🧠:[<think>Okay, so I need to prove that in triangle ABC, if the ratio of BC to AC is less than 1/2, then angle A is less than 30 degrees. Hmm, let's start by visualizing the triangle. Let me sketch triangle ABC with points A, B, and C. BC is one side, AC is another side, and AB is the third side. The given condition is that BC/AC < 1/2. So BC is less than half of AC. And we need to show that angle at A, which is ∠BAC, is less than 30 degrees.First, I should recall some triangle theorems that relate side lengths to angles. I remember that in any triangle, the larger angle is opposite the longer side. So, if BC is shorter than half of AC, then maybe angle opposite to BC, which is angle A, is smaller? Wait, angle opposite to BC would be angle A. Wait, no, in triangle ABC, side opposite angle A is BC. Wait, actually, angle A is at vertex A, so the side opposite angle A is BC. So, the side BC is opposite angle A. So, if BC is shorter, then angle A is smaller. But how does that help here?But here, BC is not just shorter, but shorter than half of AC. So, maybe I need to use the Law of Sines or the Law of Cosines here. Let me think. The Law of Sines states that in any triangle, (a/sin A) = (b/sin B) = (c/sin C), where a, b, c are the sides opposite angles A, B, C respectively. So, in this case, BC is side opposite angle A, so BC = a, AC is side opposite angle B, so AC = b. Then according to the Law of Sines, BC/sin A = AC/sin B. So, BC/AC = sin A / sin B. The given ratio BC/AC is less than 1/2, so sin A / sin B < 1/2. Therefore, sin A < (1/2) sin B.But how does that help me? I need to relate angle A to 30 degrees. Maybe I can use some inequality here. Alternatively, maybe the Law of Cosines would be better. Let's try that. The Law of Cosines states that c² = a² + b² - 2ab cos C. Hmm. So, if I consider angle A, then:BC² = AB² + AC² - 2*AB*AC*cos A.But I don't know AB. Maybe that's not helpful. Alternatively, let's try to express angle A in terms of sides BC and AC.Wait, perhaps if I fix AC and BC, and consider angle A, maybe I can find the maximum possible angle A given the ratio BC/AC < 1/2. Hmm.Alternatively, suppose we construct triangle ABC where BC = (1/2) AC. Then, in this case, angle A would be maximum when BC is exactly half of AC, and maybe angle A is 30 degrees here. Then, if BC is less than half of AC, angle A would be less than 30 degrees. Is that possible?Let me test this idea. Suppose I have triangle ABC where AC is some length, say 2 units. Then BC is 1 unit. Let's see what angle A would be. Using the Law of Sines: BC/sin A = AC/sin B. So, 1/sin A = 2/sin B => sin B = 2 sin A.But in any triangle, the sum of angles is 180 degrees. So, angle B = 180 - angle A - angle C. Hmm. Also, in this case, with sides AC=2, BC=1, and AB is unknown.Wait, maybe if I consider a right triangle. Suppose angle A is 30 degrees. Then, in a 30-60-90 triangle, the sides are in the ratio 1 : √3 : 2. So, if angle A is 30 degrees, then the side opposite 30 degrees (which would be BC) is half the hypotenuse. Wait, but here AC is the side adjacent to angle A. Wait, maybe I need to adjust.Wait, let me clarify. In triangle ABC, angle A is at vertex A. So sides: BC is opposite angle A, AC is opposite angle B, AB is opposite angle C.Wait, no. Wait, in triangle notation, side a is opposite angle A, side b is opposite angle B, side c is opposite angle C. So, if we have triangle ABC, then side BC is opposite angle A, side AC is opposite angle B, side AB is opposite angle C. Therefore, the Law of Sines is: BC/sin A = AC/sin B = AB/sin C.Therefore, given that BC/AC < 1/2, then (BC/AC) = (sin A)/(sin B) < 1/2. So, sin A < (1/2) sin B.But angle A and angle B are related since in a triangle, angles sum to 180 degrees. So angle B = 180 - angle A - angle C. But maybe we can find a relation here.Alternatively, perhaps if we consider triangle ABC and fix side AC, and vary BC such that BC < (1/2) AC. Then, angle A would be maximized when BC is equal to (1/2) AC, so maybe in that case angle A is 30 degrees, and if BC is smaller, angle A is smaller. Let's check this.Suppose AC is 2 units, and BC is 1 unit (so BC = 1/2 AC). Let's compute angle A using the Law of Cosines. Let's denote AC = 2, BC = 1, and AB is some length. Then, in triangle ABC:BC² = AB² + AC² - 2*AB*AC*cos APlugging in:1² = AB² + 2² - 2*AB*2*cos A1 = AB² + 4 - 4 AB cos AWhich simplifies to:AB² - 4 AB cos A + 3 = 0But we don't know AB or angle A. Hmm, this seems complicated. Maybe another approach.Alternatively, let's use the Law of Sines. If BC = 1, AC = 2, then:BC/sin A = AC/sin B => 1/sin A = 2/sin B => sin B = 2 sin AAlso, angle B = 180 - angle A - angle C. But without knowing angle C, it's still tricky.Wait, perhaps if we assume that angle C is 90 degrees. Let's see. If angle C is 90 degrees, then angle A + angle B = 90 degrees. Then, sin B = sin (90 - angle A) = cos angle A. Therefore, sin B = cos A. Then from above, sin B = 2 sin A => cos A = 2 sin A.Therefore, cos A = 2 sin A => tan A = 1/2 => angle A ≈ 26.565 degrees. Which is less than 30 degrees. But this is a specific case where angle C is 90 degrees. However, if angle C is not 90 degrees, perhaps angle A can be larger?Wait, let me check. Suppose instead of angle C being 90 degrees, we have angle C being acute. Let's say angle C is 60 degrees. Then angle A + angle B = 120 degrees. From sin B = 2 sin A, and angle B = 120 - angle A. So, sin (120 - angle A) = 2 sin A. Let me compute this.sin(120 - A) = sin 120 cos A - cos 120 sin A = (√3/2) cos A - (-1/2) sin A = (√3/2) cos A + (1/2) sin A.Set this equal to 2 sin A:(√3/2) cos A + (1/2) sin A = 2 sin AMultiply both sides by 2:√3 cos A + sin A = 4 sin A√3 cos A = 3 sin ADivide both sides by cos A:√3 = 3 tan Atan A = √3 / 3 = 1/√3Thus, angle A = 30 degrees.Wait, so in this case, if angle C is 60 degrees, angle A would be 30 degrees. Hmm, interesting. So when angle C is 60 degrees, angle A is 30 degrees. So in this case, BC = 1, AC = 2, angle A = 30 degrees. So if we have BC = 1/2 AC, angle A can be 30 degrees. But in the previous case, when angle C was 90 degrees, angle A was about 26.5 degrees. So, depending on angle C, angle A can vary.Therefore, when BC is 1/2 AC, angle A can be as large as 30 degrees (when angle C is 60 degrees) or less. So 30 degrees is the maximum possible angle A when BC = 1/2 AC. Then, if BC is less than 1/2 AC, angle A must be less than 30 degrees. Therefore, the maximum angle A occurs when BC is 1/2 AC and angle C is 60 degrees, giving angle A = 30 degrees. If BC is smaller, angle A becomes smaller. Hence, BC/AC < 1/2 implies angle A < 30 degrees.But how do I formalize this argument? Because in the case where BC = 1/2 AC, depending on angle C, angle A can vary. But in the example above, when angle C is 60 degrees, angle A is 30 degrees. If angle C is different, angle A could be smaller. So, the maximum angle A when BC = 1/2 AC is 30 degrees. Therefore, when BC is less than 1/2 AC, angle A must be less than 30 degrees.To make this rigorous, perhaps I can consider the maximum possible angle A given BC/AC = k < 1/2. To maximize angle A, given BC = k * AC, we need to minimize the other sides or something. Alternatively, use calculus to find the maximum angle A as a function of BC/AC.Alternatively, using the Law of Sines. Let's suppose that BC = k * AC, where k < 1/2. Then, according to Law of Sines, BC / sin A = AC / sin B => k = sin A / sin B. So sin B = sin A / k.Since the sum of angles in a triangle is 180 degrees, B = 180 - A - C. So, sin B = sin (A + C). Wait, but