C, but then angle ADC would be different.Wait, if D is on side AB, then let's see. If D is on AB, then AD + DB = AB =2. But CD=2. Let me check if that's possible.If D is on AB, then coordinates of D can be written as (2t cos θ, 2t sin θ) for some t between 0 and1. Then distance from D to C is sqrt( (2t cos θ -3)^2 + (2t sin θ)^2 ) = sqrt(4t² cos² θ -12 t cos θ +9 +4t² sin² θ ) = sqrt(4t² (cos² θ + sin² θ ) -12 t cos θ +9 ) = sqrt(4t² -12 t cos θ +9 )Set this equal to 2:sqrt(4t² -12 t cos θ +9 ) =2Square both sides:4t² -12 t cos θ +9 =4Thus:4t² -12 t cos θ +5=0Solve for t:t = [12 cos θ ± sqrt(144 cos² θ -80)] /8Discriminant:144 cos² θ -80For real solutions, discriminant must be non-negative:144 cos² θ -80 ≥0 => cos² θ ≥80/144=5/9 => cos θ ≥√5/3≈0.745 or cos θ ≤-√5/3≈-0.745But since θ is an angle in a triangle, cos θ must be positive if θ is acute, which is possible here. But if cos θ ≥√5/3≈0.745, which is higher than the upper bound we need to prove (3/4=0.75). So the maximum possible cos θ where D is on AB is around 0.745, but our upper bound is 3/4=0.75. Hmm, close. But since the problem states that D is inside the triangle, not on the edge, so maybe D can't be on AB. So this suggests that when cos θ approaches √5/3≈0.745, D approaches AB. But our upper bound is 3/4=0.75, which is slightly higher. So maybe this isn't directly helpful.Alternatively, if D is on side AC. Let me check.If D is on AC, then coordinates of D can be parametrized as (2 cos θ + t(3 -2 cos θ), 2 sin θ -2 t sin θ ) for t ∈[0,1]. Then distance from D to C is sqrt( (3 - (2 cos θ + t(3 -2 cos θ)))^2 + (0 - (2 sin θ -2 t sin θ))^2 )= sqrt( (3 -2 cos θ - t(3 -2 cos θ))^2 + (-2 sin θ +2 t sin θ )^2 )= sqrt( ( (3 -2 cos θ)(1 - t) )^2 + ( -2 sin θ (1 - t) )^2 )= sqrt( (3 -2 cos θ)^2 (1 - t)^2 +4 sin² θ (1 - t)^2 )= sqrt( [ (3 -2 cos θ)^2 +4 sin² θ ] (1 - t)^2 )= sqrt( [9 -12 cos θ +4 cos² θ +4 sin² θ ] (1 - t)^2 )= sqrt( [9 -12 cos θ +4 (cos² θ + sin² θ ) ] (1 - t)^2 )= sqrt( [9 -12 cos θ +4 ] (1 - t)^2 )= sqrt(13 -12 cos θ ) |1 - t |Set equal to 2:sqrt(13 -12 cos θ ) |1 - t | =2So |1 - t | =2 / sqrt(13 -12 cos θ )Since t ∈[0,1], 1 -t ∈[0,1], so 0 ≤1 -t ≤1, so |1 - t |=1 -t. Therefore:1 -t =2 / sqrt(13 -12 cos θ )Thus,t=1 -2 / sqrt(13 -12 cos θ )Since t ∈[0,1], we need 1 -2 / sqrt(13 -12 cos θ ) ≥0 => sqrt(13 -12 cos θ ) ≥2 =>13 -12 cos θ ≥4 => -12 cos θ ≥-9 =>cos θ ≤9/12=3/4. Which is our upper bound. So if cos θ=3/4, then sqrt(13 -12*(3/4))=sqrt(13 -9)=sqrt(4)=2, so t=1 -2/2=0. Therefore, when cos θ=3/4, D is at point C. But D is supposed to be inside the triangle, so t must be greater than0. Therefore, cos θ must be less than3/4 to have t>0, which matches our previous result.But this doesn't give a lower bound. For the lower bound, maybe when D is on another edge.If D is on side BC. Since BC is from (0,0) to (3,0), and CD=2. If D is on BC, then coordinates of D are (3 -2,0)=(1,0). So D=(1,0). But D=(1,0) is on BC. Then, we need to check if angle ADC=180° - θ.Compute angle ADC when D=(1,0). Point A=(2 cos θ, 2 sin θ), D=(1,0), C=(3,0). So angle ADC is the angle at D between A, D, C.Vectors DA=(2 cos θ -1, 2 sin θ -0)=(2 cos θ -1, 2 sin θ )Vectors DC=(3 -1,0 -0)=(2,0 )The angle between DA and DC is angle ADC. Let's compute it.cos(angle ADC)= (DA · DC)/( |DA| |DC| )DA · DC= (2 cos θ -1)*2 +2 sin θ *0=4 cos θ -2|DA|=sqrt( (2 cos θ -1)^2 + (2 sin θ )^2 )=sqrt(4 cos² θ -4 cos θ +1 +4 sin² θ )=sqrt(4 (cos² θ + sin² θ ) -4 cos θ +1 )=sqrt(4 -4 cos θ +1 )=sqrt(5 -4 cos θ )|DC|=2Therefore,cos(angle ADC)=(4 cos θ -2)/(2 sqrt(5 -4 cos θ ))=(2 cos θ -1)/sqrt(5 -4 cos θ )We are given that angle ADC=180° -θ, so cos(angle ADC)= -cos θTherefore,(2 cos θ -1)/sqrt(5 -4 cos θ ) = -cos θMultiply both sides by sqrt(5 -4 cos θ ):2 cos θ -1 = -cos θ sqrt(5 -4 cos θ )Square both sides:(2 cos θ -1)^2 = cos² θ (5 -4 cos θ )Expand left side:4 cos² θ -4 cos θ +1 =5 cos² θ -4 cos³ θBring all terms to left:4 cos² θ -4 cos θ +1 -5 cos² θ +4 cos³ θ=0Simplify:- cos² θ -4 cos θ +1 +4 cos³ θ=0Rearranged:4 cos³ θ -cos² θ -4 cos θ +1=0Factor this cubic equation. Let's try possible rational roots using Rational Root Theorem. Possible roots: ±1, ±1/2, ±1/4.Test cos θ=1: 4 -1 -4 +1=0 → 0=0. So cos θ=1 is a root.Perform polynomial division or factor out (cos θ -1):Using synthetic division:Coefficients:4 | -1 | -4 |1Divide by (cos θ -1):1 | 4 | -1 | -4 |1Bring down 4.Multiply by1:4Add to next term: -1 +4=3Multiply by1:3Add to next term:-4 +3=-1Multiply by1:-1Add to last term:1 +(-1)=0So the cubic factors as (cos θ -1)(4 cos² θ +3 cos θ -1)=0Set each factor to zero:cos θ=1 → θ=0°, impossible in a triangle.Solve 4 cos² θ +3 cos θ -1=0:Discriminant=9 +16=25Solutions:cos θ=(-3 ±5)/8cos θ=(2)/8=1/4 or cos θ=(-8)/8=-1cos θ=-1 is impossible, so cos θ=1/4. So if D is on BC, then cos θ=1/4. But D is supposed to be inside the triangle, so when cos θ=1/4, D is on the boundary (BC). Therefore, for D to be strictly inside, cos θ must be greater than1/4. Hence, the lower bound is 1/4.Therefore, combining both results, we have 1/4 < cos θ <3/4.So that's the reasoning. When D approaches the edge BC, cos θ approaches1/4, and when D approaches the vertex C, cos θ approaches3/4. Since D must be inside the triangle, cos θ must be strictly between1/4 and3/4.Final Answerboxed{left( frac{1}{4}, frac{3}{4} right)}Final Answerboxed{frac{1}{4} < cos B < frac{3}{4}}</think>Given triangle ( triangle ABC ) with ( AB = 2 ), ( BC = 3 ), and a point ( D ) inside the triangle such that ( CD = 2 ) and ( angle ADC = 180^circ - angle B ). We need to prove ( frac{1}{4} < cos B < frac{3}{4} ).1. Using Law of Cosines in ( triangle ABC ): [ AC^2 = AB^2 + BC^2 - 2 cdot AB cdot BC cdot cos B ] [ AC^2 = 2^2 + 3^2 - 2 cdot 2 cdot 3 cdot cos B = 13 - 12 cos B ] [ AC = sqrt{13 - 12 cos B} ]2. Using Law of Cosines in ( triangle ADC ): [ AC^2 = AD^2 + CD^2 - 2 cdot AD cdot CD cdot cos(180^circ - B) ] [ 13 - 12 cos B = AD^2 + 4 + 4 AD cos B ] [ AD^2 + 4 AD cos B + 12 cos B - 9 = 0 ]3. Solving the quadratic equation for ( AD ): [ AD = frac{-4 cos B pm sqrt{(4 cos B)^2 - 4 cdot 1 cdot (12 cos B - 9)}}{2} ] [ text{Discriminant} = 16 cos^2 B - 48 cos B + 36 = (4 cos B - 6)^2 ] [ AD = frac{-4 cos B pm (6 - 4 cos B)}{2} ] [ AD = 3 - 4 cos B quad (text{since } AD text{ must be positive}) ]4. Establishing upper bound: [ 3 - 4 cos B > 0 implies cos B < frac{3}{4} ]5. Establishing lower bound: - If ( D ) is on the boundary ( BC ), solving the cubic equation derived from the angle condition gives ( cos B = frac{1}{4} ). - Since ( D ) must be inside the triangle, ( cos B ) must be strictly greater than ( frac{1}{4} ).Thus, combining both bounds, we get:[boxed{frac{1}{4} < cos B < frac{3}{4}}],<think>Okay, so I need to prove that in triangle ABC, where AB=2, BC=3, and there's a point D inside the triangle such that CD=2 and angle ADC is equal to 180 degrees minus angle B. Then I have to show that the cosine of angle B is between 1/4 and 3/4. Hmm, let's start by drawing a diagram to visualize the problem. First, triangle ABC with AB=2 and BC=3. Let me note that the length of AC isn't given, so maybe I need to express AC in terms of angle B. Also, point D is inside the triangle with CD=2 and angle ADC=180° - angle B. That seems important. Maybe some trigonometric identities or the Law of Cosines will come into play here.Let me recall the Law of Cosines. For any triangle, c² = a² + b² - 2ab cosθ, where θ is the angle opposite side c. So in triangle ABC, if we let angle B be at vertex B, then sides AB=2, BC=3, so AC² = AB² + BC² - 2*AB*BC*cos B. Wait, actually, in triangle ABC, angle B is at vertex B, so sides adjacent to angle B are AB and BC. Wait, AB is adjacent to angle B, and BC is adjacent to angle B as well. Wait, no, in triangle ABC, angle B is between sides AB and BC. So the Law of Cosines for angle B would be AC² = AB² + BC² - 2*AB*BC*cos B. So AC² = 2² + 3² - 2*2*3*cos B = 4 + 9 - 12 cos B = 13 - 12 cos B. So AC = sqrt(13 - 12 cos B). Hmm, that might be useful later.Now, point D is inside the triangle such that CD=2 and angle ADC=180° - angle B. Let me try to represent this. So in triangle ADC, angle at D is angle ADC, which is 180° - angle B. So maybe I can use the Law of Sines or Cosines in triangle ADC as well.Since CD=2, and we know angle ADC, perhaps we can relate the sides of triangle ADC. Let's denote angle ADC as 180° - angle B. Let me denote angle B as θ for simplicity. Then angle ADC = 180° - θ. Also, in triangle ABC, angle B is θ, so angle at B is θ. So in triangle ADC, angle at D is 180° - θ. In triangle ADC, the sum of angles should be 180°, so angle at A plus angle at C plus angle at D equals 180°. Therefore, angle at A plus angle at C plus (180° - θ) = 180°, so angle at A + angle at C = θ. Hmm, not sure if that helps immediately.Alternatively, maybe using the Law of Sines in triangle ADC. Let's denote some variables. Let me suppose in triangle ADC, sides are AD, DC, AC. We know DC=2, angle at D is 180° - θ. Let's denote AD as x and AC as y (we already have AC expressed in terms of θ as sqrt(13 - 12 cos θ)). Then, using the Law of Sines in triangle ADC: AD/sin(angle at C) = DC/sin(angle at A) = AC/sin(angle D). But angle D is 180° - θ, so sin(angle D) = sin(θ). So AC / sin(θ) = DC / sin(angle A). Therefore, y / sin θ = 2 / sin(angle A). Therefore, sin(angle A) = (2 sin θ) / y.Similarly, AD / sin(angle C) = DC / sin(angle A), so x / sin(angle C) = 2 / sin(angle A). So x = 2 sin(angle C) / sin(angle A). But angle A + angle C = θ, so angle C = θ - angle A. Therefore, x = 2 sin(θ - angle A) / sin(angle A). Hmm, getting a bit complicated here.Alternatively, maybe using the Law of Cosines in triangle ADC. Let's consider triangle ADC with sides AD=x, DC=2, AC=y, and angle at D is 180° - θ. Then, by Law of Cosines: AC² = AD² + DC² - 2*AD*DC*cos(angle ADC). But angle ADC is 180° - θ, so cos(angle ADC) = cos(180° - θ) = -cos θ. Therefore:y² = x² + 2² - 2*x*2*(-cos θ) y² = x² + 4 + 4x cos θBut we also have from triangle ABC, y² = 13 - 12 cos θ. Therefore:13 - 12 cos θ = x² + 4 + 4x cos θ Rearranging:x² + 4x cos θ + 4 - 13 + 12 cos θ = 0 x² + 4x cos θ + (12 cos θ - 9) = 0So we have an equation involving x and cos θ. But we need another equation to relate x and cos θ. Maybe using triangle ABD or triangle BDC?Wait, point D is inside triangle ABC, so perhaps we can consider triangles ABD and CBD. Let me check. Since CD=2, BC=3, then BD must be 3 - CD? Wait, no, because D is inside the triangle, so BD is not necessarily BC - CD. It could be part of some other configuration.Alternatively, maybe using coordinates. Let me assign coordinates to the triangle ABC. Let me place point B at the origin (0,0). Then, since AB=2, let me place point A at (2,0). Then, point C is somewhere in the plane such that BC=3. Let me denote coordinates: point B (0,0), point A (2,0), and point C (x,y) such that the distance from B to C is 3, so x² + y² = 9. Also, AC can be found as sqrt( (x-2)^2 + y² ), which we already know is sqrt(13 - 12 cos θ). But θ is angle at B, so angle between BA and BC. Since BA is along the x-axis from B to A (2,0), and BC is from B to C (x,y). So the angle θ at B can be found via the dot product. The vectors BA and BC are (2,0) and (x,y), respectively. The cosine of angle θ is (BA · BC) / (|BA||BC|) = (2x + 0*y)/(2*3) = (2x)/6 = x/3. Therefore, cos θ = x/3. So x = 3 cos θ, and since x² + y² = 9, then y = 3 sin θ. Therefore, coordinates of C are (3 cos θ, 3 sin θ). That might be helpful.Now, point D is inside triangle ABC such that CD=2 and angle ADC=180° - θ. Let me assign coordinates to D as well. Let me denote D as (p,q). Then, CD=2, so the distance from C to D is sqrt( (p - 3 cos θ)^2 + (q - 3 sin θ)^2 ) = 2. Also, angle ADC=180° - θ. Let's see, angle at D in triangle ADC is 180° - θ. So angle ADC is the angle at D between points A, D, C. So vectors DA and DC. The angle between DA and DC is 180° - θ. So using coordinates, we can express vectors DA and DC.Vector DA would be from D to A: (2 - p, 0 - q). Vector DC would be from D to C: (3 cos θ - p, 3 sin θ - q). The angle between vectors DA and DC is 180° - θ. So the cosine of that angle is equal to cos(180° - θ) = -cos θ. Therefore, the dot product of DA and DC divided by the product of their magnitudes equals -cos θ.So:[(2 - p)(3 cos θ - p) + (0 - q)(3 sin θ - q)] / (|DA| |DC|) = -cos θBut |DC| is given as 2, and |DA| is the distance from D to A: sqrt( (2 - p)^2 + q^2 ). Let's denote |DA| as m. Then:[(2 - p)(3 cos θ - p) + (-q)(3 sin θ - q)] / (m * 2) = -cos θMultiply both sides by 2m:(2 - p)(3 cos θ - p) - q(3 sin θ - q) = -2m cos θBut m = sqrt( (2 - p)^2 + q^2 ). So this seems complicated. Maybe expand the left side:First term: (2 - p)(3 cos θ - p) = 6 cos θ - 2p - 3p cos θ + p² Second term: -q(3 sin θ - q) = -3 q sin θ + q² So combined: 6 cos θ - 2p - 3p cos θ + p² - 3 q sin θ + q²So left side becomes: p² + q² - 2p - 3p cos θ - 3 q sin θ + 6 cos θRight side: -2 sqrt( (2 - p)^2 + q^2 ) cos θThis is getting quite involved. Maybe there's a better approach. Alternatively, since point D is inside triangle ABC, perhaps we can use coordinate geometry with the coordinates assigned as above.Given that point D is inside triangle ABC, coordinates (p, q), CD=2, angle ADC=180° - θ. Let me write the two equations:1. (p - 3 cos θ)^2 + (q - 3 sin θ)^2 = 4 (distance CD=2)2. The angle condition: angle at D is 180° - θ. Expressed via the dot product:[(2 - p)(3 cos θ - p) + (-q)(3 sin θ - q)] = -2 cos θ sqrt( (2 - p)^2 + q^2 )This seems very complicated. Maybe I can parametrize θ and find constraints?Alternatively, since we need to find bounds on cos θ, maybe we can use some inequalities.Wait, the problem states that D is inside triangle ABC. So point D must lie within the triangle. So perhaps we can use the fact that D is inside the triangle to impose constraints on θ.Alternatively, consider triangle ABC with AB=2, BC=3, angle B=θ. Then, using the Law of Cosines, AC= sqrt(13 - 12 cos θ). Then, point D is inside ABC with CD=2 and angle ADC=180° - θ. Maybe use the Law of Cosines in triangle ADC and triangle ABD?Wait, in triangle ADC, we have sides AD, DC=2, AC, angle at D=180° - θ. So as before:AC² = AD² + DC² - 2*AD*DC*cos(180° - θ) => 13 - 12 cos θ = AD² + 4 + 4 AD cos θ => AD² + 4 AD cos θ + 4 - 13 + 12 cos θ = 0 => AD² + 4 AD cos θ + (12 cos θ - 9) = 0This is a quadratic equation in AD:AD² + (4 cos θ) AD + (12 cos θ - 9) = 0For real solutions, the discriminant must be non-negative:(4 cos θ)^2 - 4*1*(12 cos θ - 9) ≥ 0 16 cos² θ - 48 cos θ + 36 ≥ 0 Divide by 4: 4 cos² θ - 12 cos θ + 9 ≥ 0 This factors as (2 cos θ - 3)^2 ≥ 0, which is always true since squares are non-negative. So the quadratic has real roots for any θ, but since AD is a length, it must be positive.Therefore, the solutions for AD are:AD = [ -4 cos θ ± sqrt(16 cos² θ - 48 cos θ + 36) ] / 2 But sqrt(16 cos² θ -48 cos θ +36) = sqrt( (4 cos θ)^2 - 48 cos θ + 36 )Wait, earlier we saw discriminant is (2 cos θ - 3)^2, so sqrt of discriminant is |2 cos θ - 3|. Therefore:AD = [ -4 cos θ ± |2 cos θ - 3| ] / 2We need AD positive, so let's consider both cases.Case 1: 2 cos θ - 3 ≥ 0 => cos θ ≥ 3/2. But cos θ can't exceed 1, so this case is impossible.Case 2: 2 cos θ - 3 < 0 => cos θ < 3/2 (which is always true). Therefore, |2 cos θ -3| = 3 - 2 cos θ. Then,AD = [ -4 cos θ + (3 - 2 cos θ) ] / 2 = (-4 cos θ +3 -2 cos θ)/2 = (3 -6 cos θ)/2 Or AD = [ -4 cos θ - (3 - 2 cos θ) ] /2 = (-4 cos θ -3 +2 cos θ)/2 = (-2 cos θ -3)/2But AD must be positive, so the second solution gives (-2 cos θ -3)/2 >0 => -2 cos θ -3 >0 => -2 cos θ >3 => cos θ < -3/2. Which is impossible since cos θ ≥ -1. Therefore, only the first solution is valid: AD=(3 -6 cos θ)/2. But AD must be positive, so 3 -6 cos θ >0 => 3 >6 cos θ => cos θ < 1/2. Wait, but the upper bound we need to prove is cos θ <3/4, which is weaker than cos θ <1/2 here. This seems contradictory. Wait, maybe I made a mistake here.Wait, from the quadratic equation for AD:AD² + 4 cos θ AD +12 cos θ -9 =0Solutions:AD = [-4 cos θ ± sqrt( (4 cos θ)^2 -4*1*(12 cos θ -9) )]/2 = [ -4 cos θ ± sqrt(16 cos² θ -48 cos θ +36) ] /2 As we saw, sqrt( (2 cos θ -3)^2 ) = |2 cos θ -3| =3 -2 cos θ (since 2 cos θ -3 is negative). So,AD = [ -4 cos θ ± (3 -2 cos θ) ] /2So two solutions:1. [ -4 cos θ +3 -2 cos θ ] /2 = (3 -6 cos θ)/22. [ -4 cos θ -3 +2 cos θ ] /2 = (-2 cos θ -3)/2As before, the second solution is negative. So only the first solution is valid, so AD=(3 -6 cos θ)/2. Since AD must be positive, (3 -6 cos θ)/2 >0 => 3 -6 cos θ >0 => cos θ <0.5. But the problem states that we need to prove cos θ <3/4. So this suggests that cos θ must be less than 1/2. But the problem says upper bound is 3/4, which is higher. So this suggests that maybe there's a mistake here. Wait, but how?Wait, point D is inside the triangle. So even though mathematically AD=(3 -6 cos θ)/2 gives AD positive as long as cos θ <1/2, but if D is inside the triangle, perhaps there are additional constraints.Alternatively, maybe the problem statement allows D to be inside the triangle even if AD is negative? No, AD is a length, so it must be positive, but the coordinate system might have different implications.Wait, perhaps my approach here is missing something. Let's double-check. From triangle ADC, using the Law of Cosines, I derived that AD=(3 -6 cos θ)/2. But AD must be a positive length, so cos θ must be less than 1/2. But the problem says to prove cos θ <3/4. So maybe there's an error in the reasoning here.Alternatively, perhaps there are other constraints from the fact that D is inside the triangle. For instance, maybe point D must lie within certain regions of the triangle, which would impose additional inequalities.Alternatively, maybe using triangle inequality in triangle ADC. Since D is inside ABC, then in triangle ADC, the sum of any two sides must be greater than the third. So AD + DC > AC, AD + AC > DC, DC + AC > AD.Given DC=2, AC= sqrt(13 -12 cos θ), AD=(3 -6 cos θ)/2.So first inequality: AD + DC > AC (3 -6 cos θ)/2 +2 > sqrt(13 -12 cos θ) Multiply 2: 3 -6 cos θ +4 > 2 sqrt(13 -12 cos θ) 7 -6 cos θ > 2 sqrt(13 -12 cos θ) Let me square both sides (keeping in mind that left side must be positive, which it is as cos θ <1/2, so 7 -6*(1/2)=7-3=4>0):(7 -6 cos θ)^2 >4*(13 -12 cos θ) 49 -84 cos θ +36 cos² θ >52 -48 cos θ 49 -84 cos θ +36 cos² θ -52 +48 cos θ >0 -3 -36 cos θ +36 cos² θ >0 Divide by 3: -1 -12 cos θ +12 cos² θ >0 12 cos² θ -12 cos θ -1 >0 Let me solve 12x² -12x -1 >0 where x=cos θ. Quadratic equation: 12x² -12x -1=0 Discriminant=144 +48=192 Solutions: [12 ± sqrt(192)]/(24)= [12 ±8*sqrt(3)]/24= [3 ±2*sqrt(3)]/6 Approximately, sqrt(3)=1.732, so 2*sqrt(3)=3.464. So [3 +3.464]/6≈6.464/6≈1.077, which is more than 1, impossible for cos θ. [3 -3.464]/6≈-0.464/6≈-0.077. Therefore, the inequality 12x² -12x -1 >0 holds when x < [3 -2√3]/6≈-0.077 or x > [3 +2√3]/6≈1.077. But cos θ is between -1 and 1, so the inequality holds when cos θ < [3 -2√3]/6≈-0.077. But earlier we had cos θ <1/2≈0.5. Therefore, this inequality would require cos θ <≈-0.077. But in our previous result, cos θ <0.5. Therefore, combining both, we have cos θ <≈-0.077. But this contradicts the lower bound we need to prove, which is 1/4. So clearly, something is wrong here.Wait, this suggests that my approach is flawed. Let me check the steps again.Starting from triangle ADC: AC² = AD² + DC² -2*AD*DC*cos(angle ADC). Since angle ADC=180° -θ, so cos(angle ADC)= -cos θ. Therefore, AC² = AD² +4 +4 AD cos θ. Then AC² is also 13 -12 cos θ. Therefore:AD² +4 AD cos θ +4 =13 -12 cos θ AD² +4 AD cos θ =9 -12 cos θ AD² +4 AD cos θ +12 cos θ -9=0 Wait, perhaps a miscalculation here. Let me check again:Original equation from Law of Cosines:AC² = AD² + DC² - 2*AD*DC*cos(angle ADC) 13 -12 cos θ = AD² +4 -2*AD*2*(-cos θ) So 13 -12 cos θ = AD² +4 +4 AD cos θ Then moving 4 to left: 13 -12 cos θ -4 = AD² +4 AD cos θ So 9 -12 cos θ = AD² +4 AD cos θ Therefore, AD² +4 AD cos θ +12 cos θ -9=0? Wait, no, moving terms to left:AD² +4 AD cos θ =9 -12 cos θ So AD² +4 AD cos θ +12 cos θ -9=0? Wait, no. If AD² +4 AD cos θ =9 -12 cos θ, then bringing all terms to left:AD² +4 AD cos θ +12 cos θ -9=0. Wait, no:Original equation: AD² +4 AD cos θ =9 -12 cos θ Therefore, moving all terms to left: AD² +4 AD cos θ +12 cos θ -9=0. Wait, no, that would be AD² +4 AD cos θ -9 +12 cos θ=0. Hmm, that is correct. So it's AD² +4 AD cos θ +12 cos θ -9=0.Then, factor terms with cos θ: AD² + cos θ*(4 AD +12) -9=0. Hmm, but perhaps I should treat this as a quadratic in AD:AD² + (4 cos θ) AD + (12 cos θ -9)=0Which is the same as before. Then, solving for AD:AD = [-4 cos θ ± sqrt(16 cos² θ -4*1*(12 cos θ -9))]/2 Discriminant:16 cos² θ -48 cos θ +36= (4 cos θ)^2 -2*4 cos θ*6 +6^2= (4 cos θ -6)^2Wait, that's different from before. Wait, 16 cos² θ -48 cos θ +36= (4 cos θ)^2 -48 cos θ +36. Let me check if that's a perfect square:Suppose (a cos θ +b)^2= a² cos² θ +2ab cos θ +b². Comparing to 16 cos² θ -48 cos θ +36:a²=16 =>a=42ab= -48 =>2*4*b= -48 =>8b= -48 =>b= -6Then b²=36. So yes, (4 cos θ -6)^2=16 cos² θ -48 cos θ +36. Therefore, sqrt(discriminant)=|4 cos θ -6|=6 -4 cos θ (since 4 cos θ -6 is always negative because cos θ ≤1, so 4*1 -6= -2 <0). Therefore, the solutions are:AD= [ -4 cos θ ± (6 -4 cos θ) ] /2First solution:[ -4 cos θ +6 -4 cos θ ] /2= (6 -8 cos θ)/2=3 -4 cos θSecond solution:[ -4 cos θ -6 +4 cos θ ] /2= (-6)/2= -3But AD can't be negative, so only the first solution is valid: AD=3 -4 cos θ.Wait, this contradicts my previous result where I thought AD=(3 -6 cos θ)/2. Where did I go wrong earlier?Ah! Wait, in the discriminant step. Earlier, I thought sqrt( (2 cos θ -3)^2 ), but actually, the discriminant simplifies to (4 cos θ -6)^2, so sqrt of that is |4 cos θ -6|=6 -4 cos θ. Therefore, solutions are [ -4 cos θ ± (6 -4 cos θ) ] /2. So first solution is [ -4 cos θ +6 -4 cos θ ]/2= (6 -8 cos θ)/2=3 -4 cos θ. Second solution is [ -4 cos θ -6 +4 cos θ ]/2= (-6)/2= -3. Therefore, AD=3 -4 cos θ. Since AD must be positive, 3 -4 cos θ>0 =>4 cos θ <3 =>cos θ <3/4. Which aligns with the upper bound we need to prove: cos B <3/4. Great! So that was a miscalculation earlier. So AD=3 -4 cos θ must be positive, so cos θ <3/4. Which gives the upper bound required. Now, for the lower bound.Additionally, since D is inside the triangle ABC, other conditions must hold. For instance, in triangle ABD, we can apply the triangle inequality. Also, perhaps considering the position of D relative to the triangle.Alternatively, since D is inside ABC, the length CD=2 must be less than BC=3, which it is. But maybe more precise conditions.Alternatively, let's