angle C is related to sides as well.Alternatively, perhaps using the area of the triangle. The area can be expressed as (1/2)*AB*AC*sin A. But not sure.Wait, another idea. Suppose we fix AC and point A, and vary point B such that BC < (1/2) AC. Then, the locus of points B such that BC = (1/2) AC is a circle with center at C and radius (1/2) AC. Then, the set of all possible B such that BC < (1/2) AC is inside this circle. Then, angle A is maximized when B is located such that AB is tangent to this circle. Because when B is on the circle, angle A is maximized.Wait, this might be a geometric approach. Let me elaborate.Let’s fix points A and C such that AC is a fixed length. Let’s place point A at the origin and point C at (c, 0) on the coordinate plane. Then, the set of points B such that BC < (1/2) AC is the interior of a circle with center at C and radius (1/2) AC. Let’s denote AC = 2 units for simplicity, so the circle has radius 1 unit.We need to find the maximum possible angle at A (origin) formed by points B inside this circle. The angle at A is ∠BAC. To maximize this angle, point B should be positioned such that AB is tangent to the circle. Because the angle is maximized when B is as far as possible from the line AC, but still within the circle.Wait, when you have a circle and a fixed point outside, the maximum angle subtended by the circle at that point is achieved when the lines are tangent to the circle. So, similarly, here, point A is outside the circle centered at C with radius 1 (since AC = 2, distance from A to C is 2, radius is 1). The maximum angle ∠BAC would be achieved when AB is tangent to the circle.So, let's calculate that angle. The tangent from point A to the circle centered at C with radius 1. The distance from A to C is 2 units. The tangent of angle at A would be equal to the ratio of the radius to the distance from A to C. Wait, the angle between the tangent and the line AC can be found using trigonometry.In the right triangle formed by A, the center C, and the tangent point T. Then, the length AC is 2, CT is 1 (radius), and AT is the tangent. By Pythagoras, AT = sqrt(AC² - CT²) = sqrt(4 - 1) = sqrt(3). Then, the angle at A, which is ∠TAC, has sin(theta) = opposite/hypotenuse = CT / AC = 1/2. Wait, but wait, no. Wait, in the right triangle, sin(theta) is opposite over hypotenuse. The angle theta is angle at A in the right triangle, so opposite side is CT (length 1), hypotenuse is AC (length 2). So sin(theta) = 1/2, so theta = 30 degrees. Therefore, the angle between the tangent line AB and AC is 30 degrees. Therefore, the maximum possible angle at A is 30 degrees when BC = 1 (since the tangent point is on the circle). Therefore, if BC is less than 1, the angle at A must be less than 30 degrees.This seems to align with our previous examples. Therefore, this geometric approach shows that when BC is equal to 1/2 AC, the maximum angle at A is 30 degrees, achieved when AB is tangent to the circle. If BC is less than 1/2 AC, point B is inside the circle, so angle A is less than 30 degrees. Therefore, this proves that ∠A < 30° when BC/AC < 1/2.So, summarizing the steps:1. Place points A and C on a coordinate system with AC = 2 units for simplicity.2. The locus of points B such that BC = 1 unit (1/2 of AC) is a circle centered at C with radius 1.3. The maximum angle ∠BAC occurs when AB is tangent to this circle.4. Using trigonometry in the right triangle formed by A, C, and the tangent point T, we find that ∠TAC = 30°.5. Therefore, if BC < 1 unit (i.e., BC/AC < 1/2), point B is inside the circle, resulting in ∠BAC < 30°.This completes the proof.Final Answerboxed{angle A < 30^circ}</think>Given a triangle (ABC) where (frac{BC}{AC} < frac{1}{2}), we need to prove that (angle A < 30^circ).1. Using the Law of Sines: - The Law of Sines states (frac{BC}{sin A} = frac{AC}{sin B}). - Given (frac{BC}{AC} < frac{1}{2}), we have (frac{sin A}{sin B} < frac{1}{2}). - This implies (sin A < frac{1}{2} sin B).2. Geometric Approach: - Fix points (A) and (C) such that (AC) is a fixed length. Place (A) at the origin and (C) at ((c, 0)) on the coordinate plane. - The set of points (B) such that (BC < frac{1}{2} AC) forms a circle centered at (C) with radius (frac{1}{2} AC). - For simplicity, assume (AC = 2) units, making the radius of the circle 1 unit.3. Maximizing (angle BAC): - The maximum angle (angle BAC) occurs when (B) is on the tangent to the circle centered at (C). - The tangent from (A) to the circle forms a right triangle with (AC = 2), radius (CT = 1), and hypotenuse (AT = sqrt{AC^2 - CT^2} = sqrt{3}). - The angle (theta = angle TAC) is calculated using (sin theta = frac{CT}{AC} = frac{1}{2}), giving (theta = 30^circ).4. Conclusion: - When (BC = frac{1}{2} AC), the maximum (angle BAC) is (30^circ). - If (BC < frac{1}{2} AC), point (B) is inside the circle, resulting in (angle BAC < 30^circ).Thus, we conclude that (angle A < 30^circ) when (frac{BC}{AC} < frac{1}{2}).[boxed{angle A < 30^circ}],<think>Okay, so I need to prove that angle A is less than 30 degrees in triangle ABC given that the ratio of BC to AC is less than half. Hmm, let's see. Let me try to visualize the triangle first. Triangle ABC, with BC opposite angle A, AC opposite angle B, and AB opposite angle C. Wait, actually, in standard notation, side a is BC, side b is AC, and side c is AB. So angle A is opposite side a, which is BC. So the problem states that BC/AC < 1/2, which is equivalent to a/b < 1/2, and we need to prove that angle A < 30 degrees.Alright, so angle A is opposite side BC, which is shorter than half of AC. So maybe I can use the Law of Sines here. The Law of Sines says that a/sin A = b/sin B = c/sin C. Since we know a/b < 1/2, let's write that ratio in terms of the Law of Sines. So a/b = sin A / sin B < 1/2. Therefore, sin A / sin B < 1/2. Which implies that sin A < (1/2) sin B. Hmm, but how does that help me? I need to relate angle A to angle B or something else.Alternatively, maybe the Law of Cosines could help. If I can express angle A in terms of sides BC, AC, and AB. The Law of Cosines states that a² = b² + c² - 2bc cos A. But without knowing side AB or angle B or C, this might not be straightforward. Wait, but perhaps we can find some relationship here.Alternatively, maybe consider constructing a triangle where BC/AC is exactly 1/2 and see what angle A would be in that case, then argue that if BC is even smaller, angle A must be less than that. Let me explore that idea.Suppose BC = (1/2) AC. Let me set AC = 2x, so BC = x. Then, in this triangle, if angle A is 30 degrees, does that hold? Let's check with the Law of Sines. If angle A is 30 degrees, then a/sin A = x / sin 30° = x / 0.5 = 2x. And AC is 2x, so b = 2x. Then b/sin B = 2x / sin B. According to the Law of Sines, these should be equal. So 2x / sin B = 2x, which implies sin B = 1, so angle B is 90 degrees. Wait, so if BC is half of AC, and angle A is 30 degrees, then angle B has to be 90 degrees. Is that possible? Let me draw such a triangle. If angle B is 90 degrees, then triangle ABC is right-angled at B, with AC as the hypotenuse. Then BC is one leg, AB is the other leg. If BC = x and AC = 2x, then by Pythagoras, AB² + BC² = AC². So AB² + x² = (2x)² = 4x², so AB² = 3x², so AB = x√3. Therefore, such a triangle exists where angle A is 30 degrees when BC/AC = 1/2.But the problem states that BC/AC < 1/2. So if BC is smaller than x in this case, then maybe angle A becomes smaller than 30 degrees. Let me see. Suppose BC is less than x. Let's take BC = x - ε, where ε is a positive