consider coordinates again. Since we have coordinates for points A(2,0), B(0,0), C(3 cos θ,3 sin θ), and D somewhere inside with CD=2 and angle ADC=180° -θ. But perhaps using the expression we found for AD=3 -4 cos θ. Then AD=3 -4 cos θ must be less than AB + BD? Not sure.Alternatively, since D is inside the triangle, the distance from D to all sides must be positive. But maybe more effective to use the fact that in triangle ADC, AD=3 -4 cos θ, DC=2, and angle at D is 180° -θ. Then perhaps using the Law of Sines in triangle ADC.Law of Sines: AD / sin(angle at C) = DC / sin(angle at A) = AC / sin(angle D)Angle D is 180° -θ, so sin(angle D)=sin θ. So AC / sin θ = DC / sin(angle A). Therefore:AC / sin θ = 2 / sin(angle A) => sin(angle A) = 2 sin θ / AC But AC= sqrt(13 -12 cos θ). So sin(angle A)= 2 sin θ / sqrt(13 -12 cos θ)Also, in triangle ABC, by the Law of Sines:AB / sin(angle C) = BC / sin(angle A) = AC / sin θ So 2 / sin(angle C) = 3 / sin(angle A) = sqrt(13 -12 cos θ) / sin θFrom this, sin(angle A)=3 sin θ / sqrt(13 -12 cos θ)But from triangle ADC, we also have sin(angle A)=2 sin θ / sqrt(13 -12 cos θ). Therefore:3 sin θ / sqrt(13 -12 cos θ) = 2 sin θ / sqrt(13 -12 cos θ)Wait, this implies 3 sin θ =2 sin θ, which implies sin θ=0. Which would mean θ=0 or π, which is impossible in a triangle. Contradiction. So this suggests that my previous approach has an error. Where did I go wrong here?Wait, in triangle ABC, by Law of Sines:AB / sin(angle C) = BC / sin(angle A) = AC / sin θTherefore, sin(angle A)= BC * sin θ / AC = 3 sin θ / sqrt(13 -12 cos θ)But from triangle ADC, angle at A is the same as angle at A in triangle ABC. Wait, no. Wait, in triangle ADC, angle at A is the same as in triangle ABC. Wait, actually, no. Because in triangle ADC, angle at A is formed by sides AD and AC, whereas in triangle ABC, angle at A is formed by sides AB and AC. Unless D lies on AB, which it doesn't necessarily. Therefore, angle at A in triangle ADC is different from angle at A in triangle ABC. Therefore, the earlier step where I equated sin(angle A) from both triangles was incorrect. They are different angles. Therefore, that approach is invalid. So need to discard that.Alternative approach: since we have AD=3 -4 cos θ. Since D is inside triangle ABC, the length AD must be less than AB + BD, but BD is unknown. Alternatively, considering coordinates.We can parametrize point D. Let me use coordinates again. Earlier, we had coordinates for A(2,0), B(0,0), C(3 cos θ,3 sin θ). Point D is (p,q) such that CD=2 and angle ADC=180° -θ.From CD=2: (p -3 cos θ)^2 + (q -3 sin θ)^2=4.From angle ADC=180° -θ: the angle between DA and DC is 180° -θ. Using vectors DA=(2 -p, -q) and DC=(3 cos θ -p, 3 sin θ -q). The dot product is (2 -p)(3 cos θ -p) + (-q)(3 sin θ -q). The magnitudes are |DA|=sqrt((2 -p)^2 + q²) and |DC|=2. The cosine of the angle between them is -cos θ (since the angle is 180° -θ). Therefore:[(2 -p)(3 cos θ -p) - q(3 sin θ -q)] / [sqrt((2 -p)^2 + q²) * 2] = -cos θThis is a complex equation, but perhaps substituting AD=3 -4 cos θ. Since AD is the distance from A to D, sqrt( (2 -p)^2 + q² )=3 -4 cos θ. Therefore:(2 -p)^2 + q²=(3 -4 cos θ)^2=9 -24 cos θ +16 cos² θAlso, from CD=2:(p -3 cos θ)^2 + (q -3 sin θ)^2=4Now, we have two equations:1. (2 -p)^2 + q²=9 -24 cos θ +16 cos² θ2. (p -3 cos θ)^2 + (q -3 sin θ)^2=4Let me expand both equations.Equation 1:(2 -p)^2 + q²=9 -24 cos θ +16 cos² θ Expanding (2 -p)^2: 4 -4p +p² So 4 -4p +p² + q²=9 -24 cos θ +16 cos² θ Thus: p² + q² -4p +4=9 -24 cos θ +16 cos² θ Simplify: p² + q² -4p=5 -24 cos θ +16 cos² θ ...(1)Equation 2:(p -3 cos θ)^2 + (q -3 sin θ)^2=4 Expanding: p² -6 p cos θ +9 cos² θ + q² -6 q sin θ +9 sin² θ=4 Combine terms: p² + q² -6 p cos θ -6 q sin θ +9(cos² θ + sin² θ)=4 Since cos² θ + sin² θ=1: p² + q² -6 p cos θ -6 q sin θ +9=4 Thus: p² + q² -6 p cos θ -6 q sin θ= -5 ...(2)Now, subtract equation (2) from equation (1):(p² + q² -4p) - (p² + q² -6 p cos θ -6 q sin θ)= [5 -24 cos θ +16 cos² θ] - [-5]Simplify left side:-4p - (-6 p cos θ -6 q sin θ)= -4p +6 p cos θ +6 q sin θRight side:5 -24 cos θ +16 cos² θ +5=10 -24 cos θ +16 cos² θTherefore:-4p +6 p cos θ +6 q sin θ=10 -24 cos θ +16 cos² θ ...(3)Now, this is getting quite involved, but perhaps we can express p and q in terms of θ. Alternatively, since we have AD=3 -4 cos θ, which is sqrt( (2 -p)^2 + q²)=3 -4 cos θ. From equation (1):p² + q² -4p=5 -24 cos θ +16 cos² θBut (2 -p)^2 + q²=9 -24 cos θ +16 cos² θ So expanding, 4 -4p +p² + q²=9 -24 cos θ +16 cos² θ Therefore, p² + q² -4p=5 -24 cos θ +16 cos² θ, which is exactly equation (1). So equation (1) is redundant as it comes directly from the expression for AD.Therefore, we need another equation. Equation (2):p² + q² -6 p cos θ -6 q sin θ= -5But from equation (1):p² + q²=4p +5 -24 cos θ +16 cos² θSubstitute into equation (2):4p +5 -24 cos θ +16 cos² θ -6 p cos θ -6 q sin θ= -5 Simplify:4p +5 -24 cos θ +16 cos² θ -6 p cos θ -6 q sin θ= -5 Bring 5 to the right:4p -24 cos θ +16 cos² θ -6 p cos θ -6 q sin θ= -10 Factor terms:p(4 -6 cos θ) + q(-6 sin θ) +16 cos² θ -24 cos θ= -10 ...(4)This is one equation relating p and q. Additionally, we have equation from AD:sqrt( (2 -p)^2 + q² )=3 -4 cos θLet me denote equation (4) as:p(4 -6 cos θ) -6 q sin θ= -10 -16 cos² θ +24 cos θ ...(4a)Let me solve for one variable in terms of the other. For example, solve for q from equation (4a):-6 q sin θ= -10 -16 cos² θ +24 cos θ -p(4 -6 cos θ)Therefore,q= [10 +16 cos² θ -24 cos θ +p(4 -6 cos θ)]/(6 sin θ)Now, substitute this into the AD equation:(2 -p)^2 + [10 +16 cos² θ -24 cos θ +p(4 -6 cos θ)]²/(36 sin² θ)= (3 -4 cos θ)^2This is extremely complex, but maybe expanding and simplifying. However, this seems impractical. Perhaps another approach.Alternatively, since we have AD=3 -4 cos θ and we need D to be inside the triangle, maybe considering the ranges for AD. Since D is inside the triangle, AD must be less than AC and less than AB + BD. But I don't know BD. Alternatively, using coordinates.Since point D is inside triangle ABC, its coordinates (p, q) must satisfy certain inequalities. For instance, the barycentric coordinates must be positive, or in the coordinate system we set up earlier, point D must lie within the convex hull of A, B, C. But this might not be straightforward.Alternatively, since we have expressions involving cos θ, perhaps we can find the range of θ for which such a point D exists inside the triangle.Given that AD=3 -4 cos θ must be positive, we have cos θ <3/4. For the lower bound, we need cos θ >1/4. How do we derive that?Perhaps from the requirement that point D lies inside the triangle. Let's think about when θ is very small (cos θ approaches 1). If θ approaches 0, then angle ADC=180° -θ approaches 180°, meaning points A, D, C are almost colinear with D between A and C. But CD=2, and AC= sqrt(13 -12 cos θ). As θ approaches 0, AC approaches sqrt(13 -12*1)=sqrt(1)=1. So AC approaches 1, which would mean CD=2 cannot be possible since CD must be less than AC. Hence, θ cannot be too small. Similarly, when θ approaches 180°, but θ is an angle in a triangle, so θ must be less than 180°, and more specifically, in triangle ABC, angles must satisfy triangle inequalities.Alternatively, consider the lower bound. Suppose cos θ=1/4. Then AD=3 -4*(1/4)=3 -1=2. So AD=2. Now, in triangle ABC, AB=2. So point D would be 2 units from A and 2 units from C. If AD=2 and CD=2, then triangle ADC would have sides AD=2, DC=2, AC= sqrt(13 -12*(1/4))=sqrt(13 -3)=sqrt(10). So in this case, triangle ADC would have sides 2,2,sqrt(10). The angle at D would be angle ADC=180° -θ. If cos θ=1/4, then θ= arccos(1/4), so angle ADC=180° -arccos(1/4). Now, does such a triangle ADC exist? Yes, since 2+2>sqrt(10)≈3.16, which is not true, 2+2=4>3.16, so the triangle inequality holds. Wait, 2+2=4>sqrt(10). So triangle ADC exists. But does point D lie inside triangle ABC?This is more complex. Maybe when cos θ=1/4, D is on the boundary of the triangle, and for cos θ>1/4, D is inside. Hence, the lower bound.Alternatively, consider the Law of Cosines in triangle ABC. If cos θ=1/4, then AC= sqrt(13 -12*(1/4))=sqrt(13 -3)=sqrt(10)≈3.16. Then, in triangle ADC with AD=2, DC=2, AC=sqrt(10), the triangle is valid.But to ensure D is inside triangle ABC, perhaps when cos θ=1/4, point D coincides with a vertex or lies on an edge. Alternatively, use coordinates to check.Let’s try θ where cos θ=1/4. Then, coordinates of C would be (3*(1/4), 3*sqrt(1 - (1/4)^2 ))=(3/4, 3*sqrt(15)/4). Then AC= sqrt( (3/4 -2)^2 + (3 sqrt(15)/4)^2 )= sqrt( (-5/4)^2 + (9*15)/16 )= sqrt(25/16 +135/16)=sqrt(160/16)=sqrt(10), which matches. Point D has AD=2, which in coordinates would mean sqrt( (2 -p)^2 + q² )=2. So (2 -p)^2 + q²=4. Also, CD=2: (p -3/4)^2 + (q -3 sqrt(15)/4)^2=4.Let me solve these equations for p and q.Equation 1: (2 -p)^2 + q²=4 Equation 2: (p -3/4)^2 + (q -3 sqrt(15)/4)^2=4Expand equation 1:4 -4p +p² + q²=4 =>p² + q² -4p=0 Equation 2:p² - (3/2)p +9/16 + q² - (3 sqrt(15)/2)q + (9*15)/16=4 Simplify equation 2:p² + q² - (3/2)p - (3 sqrt(15)/2)q +9/16 +135/16=4 Combine constants: (9+135)/16=144/16=9. So equation 2 becomes:p² + q² - (3/2)p - (3 sqrt(15)/2)q +9=4 =>p² + q² - (3/2)p - (3 sqrt(15)/2)q +5=0From equation 1:p² + q²=4pSubstitute into equation 2:4p - (3/2)p - (3 sqrt(15)/2)q +5=0 => (5/2)p - (3 sqrt(15)/2)q +5=0 Multiply by 2:5p -3 sqrt(15) q +10=0 Solve for q:3 sqrt(15) q=5p +10 =>q=(5p +10)/(3 sqrt(15))Now, substitute q into equation 1: p² + q²=4pSo p² + [ (5p +10)^2 ] / (9*15) =4p Calculate denominator:9*15=135 Numerator:(25p² +100p +100)/135 Thus:p² + (25p² +100p +100)/135 =4p Multiply through by 135:135p² +25p² +100p +100=540p Combine terms:160p² +100p +100 -540p=0 160p² -440p +100=0 Divide by 20:8p² -22p +5=0 Solve quadratic: p=[22 ±sqrt(484 -160)]/16=[22 ±sqrt(324)]/16=[22 ±18]/16Solutions:(22+18)/16=40/16=2.5 (22-18)/16=4/16=0.25So p=2.5 or p=0.25Check p=2.5: Then from equation 1: (2 -2.5)^2 + q²=4 =>0.25 + q²=4 =>q²=3.75 =>q=±sqrt(15)/2≈±1.936But point D must be inside triangle ABC. Coordinates of C are (3/4,3 sqrt(15)/4≈2.904). If p=2.5, which is greater than the x-coordinate of point A (2,0), so outside the triangle. Hence, p=2.5 is invalid.For p=0.25: q=(5*0.25 +10)/(3 sqrt(15))=(1.25 +10)/3 sqrt(15)=11.25/(3 sqrt(15))=3.75/sqrt(15)=3.75 sqrt(15)/15=0.25 sqrt(15)≈0.25*3.872≈0.968So coordinates of D are (0.25,0.968). Check if this is inside triangle ABC. Points A(2,0), B(0,0), C(3/4,2.904). The triangle ABC has vertices at (0,0), (2,0), and (0.75,2.904). The point D(0.25,0.968) is inside this triangle. To confirm, we can check the barycentric coordinates or see if it's within the edges.Alternatively, since D has x-coordinate 0.25, which is between 0 and 2, and y-coordinate 0.968, which is positive and below the line from B to C. Let me compute the equation of edge BC. Points B(0,0) to C(0.75,2.904). The slope is (2.904 -0)/(0.75 -0)=2.904/0.75≈3.872. The equation is y=3.872x. At x=0.25, y=3.872*0.25≈0.968, which is exactly the y-coordinate of D. Therefore, point D lies on the edge BC. But the problem states that D is inside the triangle. Therefore, when cos θ=1/4, point D lies on the edge BC, which is the boundary. Hence, for D to be strictly inside the triangle, cos θ must be greater than 1/4.Therefore, the lower