small number. Then, how would angle A change? In the previous scenario, angle A was 30 degrees when BC = x. If BC is smaller, perhaps angle A becomes smaller.Wait, let's think about this. If BC decreases while AC stays the same, then the side opposite angle A is getting shorter. So angle A should decrease. Is that always true? If we fix AC and decrease BC, then angle A is opposite BC, so it should get smaller. But also, the other sides and angles might adjust. Hmm, maybe I need to formalize this.Alternatively, use the Law of Sines again. Let's assume that in the case when BC/AC = 1/2, angle A is 30 degrees. Then if BC/AC is less than 1/2, angle A must be less than 30 degrees. But is that rigorous enough? Maybe not. Let's try to make it more precise.Suppose in triangle ABC, BC = k * AC, where k < 1/2. Then by the Law of Sines, BC / sin A = AC / sin B. So, k * AC / sin A = AC / sin B. Simplify, k / sin A = 1 / sin B. So sin B = sin A / k. But since sin B can't exceed 1, we have sin A / k ≤ 1 => sin A ≤ k. Since k < 1/2, this gives sin A < 1/2. Therefore, angle A must be less than 30 degrees or greater than 150 degrees. But angle A can't be greater than 150 degrees in a triangle because the sum of angles is 180. If angle A were greater than 150, then the other two angles would have to sum to less than 30, which would mean that side BC (opposite angle A) would be the largest side, but BC is less than half of AC, which is a contradiction. Therefore, angle A must be less than 30 degrees.Wait, let me check that again. If angle A is greater than 150 degrees, the other two angles sum to less than 30, so they are each less than 30. Then the sides opposite those angles would be shorter. But AC is opposite angle B, which would be less than 30, so AC would be shorter than BC? Wait, but BC is less than half of AC. If angle B is less than 30, then side AC (opposite angle B) is less than BC (opposite angle A). But BC is less than half of AC. So if AC is less than BC, but BC is less than half of AC, that would imply BC < (1/2) AC < (1/2) BC, which would mean BC < (1/2) BC, which implies BC is negative, which is impossible. Therefore, angle A cannot be greater than 150 degrees. Therefore, angle A must be less than 30 degrees. Therefore, proven.Alternatively, perhaps another approach. Let's assume angle A is 30 degrees and show that BC/AC cannot be less than 1/2. Then, since we have BC/AC < 1/2, angle A must be less than 30 degrees.So, suppose angle A is 30 degrees. Then by the Law of Sines, BC / sin A = AC / sin B. So BC / AC = sin A / sin B = sin 30° / sin B = 0.5 / sin B. Since in any triangle, angles B and C must sum to 150 degrees (since angle A is 30). The maximum value of sin B occurs when angle B is 90 degrees, which would make sin B = 1. Then BC / AC = 0.5 / 1 = 0.5. If angle B is less than 90 degrees, sin B is less than 1, so BC / AC would be greater than 0.5. Wait, so if angle A is 30 degrees, BC / AC is at least 0.5. Therefore, if BC / AC is less than 0.5, angle A must be less than 30 degrees.That seems like a solid argument. Because if angle A were 30 degrees, BC / AC would be at least 0.5, but since it's given as less than 0.5, angle A must be smaller. Therefore, angle A < 30 degrees. That's a proof by contradiction.Alternatively, directly, from the Law of Sines: BC / AC = sin A / sin B < 1/2. So sin A < (1/2) sin B. But in any triangle, angle B = 180° - angle A - angle C. So angle B is at least ... Hmm, maybe not straightforward. Let's think.But since sin B <= 1, then sin A < 1/2, so angle A < 30° or >150°, but angle A can't be >150° because then the other two angles would be <30°, which would make sides opposite them smaller, but BC is already less than half of AC. If angle A were obtuse, then BC would be the largest side, but BC is less than half of AC, which is a contradiction. Therefore, angle A must be acute and less than 30°.So combining these two observations: sin A < 1/2 and angle A can't be obtuse. Therefore, angle A must be less than 30 degrees. That works.Alternatively, using the Law of Cosines. Let's consider angle A. The Law of Cosines says:BC² = AB² + AC² - 2 * AB * AC * cos ABut we have BC < (1/2) AC. Let's denote BC = k * AC, where k < 1/2.So (k AC)² = AB² + AC² - 2 * AB * AC * cos ADivide both sides by AC²:k² = (AB/AC)² + 1 - 2 * (AB/AC) * cos ALet me set AB/AC = m, where m is some positive real number. Then:k² = m² + 1 - 2 m cos AWe need to find the maximum possible angle A given k < 1/2.But this might be more complicated. Alternatively, consider the case when angle A is 30°, then find BC/AC.If angle A is 30°, then BC² = AB² + AC² - 2 AB * AC cos 30°Assuming BC = (1/2) AC, let's set AC = 2, so BC = 1.Then:1² = AB² + 2² - 2 * AB * 2 * cos 30°1 = AB² + 4 - 4 AB (√3 / 2)1 = AB² + 4 - 2√3 ABThen AB² - 2√3 AB + 3 = 0Solving quadratic equation: AB = [2√3 ± sqrt(12 - 12)] / 2 = [2√3]/2 = √3So AB = √3. Therefore, the triangle with AC = 2, BC = 1, AB = √3 has angle A = 30°, and indeed, BC/AC = 1/2.Therefore, if BC/AC < 1/2, then angle A must be less than 30°, because when BC/AC decreases, angle A decreases as well, as the side opposite is getting shorter relative to AC. So this case shows that the maximum possible angle A when BC/AC = 1/2 is 30°, so any smaller ratio would result in a smaller angle.Thus, combining the Law of Sines approach and the specific case where angle A is 30°, we can conclude that angle A must be less than 30°.Final Answerboxed{angle A < 30^circ}</think>Given a triangle (ABC) where (frac{BC}{AC} < frac{1}{2}), we need to prove that (angle A < 30^circ).1. Using the Law of Sines: - The Law of Sines states (frac{BC}{sin A} = frac{AC}{sin B}). - Given (frac{BC}{AC} < frac{1}{2}), we can write (frac{sin A}{sin B} < frac{1}{2}). - This implies (sin A < frac{1}{2} sin B).2. Bounding (sin A): - Since (sin B leq 1), we have (sin A < frac{1}{2}). - The sine function being less than (frac{1}{2}) implies that the angle (A) must be less than (30^circ) or greater than (150^circ).3. Excluding obtuse angle: - If (angle A) were greater than (150^circ), the other two angles would sum to less than (30^circ), making the sides opposite those angles very small. - However, (BC) (opposite (angle A)) is given to be less than half of (AC), which contradicts the triangle inequality if (angle A) were obtuse. Thus, (angle A) cannot be greater than (150^circ).4. Constructing a specific case: - If (angle A = 30^circ), then by the Law of Sines, (frac{BC}{AC} = frac{sin 30^circ}{sin B} = frac{0.5}{sin B}). - The maximum value of (sin B) is 1 (when (angle B = 90^circ)), leading to (frac{BC}{AC} = 0.5). - Since (frac{BC}{AC} < 0.5), (angle A) must be less than (30^circ).Thus, combining these arguments, we conclude that (angle A < 30^circ).[boxed{angle A < 30^circ}]]

🔑:Solution. Let a=BC, b=AC. The first method of solving. Construct a circle of radius a passing through points B and C such that its center O lies on the same side of the line BC as point A. Then A cannot lie inside this circle or on the circle itself, since AC is greater than its diameter. Therefore, angle A is less than half of the central angle BOC, which is 60^{circ}.The second method of solving is based on the cosine theorem. Let x=AB, t=cos angle A, then a^{2}=x^{2}+b^{2}-2bx cdot t Leftrightarrow x^{2}-2btx+b^{2}-a^{2}=0. quad Consider the discriminant of this equation frac{D}{4}=b^{2}t^{2}-b^{2}+a^{2} geq 0 Rightarrow t^{2} geq frac{b^{2}-a^{2}}{b^{2}}=1-left(frac{a}{b}right)^{2}>1-frac{1}{4}. Hence, cos angle A>frac{sqrt{3}}{2} and thus, angle A<30^{circ}.

🧠:[<think>Okay, let's see. I need to find the largest number among a, b, c, and d. Each of them is defined with exponents, some negative exponents and fractional bases. Hmm. Let me start by recalling how negative exponents work. I think a negative exponent means the reciprocal, right? So, like, x^{-n} is the same as 1/(x^n). And also, fractional exponents: x^{1/n} is the nth root of x. So, for example, x^{1/2} is the square root of x. Alright, let's tackle each variable one by one. Starting with a: a = (-1/2)^{-1}. Hmm. Negative exponent here. So that should be the reciprocal of (-1/2). Let me compute that. The reciprocal of -1/2 is -2, right? Because if you flip the numerator and denominator, 1 divided by (-1/2) is -2/1, which is -2. So a = -2. Okay, that seems straightforward.Next, b = 2^{-1/2}. Again, negative exponent. So that's 1 over 2^{1/2}. 2^{1/2} is the square root of 2, which is approximately 1.414. So 1 divided by 1.414 is roughly 0.707. So b ≈ 0.707. Then, c = (1/2)^{-1/2}. Let's see. First, the base here is 1/2, and the exponent is -1/2. So again, applying the negative exponent rule: that's 1 divided by (1/2)^{1/2}. Let me compute (1/2)^{1/2}. That's the square root of 1/2. The square root of 1 is 1, the square root of 2 is approximately 1.414, so sqrt(1/2) is 1 / sqrt(2) ≈ 0.707. Therefore, (1/2)^{-1/2} is 1 / 0.707 ≈ 1.414. So c ≈ 1.414. Wait, that's the same as sqrt(2). Because 1 divided by sqrt(2) is sqrt(2)/2 ≈ 0.707, but here we're taking the reciprocal, so it would be sqrt(2). Let me confirm that. (1/2)^{-1/2} is equal to (2/1)^{1/2} because (a/b)^{-n} = (b/a)^n. So yes, (1/2)^{-1/2} is the same as 2^{1/2}, which is sqrt(2) ≈ 1.414. So c = sqrt(2) ≈ 1.414.Lastly, d = 2^{-1}. That's 1 over 2^1, which is 1/2, so 0.5. So d = 0.5.Alright, so summarizing:a = -2b ≈ 0.707c ≈ 1.414d = 0.5Now, we need to find the largest among these four numbers. Let's list them out:a is negative, so that's the smallest. Then d is 0.5, which is less than b's 0.707, which is less than c's approximately 1.414. So the order from largest to smallest is c > b > d > a.Therefore, the largest number is c, which is option C. Wait, but let me double-check my calculations to make sure I didn't mess up any signs or reciprocals.Starting with a: (-1/2)^{-1} = 1 / (-1/2) = -2. Yep, that's correct.b: 2^{-1/2} = 1 / 2^{1/2} = 1 / sqrt(2) ≈ 0.707. Correct.c: (1/2)^{-1/2} = (2/1)^{1/2} = sqrt(2) ≈ 1.414. Correct.d: 2^{-1} = 1/2 = 0.5. Correct.So yes, c is the largest. So the answer should be C. Hmm, but let me check again if any of these could be miscalculated, especially with the negative exponents and fractional bases.Wait, for c, (1/2)^{-1/2} is equivalent to (2)^{1/2} because when you have a negative exponent, you can invert the base. So (a/b)^{-n} = (b/a)^n. Therefore, (1/2)^{-1/2} is (2/1)^{1/2} which is 2^{1/2} or sqrt(2). That's correct.Another way to compute c: (1/2)^{-1/2} = e^{(-1/2) * ln(1/2)} = e^{(-1/2)*(-ln 2)} = e^{(ln 2)/2} = sqrt(e^{ln2}) = sqrt(2). Yep, same result. So c is indeed sqrt(2).So the numbers are a=-2, b≈0.707, c≈1.414, d=0.5. So clearly, c is the largest. Therefore, the answer is C.Final Answerboxed{C}</think>Given ( a = left( -frac{1}{2} right)^{-1} ), ( b = 2^{-frac{1}{2}} ), ( c = left( frac{1}{2} right)^{-frac{1}{2}} ), and ( d = 2^{-1} ), we need to find the largest among these four numbers.1. Calculating ( a ): [ a = left( -frac{1}{2} right)^{-1} = -2 ]2. Calculating ( b ): [ b = 2^{-frac{1}{2}} = frac{1}{sqrt{2}} approx 0.707 ]3. Calculating ( c ): [ c = left( frac{1}{2} right)^{-frac{1}{2}} = 2^{frac{1}{2}} = sqrt{2} approx 1.414 ]4. Calculating ( d ): [ d = 2^{-1} = frac{1}{2} = 0.5 ]Summarizing the values:- ( a = -2 )- ( b approx 0.707 )- ( c approx 1.414 )- ( d = 0.5 )The largest value among these is ( c approx 1.414 ).[boxed{C}],<think>Okay, so I need to figure out which of these four numbers—a, b, c, or d—is the largest. Let me start by understanding each one. The problem gives me:a = (-1/2)^-1b = 2^(-1/2)c = (1/2)^(-1/2)d = 2^-1Hmm, negative exponents can be tricky, but I remember that a negative exponent means taking the reciprocal of the base raised to the positive exponent. Let me go through each one step by step.Starting with a: (-1/2)^-1. The exponent is -1, so I take the reciprocal of the base. The reciprocal of -1/2 is -2. So a = -2. Wait, negative number? So a is -2, which is negative. So that's going to be the smallest? Maybe, but let me check the others.Next, b: 2^(-1/2). This is the same as 1 over 2^(1/2). Since 2^(1/2) is the square root of 2, which is approximately 1.414. So 1 divided by 1.414 is about 0.707. So b is approximately 0.707. Positive number, okay.Then c: (1/2)^(-1/2). Again, negative exponent, so reciprocal. So this becomes (2/1)^(1/2), which is sqrt(2). Wait, is that right? Let me double-check. The reciprocal of 1/2 is 2, so (1/2)^(-1/2) = (2)^(1/2) = sqrt(2), which is also approximately 1.414. So c is around 1.414. Positive number, larger than b.Finally, d: 2^-1. That's 1/2, so 0.5. So d is 0.5. So putting it all together:a = -2b ≈ 0.707c ≈ 1.414d = 0.5Comparing these, the largest is c, which is approximately 1.414, followed by b, then d, and a is the smallest. So the answer should be option C: c.Wait, but let me make sure I didn't make any mistakes with the exponents, especially with the negative signs and fractional exponents.For a: (-1/2)^-1. The reciprocal of -1/2 is indeed -2. Correct.For b: 2^(-1/2). That is 1/(2^(1/2)) = 1/sqrt(2) ≈ 0.707. Correct.For c: (1/2)^(-1/2). The reciprocal of 1/2 is 2, so it's 2^(1/2) = sqrt(2). Yes, that's right.For d: 2^-1 = 1/2 = 0.5. Correct.So comparing all four, c is the largest. Therefore, the answer is C.Final Answerboxed{C}</think>Given ( a = left( -frac{1}{2} right)^{-1} ), ( b = 2^{-frac{1}{2}} ), ( c = left( frac{1}{2} right)^{-frac{1}{2}} ), and ( d = 2^{-1} ), we need to find the largest among these four numbers.1. Calculating ( a ): [ a = left( -frac{1}{2} right)^{-1} = -2 ] (since the reciprocal of (-frac{1}{2}) is (-2)).2. Calculating ( b ): [ b = 2^{-frac{1}{2}} = frac{1}{sqrt{2}} approx 0.707 ]3. Calculating ( c ): [ c = left( frac{1}{2} right)^{-frac{1}{2}} = left( 2 right)^{frac{1}{2}} = sqrt{2} approx 1.414 ]4. Calculating ( d ): [ d = 2^{-1} = frac{1}{2} = 0.5 ]Comparing the values:- ( a = -2 )- ( b approx 0.707 )- ( c approx 1.414 )- ( d = 0.5 )The largest value is ( c approx 1.414 ).[boxed{C}]]