bound is cos θ>1/4. Combining with the upper bound from before, cos θ<3/4, we get 1/4 <cos θ<3/4. Hence, proved.Final Answerboxed{left( frac{1}{4}, frac{3}{4} right)}Alternatively, as per the problem statement, the answer should be boxed as boxed{frac{1}{4}<cos B<frac{3}{4}}.Final Answerboxed{frac{1}{4}<cos B<frac{3}{4}}</think>Given triangle ( triangle ABC ) with ( AB = 2 ), ( BC = 3 ), and a point ( D ) inside the triangle such that ( CD = 2 ) and ( angle ADC = 180^circ - angle B ), we need to prove ( frac{1}{4} < cos B < frac{3}{4} ).1. Law of Cosines in ( triangle ABC ): [ AC^2 = AB^2 + BC^2 - 2 cdot AB cdot BC cdot cos B = 4 + 9 - 12 cos B = 13 - 12 cos B ] Thus, ( AC = sqrt{13 - 12 cos B} ).2. Law of Cosines in ( triangle ADC ): Given ( angle ADC = 180^circ - B ), we have: [ AC^2 = AD^2 + CD^2 - 2 cdot AD cdot CD cdot cos(180^circ - B) ] Since ( cos(180^circ - B) = -cos B ), this simplifies to: [ 13 - 12 cos B = AD^2 + 4 + 4 cdot AD cdot cos B ] Rearranging gives: [ AD^2 + 4 cdot AD cdot cos B + 12 cos B - 9 = 0 ]3. Solving the Quadratic Equation: The quadratic equation in ( AD ): [ AD^2 + 4 cos B cdot AD + 12 cos B - 9 = 0 ] The discriminant must be non-negative for real solutions: [ (4 cos B)^2 - 4 cdot 1 cdot (12 cos B - 9) geq 0 implies 16 cos^2 B - 48 cos B + 36 geq 0 ] This simplifies to: [ (4 cos B - 6)^2 geq 0 ] Which is always true. Solving for ( AD ): [ AD = frac{-4 cos B pm (6 - 4 cos B)}{2} ] The valid solution is: [ AD = 3 - 4 cos B ] Since ( AD ) must be positive, ( 3 - 4 cos B > 0 implies cos B < frac{3}{4} ).4. Lower Bound for ( cos B ): Considering ( D ) inside ( triangle ABC ), when ( cos B = frac{1}{4} ), point ( D ) lies on the boundary. For ( D ) to be strictly inside, ( cos B ) must be greater than ( frac{1}{4} ).Thus, combining both bounds, we get:[boxed{frac{1}{4} < cos B < frac{3}{4}}]]

🔑:10. Construct the point D' symmetric to D with respect to AC. Thenangle A D' C = angle A D C = 180^{circ} - angle BRightarrow A, B, C, D' are concyclic.BSince C D' = C D = 2 = A B, we have B C parallel A D'.From the fact that point D is inside triangle ABC, we knowbegin{array}{l}angle D' C A = angle D C A < angle A C B = angle C A D' Rightarrow A D' < C D' = 2 .end{array}Let A D = x. Then A D' = x in (0, 2).By the Law of Cosines, we havebegin{array}{l}A B^{2} + B C^{2} - 2 A B cdot B C cos B = A C^{2} = D' A^{2} + D' C^{2} - 2 D' A cdot D' C cos angle A D' C Rightarrow 2^{2} + 3^{2} - 2 times 2 times 3 cos B quad = x^{2} + 2^{2} - 2 x cdot 2(-cos B) Rightarrow x^{2} + 4 x cos B + 12 cos B - 9 = 0 . Rightarrow x = -4 cos B + 3 text{ or } x = -3 text{ (rejected) } Rightarrow 0 < -4 cos B + 3 < 2 Rightarrow frac{1}{4} < cos B < frac{3}{4} .end{array}

🧠:[<think>Okay, let's see. I need to find the constant 'a' such that the maximum value of the function y = |sin²x - 4sinx - a| is 4. Hmm, alright. Let me start by understanding the problem step by step.First, the function given is the absolute value of another function: sin²x - 4sinx - a. So, y = |f(x)| where f(x) = sin²x - 4sinx - a. The maximum value of this absolute function is 4. That means, for all x, |f(x)| ≤ 4, and there exists some x where |f(x)| = 4.So, my goal is to find 'a' such that the maximum of |f(x)| is exactly 4. To do this, I need to analyze the function f(x) = sin²x - 4sinx - a.Let me first consider substituting t = sinx. Since sinx ranges between -1 and 1, t ∈ [-1, 1]. Then, the function f(x) can be rewritten in terms of t:f(t) = t² - 4t - aBut since t is in [-1, 1], the function f(t) is a quadratic function in t. However, since t is restricted to [-1, 1], the maximum and minimum of f(t) will occur either at the critical points within this interval or at the endpoints.So, let's analyze f(t) = t² - 4t - a over the interval t ∈ [-1, 1].First, find the derivative of f(t) with respect to t to locate critical points.f'(t) = 2t - 4Set derivative equal to zero:2t - 4 = 0 ⇒ t = 2But t = 2 is outside the interval [-1, 1], so there are no critical points within the interval. Therefore, the extrema of f(t) on [-1, 1] must occur at the endpoints t = -1 and t = 1.Let me compute f(-1) and f(1):At t = -1:f(-1) = (-1)² - 4*(-1) - a = 1 + 4 - a = 5 - aAt t = 1:f(1) = (1)² - 4*(1) - a = 1 - 4 - a = -3 - aSo, over the interval t ∈ [-1, 1], f(t) ranges from -3 - a (at t = 1) to 5 - a (at t = -1). Wait, but since there are no critical points inside the interval, these are the only two points to consider.But since the quadratic function f(t) = t² -4t -a opens upwards (since the coefficient of t² is positive), but since the vertex is at t = 2 (which is outside the interval), the function is decreasing on the interval [-1, 1]. Wait, since the derivative is 2t - 4, which for t in [-1, 1], the derivative is always negative (since 2t is at most 2*1 = 2, so 2t -4 is at most 2 -4 = -2). So, the function is decreasing on the entire interval [-1, 1]. Therefore, the maximum value is at t = -1 (which is 5 - a) and the minimum value is at t = 1 (which is -3 - a).Therefore, the original function f(t) ranges from -3 - a to 5 - a. So, when we take the absolute value |f(t)|, the maximum of |f(t)| could be either |5 - a| or |-3 - a|, whichever is larger. However, depending on the value of 'a', the maximum of |f(t)| could also occur at a point where f(t) crosses zero. Wait, but since we have a continuous function on a closed interval, the maximum of |f(t)| will occur either at the endpoints or at a point where f(t) = 0, but since f(t) is monotonic (since it's decreasing), it can cross zero at most once. So, if the function crosses zero between t = -1 and t = 1, then the maximum absolute value might be at the crossing point? Wait, no. Let me think again.Wait, if the function is continuous and monotonic (strictly decreasing here), then it crosses zero at most once. So, the function goes from 5 - a at t = -1 to -3 - a at t = 1. So, if 5 - a and -3 - a have opposite signs, then the function crosses zero somewhere in between. But in that case, the maximum of |f(t)| would still occur at one of the endpoints, since the function is linear (if it were linear). Wait, but actually, the function is quadratic here. Wait, no, in terms of t, f(t) is quadratic, but in terms of x, t is sinx. Wait, but we substituted t = sinx, so over the interval t in [-1,1], f(t) is a quadratic function. However, in our case, since f(t) is a quadratic function in t, which is being considered over the interval t ∈ [-1,1]. But since the function is quadratic but restricted to a linear path? Wait, maybe I confused something here.Wait, perhaps I need to correct myself. The function f(t) = t² -4t -a is indeed quadratic in t, but over the interval t ∈ [-1,1], and since the vertex is at t = 2, which is outside the interval, the function is decreasing over the entire interval [-1,1]. So, f(t) is decreasing from t = -1 to t = 1. So, as t increases from -1 to 1, f(t) decreases from 5 - a to -3 - a. Therefore, the function f(t) is a straight line in terms of behavior (monotonically decreasing), but it's actually a quadratic curve. However, since the interval is between -1 and 1, and the vertex is outside, the function does not change direction in this interval. Therefore, it is strictly decreasing.Therefore, the maximum of |f(t)| would occur either at t = -1 or t = 1, unless the function crosses zero within the interval, in which case the point where f(t) = 0 would have |f(t)| = 0, which is not a maximum. Wait, but if the function crosses zero, then near that point, |f(t)| would be small. However, depending on the values at the endpoints, the maximum of |f(t)| could still be at the endpoints. Let me check with some examples.Suppose 5 - a and -3 - a are both positive. Then, |f(t)| is just f(t), which is decreasing from 5 - a to -3 - a. But if both are positive, then -3 - a > 0 implies a < -3. Then, the maximum is at t = -1: 5 - a. But since a < -3, 5 - a would be greater than 5 - (-3) = 8. But the maximum value is supposed to be 4, so this case is impossible.Similarly, if both endpoints are negative, then |f(t)| would be -f(t), which is increasing from -(5 - a) to -(-3 - a) = 3 + a. So, if both are negative, then 5 - a < 0 implies a > 5, and -3 - a < 0 implies a > -3. But if a > 5, then the maximum of |f(t)| is at t = 1: 3 + a. We set this equal to 4: 3 + a = 4 ⇒ a = 1. But wait, if a = 1, which is not greater than 5. Contradiction. So, that can't be.Therefore, the function must cross zero in the interval. That is, 5 - a > 0 and -3 - a < 0. So, 5 - a > 0 implies a < 5, and -3 - a < 0 implies a > -3. So, a ∈ (-3, 5). In this case, the function f(t) starts at 5 - a (positive) and decreases to -3 - a (negative), crossing zero somewhere in between. Therefore, the maximum of |f(t)| would be the maximum of the absolute values at the endpoints, i.e., max{5 - a, | -3 - a |}. However, since the function crosses zero, there might be a point where |f(t)| reaches a maximum somewhere inside the interval? Wait, let's see.But since f(t) is continuous and strictly decreasing from positive to negative, the maximum of |f(t)| would occur at whichever endpoint is further from zero. Let me think. Suppose 5 - a > | -3 - a |. Then, the maximum is 5 - a. If | -3 - a | > 5 - a, then the maximum is | -3 - a |. So, we need to compare these two values and set the larger one equal to 4.So, the maximum of |f(t)| is max{5 - a, 3 + a} (since | -3 - a | = |a + 3| = 3 + a when a + 3 ≥ 0, i.e., a ≥ -3, which is the case here because a ∈ (-3,5)). Therefore, the maximum is the larger of 5 - a and 3 + a. To find when these two are equal: 5 - a = 3 + a ⇒ 5 -3 = a + a ⇒ 2 = 2a ⇒ a = 1.So, when a < 1, 5 - a > 3 + a, so the maximum is 5 - a.When a ≥ 1, 3 + a ≥ 5 - a, so the maximum is 3 + a.Therefore, the maximum of |f(t)| is:- 5 - a, if a ≤ 1- 3 + a, if a ≥ 1But we are given that the maximum value is 4. Therefore, we set both expressions equal to 4 and check the possible values of a.Case 1: a ≤ 1, so 5 - a = 4 ⇒ a = 1.Case 2: a ≥ 1, so 3 + a = 4 ⇒ a = 1.Therefore, in both cases, a = 1. But wait, a = 1 is the boundary point between the two cases. Let me check if a = 1 satisfies the original condition.If a = 1, then f(t) = t² -4t -1.At t = -1: f(-1) = 1 +4 -1 = 4. So, |4| = 4.At t = 1: f(1) = 1 -4 -1 = -4. So, |-4| = 4.Therefore, indeed, the maximum of |f(t)| is 4 when a =1.But wait, let's confirm that there isn't a higher value somewhere in between. For example, if a were different, maybe the function inside the absolute value reaches higher than 4. Wait, but since the maximum is set to 4, and we found a=1 by equating the endpoints, but since the function is decreasing from 5 - a to -3 -a, the maximum absolute value is at the endpoints. However, when a =1, both endpoints give |4| and |-4|, so the maximum is 4. If a were different, say a=0, then the endpoints would be 5 -0 =5 and -3 -0= -3, so |5| =5 and |-3|=3, so maximum is 5, which is higher than 4. Similarly, if a=2, which is greater than 1, then 5 -2=3 and 3 +2=5, so maximum is 5. But we need the maximum to be 4. Therefore, only when a=1, both endpoints give 4. Therefore, a=1 is the answer.But wait, just to be thorough, let's check if when a=1, is there any other point in the interval where |f(t)| could be larger than 4? For a=1, f(t)= t² -4t -1. Let's see, f(t) is a quadratic function. Let's find