🔑:Since a=(- frac {1}{2})^{-1}=-2,b=2^{- frac {1}{2}}= frac {1}{ sqrt {2}}= frac { sqrt {2}}{2},c=( frac {1}{2})^{- frac {1}{2}}= sqrt {2},d=2^{-1}= frac {1}{2},Therefore, the largest among these four numbers is c.Hence, the answer is boxed{C}.This problem is solved using the properties and rules of exponentiation of rational numbers. The problem requires a comparison of the size of four numbers, which is a basic problem. While solving, it is important to carefully read the problem and reasonably apply the properties and rules of exponentiation of rational numbers.

🧠:[<think>Alright, let's tackle this probability problem. So, we have an urn with 6 black balls, 5 red balls, and 4 white balls. That means the total number of balls is 6 + 5 + 4. Let me add that up. 6 plus 5 is 11, and 11 plus 4 is 15. So, there are 15 balls in total.The problem says three balls are drawn sequentially, and we need to find the probability that the first ball is black, the second is red, and the third is white. Hmm, okay. Since they're drawn sequentially, I guess each draw affects the next one because we're not replacing the balls, right? Unless stated otherwise, I should assume that the balls are not replaced. The problem doesn't mention replacement, so it's safe to assume that once a ball is drawn, it's not put back into the urn.So, probability problems like this often involve multiplying the probabilities of each individual event happening, considering the previous events. So, for the first draw, we want a black ball. There are 6 black balls out of 15 total. So, the probability of drawing a black ball first is 6/15.Now, after drawing one black ball, we don't replace it. So, the total number of balls left in the urn is 15 - 1 = 14. Also, the number of black balls left would be 6 - 1 = 5. But the number of red and white balls remains the same since we drew a black one. So, red is still 5, white is 4.Then, the second draw needs to be red. So, there are 5 red balls out of the remaining 14. Therefore, the probability of drawing a red ball second is 5/14.Now, after drawing a red ball, we again don't replace it. So, total balls left are 14 - 1 = 13. The number of red balls becomes 5 - 1 = 4. Black balls are still 5, and white balls are 4.For the third draw, we need a white ball. There are 4 white balls left out of 13 total. So, the probability here is 4/13.To find the overall probability of all these events happening in sequence, we multiply the individual probabilities together. So, that would be (6/15) * (5/14) * (4/13).Let me compute that step by step. First, simplify 6/15. Both 6 and 15 are divisible by 3. 6 divided by 3 is 2, and 15 divided by 3 is 5. So, 6/15 simplifies to 2/5.Then, 5/14 doesn't simplify further since 5 and 14 have no common factors besides 1. Similarly, 4/13 can't be simplified.So, multiplying 2/5 * 5/14. The 5 in the numerator cancels with the 5 in the denominator. So, that becomes 2/1 * 1/14 = 2/14. 2/14 simplifies to 1/7.Then, 1/7 * 4/13. Multiply the numerators: 1*4 = 4, and denominators: 7*13 = 91. So, the result is 4/91.Wait, let me check that again. Let's do the multiplication without simplifying step by step to ensure I didn't make a mistake.First, 6/15 * 5/14 = (6*5)/(15*14) = 30/210. Then, 30/210 simplifies. Both numerator and denominator are divisible by 30, so 30 divided by 30 is 1, and 210 divided by 30 is 7. So, that's 1/7. Then, 1/7 * 4/13 = 4/91. Yep, that's correct.Alternatively, let's compute the original fractions multiplied:6/15 * 5/14 * 4/13Multiply numerators: 6 * 5 * 4 = 120Multiply denominators: 15 * 14 * 13 = 2730So, 120/2730. Let's simplify this fraction. Divide numerator and denominator by 10: 12/273. Then, 12 and 273 are both divisible by 3: 12 ÷3=4, 273 ÷3=91. So, 4/91. Same result. So, that checks out.Therefore, the probability is 4/91.Let me think if there's another way to approach this problem. Maybe using permutations? Since the order matters here. The total number of ways to draw three balls in sequence without replacement is 15 * 14 * 13. The number of favorable outcomes is the number of ways to draw a black, then a red, then a white. So, that's 6 * 5 * 4. So, probability is (6*5*4)/(15*14*13) which is 120/2730, same as before, which reduces to 4/91. Yep, same result.So, that confirms the answer is 4/91. Let me just make sure that I didn't make any miscalculations in the multiplication or simplification steps. Let's recheck:6 * 5 = 30; 30 * 4 = 120. Correct. 15 * 14 = 210; 210 * 13 = 2730. Correct. 120 divided by 2730: divide numerator and denominator by 10 gives 12/273. 12 divided by 3 is 4, 273 divided by 3 is 91. So, 4/91. Yes, that's correct.Alternatively, 120/2730. Let's divide numerator and denominator by 30. 120 ÷30=4, 2730 ÷30=91. So, 4/91. Yep.So, I think that's solid. The key steps here are recognizing that it's a sequential probability without replacement, computing each probability step by step, and then multiplying them together. Alternatively, using permutations to compute the number of favorable outcomes over total possible outcomes. Both approaches lead to the same answer.Therefore, the probability that the first ball is black, the second red, and the third white is 4/91.Final AnswerThe probability is boxed{dfrac{4}{91}}.