its vertex. The vertex is at t = 2, which is outside the interval. Since the function is decreasing on [-1,1], the maximum value in the interval is at t=-1, which is 4, and the minimum is at t=1, which is -4. Therefore, the absolute value at both endpoints is 4, and since the function is decreasing, between t=-1 and t=1, the function crosses zero. Let's find where it crosses zero. Solve t² -4t -1 =0.Solution: t = [4 ± sqrt(16 +4)]/2 = [4 ± sqrt(20)]/2 = [4 ± 2*sqrt(5)]/2 = 2 ± sqrt(5). sqrt(5) is approximately 2.236, so 2 + sqrt(5) ≈4.236 and 2 - sqrt(5)≈-0.236. So, in the interval [-1,1], the root is at t≈-0.236. At that point, f(t)=0, so |f(t)|=0. Therefore, between t=-1 and t≈-0.236, f(t) is positive and decreasing from 4 to 0, and between t≈-0.236 and t=1, f(t) is negative and decreasing from 0 to -4. Therefore, the maximum absolute value is indeed 4 at both endpoints.Therefore, a=1 is the correct answer.But let me double-check if there is any other possible value of 'a' that could make the maximum |f(t)| equal to 4. For instance, suppose the maximum occurs not at the endpoints but somewhere inside the interval. Wait, but we saw that the function f(t) is monotonic, so the extremum of f(t) is at the endpoints, but since we take absolute value, maybe there is a case where the maximum of |f(t)| occurs inside the interval? Let's suppose that the maximum of |f(t)| occurs where the derivative of |f(t)| is zero. However, since f(t) is monotonic, |f(t)| will have its maximum at the endpoints unless the function crosses zero, in which case the maximum could still be at the endpoints or at a local extremum of |f(t)|. But since f(t) is linear? Wait, f(t) is quadratic, but over the interval t ∈ [-1,1], f(t) is quadratic but behaves like a linear function since it's strictly decreasing.Wait, let's calculate the derivative of |f(t)|. The derivative of |f(t)| is f'(t) * sign(f(t)). So, critical points occur when f(t) =0 or when f'(t)=0. But f'(t) =2t -4, which is never zero in the interval, as before. Therefore, the only critical points for |f(t)| would be where f(t)=0. At such points, |f(t)| has a minimum (zero), not a maximum. Therefore, the maximum of |f(t)| must occur at the endpoints. Therefore, our previous conclusion that a=1 is correct.Hence, the answer is a=1.Final AnswerThe constant ( a ) is boxed{1}.</think>Given the function ( y = |sin^2 x - 4 sin x - a| ), we need to find the constant ( a ) such that the maximum value of ( y ) is 4. First, we substitute ( t = sin x ), where ( t in [-1, 1] ). This transforms the function into ( y = |t^2 - 4t - a| ). We need to analyze the quadratic function ( f(t) = t^2 - 4t - a ) over the interval ( t in [-1, 1] ).The quadratic function ( f(t) = t^2 - 4t - a ) opens upwards with a vertex at ( t = 2 ), which is outside the interval ([-1, 1]). The derivative ( f'(t) = 2t - 4 ) is always negative in the interval ([-1, 1]), indicating that ( f(t) ) is strictly decreasing on this interval. Therefore, the maximum and minimum values of ( f(t) ) occur at the endpoints ( t = -1 ) and ( t = 1 ).Evaluating ( f(t) ) at the endpoints:- At ( t = -1 ): ( f(-1) = (-1)^2 - 4(-1) - a = 1 + 4 - a = 5 - a )- At ( t = 1 ): ( f(1) = (1)^2 - 4(1) - a = 1 - 4 - a = -3 - a )The function ( f(t) ) ranges from ( 5 - a ) to ( -3 - a ). The maximum value of ( |f(t)| ) must be 4. We consider the absolute values at the endpoints:- ( |5 - a| )- ( |-3 - a| )We need to find ( a ) such that the maximum of these two values is 4. This gives us two equations:1. ( 5 - a = 4 ) leading to ( a = 1 )2. ( -3 - a = -4 ) leading to ( a = 1 )Both cases result in ( a = 1 ). To verify, substituting ( a = 1 ) into the function:- At ( t = -1 ): ( |5 - 1| = 4 )- At ( t = 1 ): ( |-3 - 1| = 4 )Thus, the maximum value of ( |f(t)| ) is indeed 4 when ( a = 1 ).The constant ( a ) is boxed{1}.,<think>Okay, let's see. I need to find the constant 'a' such that the maximum value of the function y = |sin²x - 4 sinx - a| is 4. Hmm. Alright, let's break this down step by step.First, the function inside the absolute value is sin²x - 4 sinx - a. Let me call that f(x) for simplicity. So, f(x) = sin²x - 4 sinx - a. Then y = |f(x)|, and we need the maximum of y to be 4. That means the maximum value of |f(x)| over all x is 4.Since the sine function has a range of [-1, 1], sinx can take any value between -1 and 1. Let me substitute t = sinx, so t ∈ [-1, 1]. Then f(x) becomes t² - 4t - a. So, the problem reduces to finding 'a' such that the maximum of |t² - 4t - a| for t in [-1, 1] is equal to 4.So now, the problem is transformed into finding 'a' such that max_{t ∈ [-1,1]} |t² - 4t - a| = 4.Alright, so I need to analyze the function g(t) = t² - 4t - a on the interval t ∈ [-1, 1], and find 'a' such that the maximum absolute value of g(t) on this interval is 4.To find the maximum of |g(t)|, I should first find the extrema of g(t) on the interval [-1, 1], and then evaluate |g(t)| at those points as well as the endpoints, since the maximum could occur either at a critical point or at the endpoints.First, let's find the critical points of g(t). The derivative of g(t) is g'(t) = 2t - 4. Setting this equal to zero gives 2t - 4 = 0 => t = 2. However, t = 2 is not in the interval [-1, 1], so there are no critical points inside the interval. Therefore, the extrema of g(t) on [-1, 1] must occur at the endpoints.Therefore, we only need to evaluate g(t) at t = -1 and t = 1.Compute g(-1):g(-1) = (-1)^2 - 4*(-1) - a = 1 + 4 - a = 5 - a.Compute g(1):g(1) = (1)^2 - 4*(1) - a = 1 - 4 - a = -3 - a.So, on the interval [-1, 1], the function g(t) ranges from g(-1) = 5 - a to g(1) = -3 - a. Since the function is a quadratic opening upwards (coefficient of t² is positive), but since the critical point is at t=2 which is outside the interval, on the interval [-1, 1], the function is decreasing. Because the derivative is 2t - 4, which for t in [-1, 1] is always negative (since 2t is at most 2, minus 4 gives at most -2). So g(t) is decreasing on [-1, 1]. Therefore, the maximum value of g(t) is at t = -1, which is 5 - a, and the minimum is at t = 1, which is -3 - a.Thus, the function g(t) on [-1,1] is decreasing from 5 - a to -3 - a. Therefore, the range of g(t) is [ -3 - a, 5 - a ].But since we take the absolute value of g(t), we need to find the maximum of |g(t)| over this interval. Since the range of g(t) is [ -3 - a, 5 - a ], the maximum of |g(t)| could be either |5 - a| or | -3 - a |, or possibly at some point where g(t) crosses zero, but let's check.But since the function is linear? Wait, no, g(t) is quadratic, but in terms of t, but we already reduced it to the interval [-1,1], where it's decreasing. So the function is moving from 5 - a down to -3 - a as t goes from -1 to 1. So depending on the value of 'a', the function might cross zero somewhere in between. The maximum absolute value would be either at the endpoints or at the point where it crosses zero. However, if the function doesn't cross zero, then the maximum absolute value is either at t=-1 or t=1.Therefore, to find the maximum of |g(t)|, we need to consider two cases:1. The function g(t) does not cross zero in [-1,1]. Then the maximum |g(t)| is the maximum of |5 - a| and | -3 - a |.2. The function g(t) crosses zero in [-1,1]. Then the maximum |g(t)| could be at the crossing point, which would be zero, but that's not the maximum. Wait, no. If the function crosses zero, then the maximum absolute value would still be at the endpoints unless the extremum (which here is outside the interval) affects it. Wait, since the function is decreasing, if it crosses zero, then part of the function is positive and part is negative. The maximum absolute value would then be the maximum of the positive part and the negative part. But since the function is decreasing, the positive part is from t=-1 to the zero crossing, and the negative part is from the zero crossing to t=1. The maximum absolute value in this case would be the maximum of |g(-1)| and |g(1)|, unless the crossing point is such that the maximum inside the positive or negative regions is larger. Wait, but since the function is linear (it's quadratic but restricted to the interval where it's decreasing), actually, g(t) is a quadratic function in t, but over the interval [-1,1], since there's no critical point, it's strictly decreasing. Therefore, if it crosses zero, then the maximum absolute value is the maximum of |g(-1)| and |g(1)|, same as if it doesn't cross zero. Wait, but perhaps not. Let's think.Suppose the function crosses zero at some point t0 in [-1,1]. Then, the maximum of |g(t)| would be the maximum of |g(-1)| and |g(1)| because the function is decreasing. For example, if g(-1) is positive and g(1) is negative, then the maximum absolute value would be the larger of the two endpoint values. However, if one endpoint is positive and the other is negative, the maximum absolute value could be either the positive endpoint or the negative endpoint. But if the function doesn't cross zero, then both endpoints have the same sign, so the maximum absolute value is the larger of the two.Therefore, in either case, the maximum of |g(t)| is the maximum of |5 - a| and | -3 - a |. Therefore, to have the maximum of these two equal to 4.Wait, is that correct? Let me test with an example. Suppose a is such that 5 - a and -3 - a have opposite signs. Then, their absolute values would be |5 - a| and | -3 - a |. The maximum of these two would be the larger of the two magnitudes. If they have the same sign, then the maximum is the larger of the two. If they have opposite signs, maybe the maximum is still the larger of the two? Let's see. Suppose a = 1. Then 5 - 1 = 4, -3 -1 = -4. The absolute values are 4 and 4, so maximum is 4. If a = 0, then 5 -0 =5, -3 -0 = -3. Absolute values 5 and 3, so maximum is 5. If a = 6, then 5 -6 = -1, |-1|=1; -3 -6 = -9, | -9 | =9, so maximum is 9. If a = -5, then 5 - (-5)=10, |10|=10; -3 - (-5)=2, |2|=2, so maximum 10. So in these examples, the maximum is indeed the maximum of |5 -a| and |-3 -a|.But wait, in the case where the function crosses zero in between, perhaps the maximum absolute value could be higher? Wait, for instance, suppose a is such that the function crosses zero at t0. Then, if the function is positive at t=-1 and negative at t=1, then the maximum absolute value is the maximum of the positive value at t=-1 and the absolute negative value at t=1. But since the function is linear (no, it's quadratic, but over the interval, it's decreasing). Wait, g(t) = t² -4t -a. So it's a quadratic function. Its graph is a parabola opening upwards. However, on the interval [-1,1], since the critical point is at t=2, outside the interval, the function is decreasing on [-1,1]. Therefore, it's a part of the parabola that is decreasing. So the function is convex (since the coefficient of t² is positive) and decreasing on the interval. So its graph on [-1,1] is a downward-sloping curve (since it's decreasing) that is convex.Wait, actually, if it's convex (second derivative positive), then the curve is concave upward. But since the first derivative is negative throughout the interval, the function is decreasing. So the graph would be a convex curve decreasing from t=-1 to t=1. Therefore, it might have a steeper decrease at the beginning and then flatten out, but since the derivative is 2t -4, at t=-1, derivative is -6, and at t=1, derivative is -2. So the slope becomes less negative as t increases. So the function is decreasing with decreasing steepness. Therefore, the graph is a convex curve decreasing from t=-1 to t=1.Therefore, the maximum of |g(t)| could indeed occur somewhere inside the interval if the function crosses zero. For example, suppose that g(t) crosses zero at some point t0 in [-1,1]. Then, the maximum of |g(t)| would be the maximum of |g(-1)| and |g(1)|, but is that necessarily true? Wait, consider the case where the maximum of |g(t)| occurs at t0 where g(t0)=0. But zero is not the maximum. Wait, no. If the function crosses zero, then on one side of t0, the function is positive, and on the other side, it's negative. The maximum absolute value would be the maximum of the highest positive value and the lowest negative value. But since the function is decreasing, the highest positive value is at t=-1 and the lowest negative value is at t=1. Therefore, even if the function crosses zero, the maximum |g(t)| is still the maximum of |g(-1)| and |g(1)|. Wait, but actually, maybe not. Suppose the function is positive at t=-1, crosses zero at some t0, and becomes negative. Then, the maximum positive value is at t=-1, and the maximum negative value is at t=1. So their absolute values are |g(-1)| and |g(1)|. But what if the function attains a larger absolute value somewhere inside the interval?Wait, let's take an example. Suppose a = 1. Then g(t) = t² -4t -1. Let's compute g(-1) = 1 +4 -1 =4, g(1)=1 -4 -1=-4. So the function goes from 4 to -4. Since it's a convex function decreasing on [-1,1], it will cross zero somewhere in between. The maximum of |g(t)| would be 4, achieved at both endpoints. So in this case, the maximum is indeed 4. However, if a is different, say a = 0. Then g(t) = t² -4t. Then g(-1) =1 +4=5, g(1)=1 -4=-3. The maximum absolute value is 5. The function crosses zero at t=0 and t=4, but t=4 is outside the interval. So on [-1,1], it crosses zero at t=0. The maximum |g(t)| is 5 at t=-1. So even with crossing zero, the maximum is at the endpoint.Another example, a = 5. Then g(t) = t² -4t -5. g(-1)=1 +4 -5=0, g(1)=1 -4 -5=-8. The function decreases from 0 to -8. So the maximum |g(t)| is 8 at t=1. So again, maximum at the endpoint.Wait, but if a is such that the maximum inside the interval is higher? Let's see. Let me choose a value of 'a' such that the vertex of the parabola is inside the interval. But wait, the vertex is at t=2, which is outside the interval. Therefore, the function is decreasing on the entire interval [-1,1], so its maximum on the interval is at t=-1, and minimum at t=1. Therefore, regardless of 'a', the maximum and minimum of g(t) on [-1,1] are at the endpoints. Therefore, the maximum of |g(t)| must be either |g(-1)| or |g(1)|. Therefore, the problem reduces to solving max{ |5 - a|, | -3 - a | } = 4.Therefore, we need to find 'a' such that the larger of |5 - a| and | -3 - a | is equal to 4. So, we can set up the equation:max{ |5 - a|, | -3 - a| } = 4.This can be split into cases.Case 1: |5 - a| ≥ | -3 - a |. Then, |5 - a| = 4.Case 2: | -3 - a | > |5 - a|. Then, | -3 - a | =4.So, we need to solve both possibilities.First, solve |5 - a| =4.Solutions:5 - a =4 => a=1or 5 -a = -4 => a=9Similarly, solve | -3 -a | =4.=> |-3 -a| =4=> -3 -a =4 => -a=7 => a= -7or -3 -a= -4 => -a = -1 => a=1So, the solutions are a=1, a=9, a=-7.But we need to check whether these solutions satisfy the case conditions.For example, for a=1:Check if |5 -1| ≥ | -3 -1 | => |4| ≥ |-4| =>4 ≥4, which is true. So a=1 is valid for case 1.For a=9:Check if |5 -9| ≥ | -3 -9 | =>| -4 | ≥ | -12 | =>4 ≥12? No, 4 is not ≥12. Therefore, a=9 does not satisfy case 1 condition. Therefore, even though |5 -9|=4, since 4 <12, it doesn't satisfy the case 1 condition, so we must discard a=9.Similarly, for a=-7:Check if | -3 - (-7)| > |5 - (-7)| => |4| > |12| =>4 >12? No, false. Therefore, a=-7 would be in case 2, but since 4 is not greater than 12, it's invalid. So, even though |-3 -(-7)|=4, since |5 - (-7)|=12 >4, the maximum is 12, which is not 4. Therefore, a=-7 is invalid.Wait, so only a=1 is valid?Wait, but hold on, when we solve | -3 -a | =4, we get a=1 and a=-7. Let's check for a=1:We already saw that a=1 satisfies case 1. For a=-7:Check if | -3 - (-7)| = |4|=4, which is equal to 4, but we have to check if it's the maximum. So compute |5 - (-7)|=12 and | -3 - (-7)|=4. The maximum is 12, which is not 4. Therefore, even though |-3 -a|=4, the maximum is actually |5 -a|=12, so this doesn't satisfy the condition that the maximum is 4. Therefore, a=-7 is invalid.Similarly, a=9:Compute |5 -9|=4 and | -3 -9 |=12, maximum is 12, which is invalid. So a=9 is invalid.Therefore, the only valid solution is a=1.Wait, but let's check this. If a=1, then g(t) = t² -4t -1. Then, g(-1)=1 +4 -1=4, g(1)=1 -4 -1=-4. Therefore, |g(-1)|=4, |g(1)|=4. So the maximum of |g(t)| is 4, which is exactly what we need. So a=1 works.But let me check if there are other possible solutions. Suppose there is an 'a' such that the maximum of |g(t)| is 4, but achieved somewhere inside the interval. However, earlier reasoning suggested that because the function is decreasing on [-1,1], the maximum must be at the endpoints. But let me confirm this.Suppose we set 'a' such that the function g(t) reaches 4 or -4 inside the interval. For example, suppose that g(t) =4 somewhere in t ∈ (-1,1). Let's solve t² -4t -a =4. So t² -4t - (a +4)=0. The solutions are t = [4 ± sqrt(16 + 4(a +4))]/2 = [4 ± sqrt(16 +4a +16)]/2 = [4 ± sqrt(4a +32)]/2 = [4 ± 2sqrt(a +8)]/2 = 2 ± sqrt(a +8).We need t ∈ [-1,1]. So 2 ± sqrt(a +8) must be in [-1,1]. Let's see:For t = 2 + sqrt(a +8): Since sqrt(a +8) is non-negative, 2 + sqrt(a +8) ≥2, which is outside the interval [-1,1]. So this solution is invalid.For t=2 - sqrt(a +8): We need 2 - sqrt(a +8) ≥ -1 and 2 - sqrt(a +8) ≤1.First, 2 - sqrt(a +8) ≤1 => sqrt(a +8) ≥1 => a +8 ≥1 => a ≥-7.Second, 2 - sqrt(a +8) ≥ -1 => sqrt(a +8) ≤3 => a +8 ≤9 => a ≤1.Thus, for a ∈ [-7,1], the equation g(t)=4 has a solution t=2 - sqrt(a +8) in [-1,1].Similarly, if we set g(t) = -4, then t² -4t -a = -4 => t² -4t - (a -4)=0.Solutions: t = [4 ± sqrt(16 +4(a -4))]/2 = [4 ± sqrt(16 +4a -16)]/2 = [4 ± sqrt(4a)]/2 = [4 ± 2sqrt(a)]/2 = 2 ± sqrt(a).Again, t must be in [-1,1]. For t=2 + sqrt(a), same issue as before, t ≥2, invalid. For t=2 - sqrt(a), we need:2 - sqrt(a) ≤1 => sqrt(a) ≥1 => a ≥1,and 2 - sqrt(a) ≥-1 => sqrt(a) ≤3 => a ≤9.Thus, for a ∈ [1,9], t=2 - sqrt(a) is in [-1,1].Therefore, if a ∈ [-7,1], then g(t)=4 is achieved at t=2 - sqrt(a +8) ∈ [-1,1], and if a ∈ [1,9], then g(t)=-4 is achieved at t=2 - sqrt(a) ∈ [-1,1].But wait, how does this affect the maximum of |g(t)|?If the function reaches 4 or -4 inside the interval, then the maximum |g(t)| could be 4. However, we also need to check the endpoints. For example, if a=1, then the function reaches 4 at t=-1 and -4 at t=1, so the maximum is 4. If a is such that g(t)=4 is achieved inside the interval, but the endpoint values are larger in absolute value, then the maximum would still be at the endpoints.Therefore, perhaps there are multiple solutions for 'a' where either the maximum is achieved at the endpoints or inside the interval. Let's check.Suppose a is in [-7,1]. Then g(t) reaches 4 at some t0 in [-1,1], but we need to check if |g(-1)| and |g(1)| are less than or equal to 4.Given that a ∈ [-7,1], let's compute |g(-1)| = |5 - a|. If a ∈ [-7,1], then 5 -a ∈ [4,12]. So |g(-1)|=5 -a ≥4. Similarly, |g(1)| = |-3 -a|= | - (3 + a) | = |3 + a|. If a ∈ [-7,1], then 3 + a ∈ [-4,4]. Therefore, |g(1)| ∈ [0,4].So, for a ∈ [-7,1], |g(-1)| ranges from 4 to 12, and |g(1)| ranges from 0 to4. Therefore, the maximum of |g(t)| would be |g(-1)|, which is 5 -a. For the maximum to be 4, we need 5 -a=4 => a=1. So even though the function reaches 4 inside the interval when a ∈ [-7,1], the endpoint at t=-1 would have a higher value unless 5 -a=4. Therefore, only a=1 in this interval gives the maximum |g(t)|=4.Similarly, for a ∈ [1,9], the function reaches -4 at some t0 in [-1,1]. Then, |g(t0)|=4. But also, |g(-1)|=|5 -a|, and |g(1)|=|-3 -a|=|3 +a|. For a ∈ [1,9], 5 -a ∈ [-4,4], and 3 + a ∈ [4,12]. Therefore, |g(-1)|=|5 -a| ∈ [0,4], and |g(1)|= |3 +a| ∈ [4,12]. Therefore, the maximum |g(t)| is |g(1)|=3 +a. For this to be 4, we need 3 +a=4 =>a=1. Again, only a=1 in this interval gives |g(1)|=4. So in both cases, a=1 is the only solution.Therefore, regardless of whether the 4 is achieved inside the interval or at the endpoints, the only possible value for 'a' that makes the maximum of |g(t)| equal to 4 is a=1.But wait, let me check another example. Suppose a= -7. Then, g(t)=t² -4t -(-7)=t² -4t +7. Then, compute g(-1)=1 +4 +7=12, g(1)=1 -4 +7=4. Therefore, the maximum |g(t)| is 12, which is not 4. So a=-7 is invalid.Similarly, if a=9, then g(t)=t² -4t -9. Then, g(-1)=1 +4 -9= -4, g(1)=1 -4 -9= -12. So the maximum |g(t)| is 12, again invalid.Therefore, the only valid solution is a=1.Wait, but let me check one more case. Suppose a=0. Then, g(t)=t² -4t -0=t² -4t. Then, g(-1)=1 +4=5, g(1)=1 -4=-3. The maximum |g(t)| is 5. So to get the maximum down to 4, we need a=1. Therefore, the answer should be a=1.But let me verify with the original function. If a=1, then y=|sin²x -4 sinx -1|. Let's compute the maximum of this function.Since t=sinx ∈ [-1,1], we have y=|t² -4t -1|. When t=-1, y=|1 +4 -1|=|4|=4. When t=1, y=|1 -4 -1|=|-4|=4. For t in between, let's check t=0: y=|0 -0 -1|=1. t=0.5: y=|0.25 -2 -1|=| -2.75 | =2.75. t= -0.5: y=|0.25 +2 -1|=|1.25|=1.25. So the maximum is indeed 4, achieved at t=-1 and t=1. Therefore, a=1 is correct.Therefore, the answer is a=1.Final AnswerThe constant ( a ) is (boxed{1}).</think>Given the function ( y = |sin^2 x - 4 sin x - a| ), we need to find the constant ( a ) such that the maximum value of ( y ) is 4. First, we substitute ( t = sin x ) where ( t in [-1, 1] ). This transforms the function into ( y = |t^2 - 4t - a| ). We need to find ( a ) such that the maximum value of ( |t^2 - 4t - a| ) over ( t in [-1, 1] ) is 4.The function ( g(t) = t^2 - 4t - a ) is a quadratic function. Since the coefficient of ( t^2 ) is positive, the parabola opens upwards. The critical point ( t = 2 ) is outside the interval ([-1, 1]), so ( g(t) ) is decreasing on ([-1, 1]).Evaluating ( g(t) ) at the endpoints:- At ( t = -1 ): ( g(-1) = (-1)^2 - 4(-1) - a = 1 + 4 - a = 5 - a )- At ( t = 1 ): ( g(1) = (1)^2 - 4(1) - a = 1 - 4 - a = -3 - a )Since ( g(t) ) is decreasing, the maximum value on ([-1, 1]) is ( 5 - a ) and the minimum value is ( -3 - a ). The maximum absolute value of ( g(t) ) is the maximum of ( |5 - a| ) and ( |-3 - a| ).We need:[max(|5 - a|, |-3 - a|) = 4]Solving the equations:1. ( |5 - a| = 4 ) - ( 5 - a = 4 ) gives ( a = 1 ) - ( 5 - a = -4 ) gives ( a = 9 )2. ( |-3 - a| = 4 ) - ( -3 - a = 4 ) gives ( a = -7 ) - ( -3 - a = -4 ) gives ( a = 1 )Checking the solutions:- For ( a = 9 ), ( |5 - 9| = 4 ) and ( |-3 - 9| = 12 ), which is invalid.- For ( a = -7 ), ( |5 - (-7)| = 12 ) and ( |-3 - (-7)| = 4 ), which is invalid.- For ( a = 1 ), ( |5 - 1| = 4 ) and ( |-3 - 1| = 4 ), both valid.Thus, the only valid solution is ( a = 1 ).The constant ( a ) is (boxed{1}).]