</think>Given an urn with 6 black balls, 5 red balls, and 4 white balls, we need to find the probability that the first ball drawn is black, the second is red, and the third is white. The total number of balls is 15.1. First draw (black ball): - Probability = (frac{6}{15}) - Simplifies to (frac{2}{5})2. Second draw (red ball): - After removing one black ball, 14 balls remain. - Probability = (frac{5}{14})3. Third draw (white ball): - After removing one red ball, 13 balls remain. - Probability = (frac{4}{13})Multiplying these probabilities together:[left(frac{6}{15}right) times left(frac{5}{14}right) times left(frac{4}{13}right) = left(frac{2}{5}right) times left(frac{5}{14}right) times left(frac{4}{13}right)]Simplifying step-by-step:[frac{2}{5} times frac{5}{14} = frac{2 times 5}{5 times 14} = frac{10}{70} = frac{1}{7}][frac{1}{7} times frac{4}{13} = frac{4}{91}]Thus, the probability that the first ball is black, the second is red, and the third is white is (boxed{dfrac{4}{91}}).,<think>Alright, let's see. So, the problem is about an urn with 6 black, 5 red, and 4 white balls. We need to find the probability that when we draw three balls one after another, the first is black, the second is red, and the third is white. Hmm, okay. Let me think step by step.First off, probability problems can sometimes be tricky because you have to consider whether the events are independent or not. In this case, since we're drawing balls sequentially without replacement, each draw affects the next one, right? So these are dependent events. That means the probability of each subsequent draw depends on the previous ones. Got it.Let me break it down. The first draw needs to be black. There are 6 black balls. How many total balls are there? Let's add them up. 6 black + 5 red + 4 white = 15 balls in total. So, the probability of drawing a black ball first is 6 out of 15. That makes sense.Now, assuming we successfully drew a black ball first, we need to draw a red ball second. But wait, after removing one black ball, the total number of balls left is 14. The number of red balls remains 5 because we only took out a black one. So, the probability of drawing a red ball second is 5 out of 14. Right?Then, for the third draw, we need a white ball. After removing a black and then a red ball, the total number of balls left should be 13. The number of white balls is still 4 because we haven't removed any white ones yet. So, the probability of drawing a white ball third would be 4 out of 13.To find the overall probability of all three events happening in sequence—first black, then red, then white—we multiply the probabilities of each individual event. That's because each subsequent probability is conditional on the previous event occurring. So, the formula would be:Probability = (Number of black balls / Total balls) * (Number of red balls / (Total balls - 1)) * (Number of white balls / (Total balls - 2))Plugging in the numbers:Probability = (6/15) * (5/14) * (4/13)Let me calculate that. First, simplify each fraction if possible. 6/15 can be simplified by dividing numerator and denominator by 3: 2/5. Then 5/14 is already in simplest form. 4/13 is also already simplified. So, multiplying them together:2/5 * 5/14 * 4/13Hmm, I notice that 5 in the numerator of the second fraction cancels out with the denominator 5 in the first fraction. So that cancels out. Then we have 2 * (1/14) * 4/13. Let's compute that step by step.First, 2 * (1/14) is 2/14, which simplifies to 1/7. Then, 1/7 * 4/13 is 4/91. So, the probability is 4/91.Wait, let me check that again. Let's do it without simplifying first to make sure.Original fractions: 6/15 * 5/14 * 4/13Multiply numerators: 6 * 5 * 4 = 120Multiply denominators: 15 * 14 * 13 = 2730So, 120/2730. Now, let's simplify that. Divide numerator and denominator by 10: 12/273. Then divide numerator and denominator by 3: 4/91. Yep, same result. So, 4/91 is the simplified form. So, the probability is 4/91.Let me just double-check the logic here. Since each draw is without replacement, each time the total number of balls decreases by one, and the counts of the specific colors only decrease if we've drawn them. Since we first drew a black ball, red and white counts remain the same for the second draw. Then, after drawing red, white is still at 4, but total is down by two. So, the denominators are 15, 14, 13, which is correct. Numerators are 6, 5, 4. So, that all lines up.Another way to think about it: the number of favorable outcomes over total possible outcomes. The number of favorable sequences is 6 (black) * 5 (red) * 4 (white) = 120. The total number of ways to draw three balls is 15 * 14 * 13 = 2730. So, 120/2730 reduces to 4/91. Yep, that's the same as before. So that seems right.I think that's solid. I don't see any mistakes in the reasoning here. The key was recognizing that each subsequent draw depends on the previous ones and adjusting the counts accordingly. Then multiplying the probabilities for each step. Also, simplifying the fractions properly. So, yeah, 4/91 is the probability.Final AnswerThe probability is boxed{dfrac{4}{91}}.</think>To find the probability that the first ball drawn is black, the second is red, and the third is white, we need to consider the dependent events of drawing each ball sequentially without replacement.1. First draw (Black ball): - Number of black balls = 6 - Total number of balls = 15 - Probability = (frac{6}{15} = frac{2}{5})2. Second draw (Red ball): - Number of red balls remains 5 (since a black ball was drawn first) - Total number of balls remaining = 14 - Probability = (frac{5}{14})3. Third draw (White ball): - Number of white balls remains 4 (since a red ball was drawn second) - Total number of balls remaining = 13 - Probability = (frac{4}{13})Multiplying these probabilities together gives the overall probability:[text{Probability} = left(frac{6}{15}right) times left(frac{5}{14}right) times left(frac{4}{13}right)]Simplifying the fractions:[frac{6}{15} = frac{2}{5}, quad frac{5}{14}, quad frac{4}{13}]Multiplying these simplified fractions:[frac{2}{5} times frac{5}{14} times frac{4}{13} = frac{2 times 5 times 4}{5 times 14 times 13} = frac{40}{910} = frac{4}{91}]Thus, the probability that the first ball is black, the second is red, and the third is white is (boxed{dfrac{4}{91}}).]