🔑:AnalysisThis question mainly examines the application of the properties and graphs of trigonometric functions, which is a common type of question in college entrance examinations and is considered basic.SolutionFrom the given, we have y=|sin^2x-4sin x-a|=|(sin x-2)^2-4-a|. Since |sin x| leqslant 1, when sin x=-1, the function f(x) reaches its maximum value,which means |5-a|=4. Therefore, a=1.Hence, the answer is boxed{1}.

🧠:[<think>Okay, so I need to figure out which of these four options (A, B, C, D) is the correct calculation. Let me go through each one step by step. Starting with Option A: (-7) + (-7) = 0. Hmm, adding two negative numbers... When you add a negative number to another negative number, the result should be more negative, right? So, (-7) plus (-7) would be like subtracting 7 twice, which should be -14. But the answer here says it's 0. That doesn't make sense. Maybe I made a mistake? Wait, no, if you have negative 7 and add another negative 7, you combine their magnitudes. 7 + 7 is 14, so with the negative sign, it's -14. So definitely, Option A is incorrect. The answer given here is 0, which is wrong. So A is out.Moving on to Option B: (-1/10) - (+1/10) = 0. Let's break this down. Subtracting a positive number is the same as adding a negative number. So this becomes (-1/10) + (-1/10). Adding two negative fractions. So, (-1/10) + (-1/10) would be -2/10, which simplifies to -1/5. But the answer here is 0. That's not right. So Option B is also incorrect. Maybe the person thought subtracting a positive would cancel out the negative, but no, subtracting a positive from a negative makes it more negative. So B is wrong.Next is Option C: 0 + (-101) = 101. Hmm, adding a negative number to 0. Adding a negative number is like subtracting from zero. So 0 + (-101) should be -101. The answer here says positive 101, which is the opposite sign. That's definitely wrong. So Option C is incorrect. Maybe they thought the negative would cancel, but no, adding a negative to zero just gives the negative number. So C is wrong too.Now Option D: (+1/3) + (-1/2) = -1/6. Let me calculate this. To add 1/3 and -1/2, I need a common denominator. The denominators are 3 and 2, so the least common denominator is 6. Converting each fraction: 1/3 is equivalent to 2/6, and -1/2 is equivalent to -3/6. Now, adding them together: 2/6 + (-3/6) = (2 - 3)/6 = (-1)/6. Which is -1/6. So the calculation in Option D is correct. Let me double-check that. Yes, 1/3 minus 1/2. Converting to sixths: 2/6 - 3/6 = -1/6. Exactly. So D is correct. The other options all have errors in their arithmetic operations, either adding negatives as positives, misunderstanding subtraction, or sign errors. So the correct answer must be D.Final Answerboxed{D}</think>To determine which calculation is correct, we analyze each option step by step:- Option A: ((-7) + (-7) = 0) Adding two negative numbers results in a more negative number: ((-7) + (-7) = -14). This is incorrect.- Option B: (left(-frac{1}{10}right) - left(+frac{1}{10}right) = 0) Subtracting a positive number is equivalent to adding a negative number: (left(-frac{1}{10}right) + left(-frac{1}{10}right) = -frac{2}{10} = -frac{1}{5}). This is incorrect.- Option C: (0 + (-101) = 101) Adding a negative number to zero results in that negative number: (0 + (-101) = -101). This is incorrect.- Option D: (left(+frac{1}{3}right) + left(-frac{1}{2}right) = -frac{1}{6}) Converting to a common denominator: (frac{1}{3} = frac{2}{6}) and (-frac{1}{2} = -frac{3}{6}). Adding these gives (frac{2}{6} + -frac{3}{6} = -frac{1}{6}). This is correct.Thus, the correct answer is (boxed{D}).,<think>Okay, let me try to figure out which of these calculations is correct. Hmm, there are four options: A, B, C, and D. I need to check each one step by step. Let's start with option A.Option A says (-7) + (-7) equals 0. Wait, when you add two negative numbers, you add their absolute values and keep the negative sign, right? So, -7 plus -7 should be -(7 + 7), which is -14. But the answer here is 0. That doesn't make sense. So A is definitely wrong. Maybe I made a mistake? Let me double-check. If I owe someone 7 dollars and then I owe another 7 dollars, in total, I owe 14 dollars. So yeah, it should be -14. So A is incorrect.Moving on to option B: (-1/10) minus (+1/10) equals 0. Let's compute that. Subtracting a positive is the same as adding a negative. So this becomes (-1/10) + (-1/10). Adding two negatives, similar to the first problem. So, -1/10 -1/10 is -2/10, which simplifies to -1/5. But the answer here is supposed to be 0. That's not right either. So B is also incorrect.Option C: 0 + (-101) equals 101. Hmm, adding a negative number to 0 should just give you that negative number. So 0 plus (-101) is -101. But the answer here is 101. That's the opposite sign. So C is definitely wrong. Maybe they thought subtracting instead of adding? But no, it's addition. So C is not correct.Now, option D: (+1/3) + (-1/2) equals -1/6. Let me compute this. To add fractions with different denominators, we need a common denominator. The denominators are 3 and 2, so the least common denominator is 6. Converting 1/3 to sixths, it's 2/6. Converting -1/2 to sixths, it's -3/6. So adding them together: 2/6 + (-3/6) = (2 - 3)/6 = (-1)/6. Which is -1/6. So that's exactly what option D says. Therefore, D is correct.Wait, let me confirm once more. If I have 1/3 minus 1/2, that's the same as 1/3 + (-1/2). Converting to sixths: 2/6 - 3/6 = -1/6. Yep, that's right. So D is indeed the correct answer.So, summarizing:A: Wrong, because adding two negatives can't be zero.B: Wrong, subtracting a positive from a negative gives a more negative result.C: Wrong, adding a negative to zero gives the negative number.D: Correct, as shown by converting to common denominators.Therefore, the answer should be D.Final Answerboxed{D}</think>To determine which calculation is correct, we evaluate each option step by step:- Option A: ((-7) + (-7) = 0) Adding two negative numbers results in a more negative number: ((-7) + (-7) = -14). Therefore, A is incorrect.- Option B: (left(-frac{1}{10}right) - left(+frac{1}{10}right) = 0) Subtracting a positive is equivalent to adding a negative: (left(-frac{1}{10}right) + left(-frac{1}{10}right) = -frac{2}{10} = -frac{1}{5}). Therefore, B is incorrect.- Option C: (0 + (-101) = 101) Adding a negative number to zero results in that negative number: (0 + (-101) = -101). Therefore, C is incorrect.- Option D: (left(+frac{1}{3}right) + left(-frac{1}{2}right) = -frac{1}{6}) Converting to common denominators: (frac{1}{3} = frac{2}{6}) and (-frac{1}{2} = -frac{3}{6}). Adding these: (frac{2}{6} + -frac{3}{6} = -frac{1}{6}). Therefore, D is correct.Thus, the correct answer is (boxed{D}).]

🔑:(-7) + (-7) = -14, left(-frac{1}{10}right) - left(+frac{1}{10}right) = left(-frac{1}{10}right) + left(-frac{1}{10}right) = -frac{1}{5}, 0 + (-101) = -101, left(+frac{1}{3}right) + left(-frac{1}{2}right) = -frac{1}{6}. Therefore, the correct answer is boxed{D}.

🧠:[<think>Okay, so I need to find the magnitude of a complex number z that satisfies the equation (z + 2)/(z - 2) = i. Hmm, let me think about how to approach this. First, complex numbers can sometimes be tricky, but maybe I can start by letting z be a complex number and express it in terms of its real and imaginary parts. Let me denote z as x + yi, where x and y are real numbers. Then, substitute this into the equation and solve for x and y. Once I have those, I can compute the magnitude |z| which is sqrt(x² + y²). Alternatively, maybe there's a way to solve for z algebraically without breaking it into real and imaginary parts. Let me try that approach first because it might be quicker. Starting with the equation:(z + 2)/(z - 2) = iI can multiply both sides by (z - 2) to get rid of the denominator:z + 2 = i(z - 2)Now, distribute the i on the right side:z + 2 = iz - 2iThen, bring all the terms with z to one side and the constants to the other side. Let's subtract iz and subtract 2 from both sides:z - iz = -2i - 2Factor out z from the left side:z(1 - i) = -2(i + 1)Hmm, so z = [-2(i + 1)] / (1 - i)Now, I need to simplify this expression. To do that, I can multiply the numerator and the denominator by the complex conjugate of the denominator to rationalize it. The complex conjugate of (1 - i) is (1 + i). Let's try that:Multiply numerator and denominator by (1 + i):z = [-2(i + 1)(1 + i)] / [(1 - i)(1 + i)]First, let's compute the denominator. (1 - i)(1 + i) = 1² - (i)² = 1 - (-1) = 2. So denominator is 2.Now the numerator: -2(i + 1)(1 + i). Let me compute (i + 1)(1 + i). Let's expand this:(i + 1)(1 + i) = i*1 + i*i + 1*1 + 1*i = i + i² + 1 + i = (i + i) + (i² + 1) = 2i + (-1 + 1) = 2i + 0 = 2i.Wait, let me check that again. Let me compute step by step:First term: i*1 = iSecond term: i*i = i² = -1Third term: 1*1 = 1Fourth term: 1*i = iSo adding them all up: i + (-1) + 1 + i = (i + i) + (-1 + 1) = 2i + 0 = 2i. Yes, that's correct.So the numerator becomes -2*(2i) = -4i.Therefore, z = (-4i)/2 = -2i.Wait, so z is -2i? Let me check that again.Let me verify if z = -2i satisfies the original equation.Left-hand side: (z + 2)/(z - 2) = (-2i + 2)/(-2i - 2) = (2 - 2i)/(-2 - 2i)Let me compute that. Let's factor numerator and denominator:Numerator: 2 - 2i = 2(1 - i)Denominator: -2 - 2i = -2(1 + i)So the fraction becomes [2(1 - i)] / [-2(1 + i)] = [ (1 - i) / -(1 + i) ] = - (1 - i)/(1 + i)Multiply numerator and denominator by (1 - i):- [ (1 - i)^2 ] / [ (1 + i)(1 - i) ] = - [ (1 - 2i + i²) ] / (1 - i²) )Compute numerator: 1 - 2i + (-1) = -2iDenominator: 1 - (-1) = 2So we have - [ (-2i)/2 ] = - ( -i ) = iWhich matches the original equation. So yes, z = -2i is the solution. Therefore, the magnitude |z| is |-2i| = sqrt(0² + (-2)²) = sqrt(4) = 2.Wait, that seems straightforward, but let me think again. When I solved the equation, I got z = -2i, which is a purely imaginary number. Then |z| is indeed 2. But let me confirm once more because sometimes when solving equations with complex numbers, there might be a step where a mistake could happen, especially with signs.Let me redo the algebra just to be safe.Original equation: (z + 2)/(z - 2) = iMultiply both sides by (z - 2): z + 2 = i(z - 2)Expand the right side: z + 2 = iz - 2iBring z terms to left and constants to right: z - iz = -2i - 2Factor z: z(1 - i) = -2(i + 1)So z = [-2(i + 1)] / (1 - i)Multiply numerator and denominator by (1 + i):Numerator: -2(i + 1)(1 + i) = -2[(i)(1) + i(i) + 1(1) + 1(i)] = -2[i + i² + 1 + i] = -2[2i + 0] = -4iDenominator: (1 - i)(1 + i) = 1 - i² = 1 - (-1) = 2So z = (-4i)/2 = -2i. Yep, same result. Then |z| is 2. That seems correct.Alternatively, maybe there's another method. Let's see. If (z + 2)/(z - 2) = i, then this can be interpreted as a Möbius transformation. But perhaps that's overcomplicating. Alternatively, think in terms of geometry. The equation (z + 2)/(z - 2) = i implies that the complex number (z + 2) is equal to i times (z - 2). So geometrically, this is a rotation by 90 degrees (since multiplying by i is a rotation by 90 degrees) and scaling, but since i has magnitude 1, it's just a rotation.But maybe that's not necessary here. Since the algebraic method worked and gave a straightforward answer, probably that's the right way.Alternatively, if we let z = x + yi, then substitute into the equation:(z + 2)/(z - 2) = iSo (x + yi + 2)/(x + yi - 2) = iMultiply numerator and denominator:[(x + 2) + yi]/[(x - 2) + yi] = iMultiply numerator and denominator by the conjugate of the denominator to simplify:[( (x + 2) + yi )( (x - 2) - yi ) ] / [ ( (x - 2)^2 + y^2 ) ] = iCompute numerator:First, (x + 2)(x - 2) = x² - 4Then, (x + 2)(-yi) = -yi(x + 2)Then, yi*(x - 2) = yi(x - 2)Then, yi*(-yi) = -y²i² = -y²(-1) = y²So combining all terms:Numerator: (x² - 4) - yi(x + 2) + yi(x - 2) + y²Simplify the terms:The middle terms: - yi(x + 2) + yi(x - 2) = yi[ - (x + 2) + (x - 2) ] = yi[ -x - 2 + x - 2 ] = yi[ -4 ] = -4yiSo numerator becomes: (x² - 4) -4yi + y²Therefore, numerator is (x² + y² -4) -4yiDenominator is (x - 2)^2 + y^2So the equation becomes:[ (x² + y² - 4) -4yi ] / [ (x - 2)^2 + y^2 ] = iSince the right side is i, which is 0 + 1i, we can equate real and imaginary parts.Let me write the left side as [ (x² + y² -4 ) / D ] - [ 4y / D ] i, where D is the denominator, (x - 2)^2 + y^2.Setting this equal to 0 + 1i, we have:Real part: (x² + y² -4)/D = 0Imaginary part: (-4y)/D = 1So from real part: x² + y² -4 = 0 --> x² + y² = 4From imaginary part: (-4y)/D = 1 --> -4y = DBut D is (x - 2)^2 + y^2, so:-4y = (x - 2)^2 + y^2But we already know from real part that x² + y² = 4. Let me substitute x² + y² with 4 into the second equation.So, -4y = (x - 2)^2 + y^2But (x - 2)^2 = x² -4x +4, so substituting:-4y = x² -4x +4 + y²But x² + y² =4, so replace x² + y² with 4:-4y = 4 -4x +4Simplify right side:4 +4 -4x = 8 -4xSo:-4y =8 -4xDivide both sides by 4:- y = 2 -xThus, y = x -2So now we have two equations:1. x² + y² =42. y = x -2Substitute y = x -2 into first equation:x² + (x -2)^2 =4Expand (x -2)^2: x² -4x +4So:x² + x² -4x +4 =4Combine like terms:2x² -4x +4 =4Subtract 4 from both sides:2x² -4x =0Factor:2x(x -2)=0Thus, x=0 or x=2Now, for each x, find y from y =x -2.If x=0, then y=0 -2= -2. So z=0 -2i= -2i.If x=2, then y=2 -2=0. So z=2 +0i=2.But we need to check if z=2 is a solution because in the original equation, (z +2)/(z -2) would have a denominator z-2. If z=2, the denominator becomes zero, which is undefined. So z=2 is excluded.Thus, only z= -2i is the valid solution.Therefore, |z|=| -2i | = sqrt(0² + (-2)^2)=sqrt(4)=2.Okay, so both methods lead to the same answer. First, solving algebraically gives z=-2i, then breaking into real and imaginary parts gives the same result. So, the magnitude |z| is 2.Final AnswerThe magnitude of ( z ) is boxed{2}.</think>Given a complex number ( z ) that satisfies the equation (frac{z+2}{z-2} = i), we start by multiplying both sides by ( z - 2 ):[z + 2 = i(z - 2)]Expanding the right side:[z + 2 = iz - 2i]Rearranging terms to isolate ( z ):[z - iz = -2i - 2]Factoring out ( z ):[z(1 - i) = -2(i + 1)]Solving for ( z ):[z = frac{-2(i + 1)}{1 - i}]To simplify, multiply the numerator and denominator by the conjugate of the denominator ( (1 + i) ):[z = frac{-2(i + 1)(1 + i)}{(1 - i)(1 + i)}]The denominator simplifies to:[(1 - i)(1 + i) = 1 - i^2 = 1 - (-1) = 2]The numerator simplifies to:[-2(i + 1)(1 + i) = -2(i cdot 1 + i cdot i + 1 cdot 1 + 1 cdot i) = -2(i - 1 + 1 + i) = -2(2i) = -4i]Thus, we have:[z = frac{-4i}{2} = -2i]The magnitude ( |z| ) is calculated as:[| -2i | = sqrt{0^2 + (-2)^2} = sqrt{4} = 2]Therefore, the magnitude of ( z ) is (boxed{2}).,<think>Okay, let's see. I need to find the magnitude of a complex number z that satisfies the equation (z + 2)/(z - 2) = i. Hmm. Alright, so first, maybe I should recall how to handle equations with complex numbers. Let me think. Since z is a complex number, I can represent it as z = x + yi, where x and y are real numbers. Then, substitute that into the equation and solve for x and y. Once I have z, I can find |z| by taking the square root of (x² + y²). That sounds like a plan. Let me try that.So, let's write z as x + yi. Then, the equation becomes:(z + 2)/(z - 2) = iSubstituting z:[(x + yi) + 2] / [(x + yi) - 2] = iSimplify numerator and denominator:Numerator: (x + 2) + yiDenominator: (x - 2) + yiSo, [(x + 2) + yi] / [(x - 2) + yi] = iHmm, dividing complex numbers can be tricky. Maybe I should multiply both sides by the denominator to get rid of the fraction. Let's try that.Multiply both sides by [(x - 2) + yi]:[(x + 2) + yi] = i * [(x - 2) + yi]Okay, so expanding the right-hand side. Let's distribute the i:i*(x - 2) + i*(yi) = i(x - 2) + y*i²But wait, i² is -1, so that becomes:i(x - 2) - ySo the equation is:(x + 2) + yi = i(x - 2) - yNow, let's collect like terms. Let's write all the real parts on the left and the imaginary parts on the right.Wait, actually, maybe it's better to separate the equation into real and imaginary parts. Since two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. So, let's express both sides in the form a + bi and then equate the components.Left-hand side (LHS): (x + 2) + yiRight-hand side (RHS): i(x - 2) - y = -y + i(x - 2)So, writing RHS as a complex number: (-y) + (x - 2)iTherefore, equating real and imaginary parts:Real part: x + 2 = -yImaginary part: y = x - 2So now we have a system of two equations:1) x + 2 = -y2) y = x - 2Hmm, okay. Let me write that down:From equation 2: y = x - 2Substitute equation 2 into equation 1:x + 2 = - (x - 2)So, x + 2 = -x + 2Let's solve for x:x + 2 = -x + 2Add x to both sides:2x + 2 = 2Subtract 2 from both sides:2x = 0Divide by 2:x = 0Now, substitute x = 0 into equation 2:y = 0 - 2 = -2Therefore, z = x + yi = 0 + (-2)i = -2iWait, so z is -2i? Let me check if that works in the original equation.Compute (z + 2)/(z - 2):z + 2 = (-2i) + 2 = 2 - 2iz - 2 = (-2i) - 2 = -2 - 2iSo, (2 - 2i)/(-2 - 2i). Let's compute that. Multiply numerator and denominator by the conjugate of the denominator to simplify:Denominator conjugate is -2 + 2i.So:[(2 - 2i)(-2 + 2i)] / [(-2 - 2i)(-2 + 2i)]First, compute denominator:(-2)(-2) + (-2)(2i) + (-2i)(-2) + (-2i)(2i)= 4 - 4i + 4i -4i²= 4 + 0i -4(-1) (since i² = -1)= 4 + 4 = 8Numerator:(2)(-2) + (2)(2i) + (-2i)(-2) + (-2i)(2i)= -4 + 4i + 4i -4i²= -4 + 8i -4(-1)= -4 + 8i +4= 0 + 8iTherefore, the numerator is 8i and denominator is 8, so (8i)/8 = i. Which matches the original equation. So yes, z = -2i is the solution. Therefore, |z| is the magnitude, which is sqrt(0² + (-2)²) = sqrt(4) = 2. So, the answer is 2. Wait, but let me confirm again.Alternatively, maybe there's a quicker way without representing z as x + yi. Let me try another method. Starting with (z + 2)/(z - 2) = i. Let's solve for z.Multiply both sides by (z - 2):z + 2 = i(z - 2)Then, z + 2 = iz - 2iBring all terms with z to one side:z - iz = -2i - 2Factor z:z(1 - i) = -2(i + 1)Then, z = [-2(1 + i)] / (1 - i)Multiply numerator and denominator by (1 + i):z = [-2(1 + i)(1 + i)] / [(1 - i)(1 + i)]Compute denominator: (1)^2 - (i)^2 = 1 - (-1) = 2Numerator: -2(1 + 2i + i²) = -2(1 + 2i -1) = -2(2i) = -4iTherefore, z = (-4i)/2 = -2iSame result. So |z| = |-2i| = 2. So the answer is 2. Yeah, that seems consistent. So, regardless of the method, I get z = -2i, so |z| = 2.I think that's the answer. Let me just verify once more. If z = -2i, then (z + 2)/(z - 2) = ( -2i + 2 ) / ( -2i - 2 ). Let's compute that:Multiply numerator and denominator by i to make the denominator real? Hmm, not sure. Let's compute as is:Numerator: 2 - 2iDenominator: -2 - 2iDivide them:(2 - 2i)/(-2 - 2i). Factor out 2 in numerator and denominator:2(1 - i)/[-2(1 + i)] = - (1 - i)/(1 + i)Multiply numerator and denominator by (1 - i):- [(1 - i)(1 - i)] / [(1 + i)(1 - i)] = - [1 - 2i + i²]/(1 - i²) = - [1 - 2i -1]/(1 +1) = - [(-2i)/2] = - (-i) = i. Which is correct. So, yes, z = -2i and |z| = 2. So, the answer is 2.Final AnswerThe magnitude of ( z ) is boxed{2}.</think>Given a complex number ( z ) that satisfies (frac{z+2}{z-2} = i), we need to find the magnitude (|z|).First, we start by solving the equation for ( z ). Multiply both sides by ( z - 2 ):[z + 2 = i(z - 2)]Expanding the right-hand side:[z + 2 = iz - 2i]Rearranging terms to isolate ( z ):[z - iz = -2i - 2]Factoring out ( z ):[z(1 - i) = -2(1 + i)]Solving for ( z ):[z = frac{-2(1 + i)}{1 - i}]To simplify, multiply the numerator and denominator by the conjugate of the denominator:[z = frac{-2(1 + i)(1 + i)}{(1 - i)(1 + i)}]Calculating the denominator:[(1 - i)(1 + i) = 1 - i^2 = 1 - (-1) = 2]Calculating the numerator:[-2(1 + i)^2 = -2(1 + 2i + i^2) = -2(1 + 2i - 1) = -2(2i) = -4i]Thus,[z = frac{-4i}{2} = -2i]The magnitude of ( z ) is:[|z| = |-2i| = sqrt{0^2 + (-2)^2} = sqrt{4} = 2]The magnitude of ( z ) is (boxed{2}).]

🔑:To solve the equation frac{z+2}{z-2} = i, we can start by simplifying the expression to isolate z:begin{align*}frac{z+2}{z-2} &= i (z+2) &= i(z-2) z+2 &= iz - 2i.end{align*}To get rid of the imaginary number on the right side, we add 2i to both sides:begin{align*}z+2+2i &= iz z+2(1+i) &= iz.end{align*}Now, we want to bring all terms involving z to one side of the equation:begin{align*}z - iz &= -2(1+i) z(1 - i) &= -2(1+i).end{align*}We can now solve for z by dividing both sides by (1 - i):begin{align*}z &= frac{-2(1+i)}{1-i} &= frac{-2(1+i)}{1-i} cdot frac{1+i}{1+i} &= frac{-2(1+i)^2}{1^2 - i^2} &= frac{-2(1 + 2i + i^2)}{1 + 1} &= frac{-2(1 + 2i - 1)}{2} &= frac{-4i}{2} &= -2i.end{align*}Now, we find the magnitude of z using the formula for the magnitude of a complex number |z| = sqrt{a^2 + b^2}, where z = a + bi:begin{align*}|z| &= sqrt{0^2 + (-2)^2} &= sqrt{4} &= boxed{2}.end{align*}