🔑:Solution. Consider the following events: A - the first ball drawn is black, B - the second ball is red, C - the third ball is white. Let D be the event that the balls are drawn in the sequence: black, red, white. Obviously, D=A cdot B cdot C.By formula (4) we have:P(D)=P(A) cdot P(B / A) cdot P(C / A B) .We will find the probabilities entering the right-hand side of this equality. The probability that the first ball drawn is black, P(A)=frac{6}{15}=frac{2}{5}. The probability of drawing a red ball from the urn given that a black ball was initially drawn, P(B / A)=frac{5}{14}, since after removing the black ball, 14 balls remain in the urn, 5 of which are red. The probability of drawing a white ball from the urn after a black and a red ball have been drawn, P(C / A B)=frac{4}{13} (after removing the black and red balls, 13 balls remain in the urn, 4 of which are white).Thus,P(D)=frac{2}{5} cdot frac{5}{14} cdot frac{4}{13}=frac{4}{91} text {. }

🧠:[<think>Okay, let's see. I need to find the value of sinA - cosA given that sinA + cosA is sqrt5/5 in triangle ABC. Hmm. Alright, let's start by recalling some trigonometric identities that might help here. I remember that squaring both sides of an equation involving sine and cosine can sometimes be useful. Let me try that. If I square both sides of the given equation, maybe I can relate it to sinA - cosA somehow.So, the given equation is sinA + cosA = sqrt5/5. Let me square both sides:(sinA + cosA)^2 = (sqrt5/5)^2Expanding the left side using the formula (a + b)^2 = a^2 + 2ab + b^2:sin²A + 2 sinA cosA + cos²A = (5^(1/2)/5)^2Calculating the right side:(sqrt5/5)^2 = (5^(1/2))^2 / 5^2 = 5 / 25 = 1/5So the equation becomes:sin²A + 2 sinA cosA + cos²A = 1/5Now, I know that sin²A + cos²A = 1, that's one of the Pythagorean identities. Let's substitute that in:1 + 2 sinA cosA = 1/5Subtract 1 from both sides:2 sinA cosA = 1/5 - 11/5 - 1 is the same as 1/5 - 5/5, which is -4/5. So:2 sinA cosA = -4/5Divide both sides by 2:sinA cosA = -2/5Hmm, interesting. So the product of sinA and cosA is negative. That means either sinA is negative and cosA is positive, or vice versa. But in a triangle ABC, angle A is between 0 and 180 degrees, right? Because in a triangle, angles can't be 0 or 180. So, in that range, sine is always positive because angles between 0 and 180 have positive sine values. So if sine is positive and their product is negative, cosine must be negative. Therefore, angle A must be in the second quadrant, between 90 degrees and 180 degrees. That makes sense because cosine is negative there.So angle A is obtuse. Okay, but how does that help me? I need to find sinA - cosA. Let me denote that expression as x. So let x = sinA - cosA. Then, maybe squaring that would help me find x.Let me compute x^2 = (sinA - cosA)^2. Expanding that:sin²A - 2 sinA cosA + cos²AAgain, using sin²A + cos²A = 1, so:1 - 2 sinA cosABut we already found that 2 sinA cosA = -4/5, so substituting that in:x^2 = 1 - (-4/5) = 1 + 4/5 = 9/5Wait, that can't be right. Wait, if x^2 = 9/5, then x would be sqrt(9/5) or -sqrt(9/5). But sqrt(9/5) is 3/sqrt5, which is 3 sqrt5 /5. So x is either 3 sqrt5 /5 or -3 sqrt5 /5. But we need to figure out the sign. Since angle A is in the second quadrant (between 90 and 180 degrees), sinA is positive and cosA is negative. So sinA - cosA would be positive minus a negative, which is positive plus positive. Therefore, sinA - cosA is positive. So x is positive. Therefore, x = 3 sqrt5 /5. Which is option B. Wait, let me check my steps again. So starting from sinA + cosA = sqrt5 /5. Squared gives 1 + 2 sinA cosA = 1/5, so 2 sinA cosA = -4/5. Then, sinA - cosA squared is 1 - 2 sinA cosA. Wait, wait, hold on. Wait, when I expanded (sinA - cosA)^2, it's sin²A - 2 sinA cosA + cos²A, so that's 1 - 2 sinA cosA. But 2 sinA cosA is -4/5, so substituting in:x^2 = 1 - (-4/5) = 1 + 4/5 = 9/5. Yes, that's correct. Therefore, x = sqrt(9/5) or -sqrt(9/5). Since x is positive, as we reasoned earlier, x = 3 sqrt5 /5, which is option B.But wait, let me check if there's another way this could go wrong. For instance, maybe I made a mistake in assuming the sign. Let's verify with a concrete example. Suppose angle A is in the second quadrant. Let's take an angle, say 120 degrees. Then sin(120) is sqrt3/2, cos(120) is -1/2. Then sinA + cosA would be sqrt3/2 -1/2, which is approximately 0.866 - 0.5 = 0.366. If sqrt5 /5 is approximately 0.447. So that's not exactly the same, but similar. Then sinA - cosA would be sqrt3/2 - (-1/2) = sqrt3/2 +1/2 ≈ 0.866 + 0.5 = 1.366, which is approximately 3 sqrt5 /5 (since sqrt5 ≈ 2.236, so 3*2.236/5 ≈ 6.708/5 ≈1.3416). Close enough, considering 120 degrees is just an example. Therefore, seems like the answer is B. Let me check the options again:A: -sqrt5/5B: 3 sqrt5/5C: sqrt5/5D: -3/5So, yeah, B is 3 sqrt5/5, which matches our calculation. So I think that's the correct answer. Let me just recap to make sure I didn't miss anything.Given sinA + cosA = sqrt5/5. Squared gives 1 + 2 sinA cosA = 1/5 => 2 sinA cosA = -4/5. Then, sinA - cosA squared is 1 - 2 sinA cosA = 1 - (-4/5) = 9/5. Square root is 3/sqrt5, which rationalized is 3 sqrt5/5. Since angle A is obtuse, sinA - cosA is positive. Therefore, answer is B. Yep, seems solid. I don't see any mistakes here.Final Answerboxed{B}</think>Given (sin A + cos A = frac{sqrt{5}}{5}) in (triangle ABC), we need to find the value of (sin A - cos A).First, we square both sides of the given equation:[(sin A + cos A)^2 = left(frac{sqrt{5}}{5}right)^2]Expanding the left side and simplifying the right side:[sin^2 A + 2 sin A cos A + cos^2 A = frac{5}{25} = frac{1}{5}]Using the Pythagorean identity (sin^2 A + cos^2 A = 1), we substitute:[1 + 2 sin A cos A = frac{1}{5}]Solving for (2 sin A cos A):[2 sin A cos A = frac{1}{5} - 1 = frac{1}{5} - frac{5}{5} = -frac{4}{5}]Next, we consider ((sin A - cos A)^2):[(sin A - cos A)^2 = sin^2 A - 2 sin A cos A + cos^2 A]Using the Pythagorean identity again:[sin^2 A + cos^2 A = 1]Substituting the value of (2 sin A cos A):[(sin A - cos A)^2 = 1 - (-2 sin A cos A) = 1 + frac{4}{5} = frac{9}{5}]Taking the square root of both sides:[sin A - cos A = pm frac{3sqrt{5}}{5}]Since angle (A) in (triangle ABC) must be between 0 and 180 degrees, and (sin A) is positive while (cos A) is negative (as the product (sin A cos A) is negative), angle (A) is in the second quadrant. Therefore, (sin A - cos A) is positive.Thus, the value of (sin A - cos A) is:[boxed{B}],<think>Okay, so I have this problem here: In triangle ABC, sinA + cosA equals √5 over 5, and I need to find sinA minus cosA. The options are given as A through D. Hmm, let's see. I remember there are some trigonometric identities that relate sin and cos when they're added or subtracted. Maybe if I square both expressions, I can use the Pythagorean identity? Let me try that.First, let me write down what's given: sinA + cosA = √5 / 5. They want sinA - cosA. If I square both sides of the given equation, I can expand the left side. Let's do that.So, (sinA + cosA)^2 = (√5 / 5)^2. Expanding the left side gives sin²A + 2 sinA cosA + cos²A. The right side is (5^(1/2))^2 over 5^2, which simplifies to 5 / 25, or 1/5. So, the equation becomes sin²A + 2 sinA cosA + cos²A = 1/5.Wait, but sin²A + cos²A is equal to 1, right? So substituting that in, we get 1 + 2 sinA cosA = 1/5. Then subtract 1 from both sides: 2 sinA cosA = 1/5 - 1. 1 is 5/5, so 1/5 - 5/5 is -4/5. So, 2 sinA cosA = -4/5. Therefore, sinA cosA = -2/5.Okay, so now I know that sinA cosA is -2/5. But the question is asking for sinA - cosA. Let me call that expression x. So, let x = sinA - cosA. If I square x, I get (sinA - cosA)^2 = sin²A - 2 sinA cosA + cos²A. Again, sin²A + cos²A is 1, so this becomes 1 - 2 sinA cosA.We already found that sinA cosA is -2/5. So, substituting that in, we have x² = 1 - 2*(-2/5). Let's calculate that. 2*(-2/5) is -4/5, so subtracting that is the same as adding 4/5. So, x² = 1 + 4/5. 1 is 5/5, so 5/5 + 4/5 is 9/5. Therefore, x² = 9/5, which means x is either sqrt(9/5) or -sqrt(9/5). Simplifying sqrt(9/5) gives 3/sqrt(5), which rationalized is 3√5/5. So, x could be either 3√5/5 or -3√5/5.But now, how do I know which one it is? The problem doesn't specify if angle A is acute or obtuse, but since it's a triangle, angle A has to be between 0 and 180 degrees. Let me think. If sinA + cosA is positive (which it is, since √5/5 is positive), then both sinA and cosA can't be negative. If angle A is acute (between 0 and 90 degrees), both sinA and cosA are positive. If it's obtuse (between 90 and 180 degrees), sinA is still positive, but cosA is negative.Given that sinA + cosA is √5/5, which is positive, but sinA cosA is negative (-2/5). If sinA cosA is negative, that means one of them is positive and the other is negative. But since in a triangle, angles are between 0 and 180, sinA is always positive. Therefore, cosA must be negative. So angle A is obtuse (between 90 and 180 degrees), which makes cosA negative.So, in that case, sinA is positive and cosA is negative. Then, sinA - cosA would be sinA minus a negative, which is sinA + |cosA|. Since both sinA and |cosA| are positive, the result is positive. Therefore, x is positive. So, x = 3√5/5.Looking at the options, B is 3√5/5. So, the answer should be B. Let me check the steps again to make sure I didn't make a mistake.1. Start with sinA + cosA = √5/5. Squared gives 1 + 2 sinA cosA = 1/5. Then 2 sinA cosA = -4/5, so sinA cosA = -2/5. That's correct.2. Then (sinA - cosA)^2 = 1 - 2 sinA cosA = 1 - 2*(-2/5) = 1 + 4/5 = 9/5. So x squared is 9/5, so x is ±3√5/5. Correct.3. Then reasoning about the sign. Since sinA is positive (as angle in a triangle), and cosA is negative (since sinA cosA is negative), so sinA - cosA is sinA - (negative) = sinA + positive, which is positive. Therefore, x is positive. So, 3√5/5, which is option B. That seems right.Wait, but hold on. If angle A is obtuse, cosA is negative, so sinA - cosA would be sinA minus a negative number, which is sinA + |cosA|. Since both are positive, the result is positive, so x is positive. So, 3√5/5 is the correct answer, which is option B. Therefore, I think the answer is B.But let me check if there's another way to approach this problem. Maybe using substitution? Let me assume that sinA + cosA = √5/5. Let me set sinA = s and cosA = c. So, s + c = √5/5, and s^2 + c^2 = 1. Then, we can solve for s and c. Let me try that.From s + c = √5/5, so c = √5/5 - s. Substitute into s^2 + c^2 = 1:s^2 + (√5/5 - s)^2 = 1Expanding the square:s^2 + ( (√5/5)^2 - 2*(√5/5)*s + s^2 ) = 1Simplify:s^2 + (5/25 - (2√5/5)s + s^2 ) = 1Combine like terms:2s^2 - (2√5/5)s + 1/5 = 1Subtract 1:2s^2 - (2√5/5)s + 1/5 - 1 = 0Which is:2s^2 - (2√5/5)s - 4/5 = 0Multiply both sides by 5 to eliminate denominators:10s^2 - 2√5 s - 4 = 0Divide both sides by 2:5s^2 - √5 s - 2 = 0Now, we can solve this quadratic equation for s. Let's use the quadratic formula.s = [√5 ± √( (√5)^2 - 4*5*(-2) ) ] / (2*5)Calculate discriminant:(√5)^2 - 4*5*(-2) = 5 + 40 = 45So, sqrt(45) = 3√5Thus, s = [√5 ± 3√5]/10So two solutions:s = (√5 + 3√5)/10 = 4√5/10 = 2√5/5Or s = (√5 - 3√5)/10 = (-2√5)/10 = -√5/5But since angle A is in a triangle (0 < A < π), sinA must be positive. Therefore, s = 2√5/5.Then, cosA = √5/5 - s = √5/5 - 2√5/5 = (-√5)/5Therefore, sinA - cosA = (2√5/5) - (-√5/5) = 2√5/5 + √5/5 = 3√5/5. So, that's the same answer as before, option B. So this confirms the previous result.Therefore, the answer is B: 3√5/5.Final Answerboxed{B}</think>Given in (triangle ABC), (sin A + cos A = frac{sqrt{5}}{5}), we need to find (sin A - cos A).First, we square both sides of the given equation:[(sin A + cos A)^2 = left(frac{sqrt{5}}{5}right)^2]Expanding the left side, we get:[sin^2 A + 2 sin A cos A + cos^2 A = frac{1}{5}]Using the Pythagorean identity (sin^2 A + cos^2 A = 1), we substitute:[1 + 2 sin A cos A = frac{1}{5}]Solving for (2 sin A cos A):[2 sin A cos A = frac{1}{5} - 1 = frac{1}{5} - frac{5}{5} = -frac{4}{5}]Thus,[sin A cos A = -frac{2}{5}]Next, we consider ((sin A - cos A)^2):[(sin A - cos A)^2 = sin^2 A - 2 sin A cos A + cos^2 A]Using the Pythagorean identity again:[sin^2 A + cos^2 A = 1]Substituting (sin A cos A = -frac{2}{5}):[(sin A - cos A)^2 = 1 - 2 left(-frac{2}{5}right) = 1 + frac{4}{5} = frac{9}{5}]Taking the square root of both sides:[sin A - cos A = pm frac{3sqrt{5}}{5}]Since angle (A) is in a triangle, it must be between (0) and (180^circ). Given (sin A + cos A = frac{sqrt{5}}{5}) is positive and (sin A cos A = -frac{2}{5}) is negative, (cos A) must be negative (as (sin A) is always positive in a triangle). Therefore, (sin A - cos A) is positive because it is the sum of two positive values (since (cos A) is negative).Thus,[sin A - cos A = frac{3sqrt{5}}{5}]The answer is (boxed{B}).]

🔑:Given sin A + cos A = frac{sqrt{5}}{5}, we want to find the value of sin A - cos A. We will follow the steps outlined in the solution provided.1. We start with the given equation and square both sides:(sin A + cos A)^2 = left(frac{sqrt{5}}{5}right)^2.This simplifies to:sin^2 A + 2sin A cos A + cos^2 A = frac{1}{5}.2. Knowing that sin^2 A + cos^2 A = 1, we can substitute this into our equation:1 + 2sin A cos A = frac{1}{5}.3. Rearranging the equation for 2sin A cos A, we find:2sin A cos A = frac{1}{5} - 1 = -frac{4}{5}.Dividing by 2 gives us:sin A cos A = -frac{2}{5}.4. The condition sin A cos A 0 and cos A < 0.5. Next, we want to find sin A - cos A, so we square this expression:(sin A - cos A)^2 = sin^2 A - 2sin A cos A + cos^2 A.Using sin^2 A + cos^2 A = 1 and substituting -2sin A cos A = frac{4}{5}, we get:1 - (-frac{4}{5}) = frac{9}{5}.6. Taking the square root of both sides to find sin A - cos A:sin A - cos A = sqrt{frac{9}{5}} = frac{3sqrt{5}}{5}.Therefore, the correct answer is boxed{B}, which is sin A - cos A = frac{3sqrt{5}}{5}.

🧠:[<think>Okay, let me try to work through this problem step by step. The question is: If sin³θ - cos³θ ≥ cosθ - sinθ, where 0 ≤ θ < 2π, then the range of θ is one of the given options. The options are A through D, with C being [π/4, 5π/4] and others as listed. First, I need to understand the inequality given: sin³θ - cos³θ is greater than or equal to cosθ - sinθ. Maybe I can start by rearranging the inequality to bring all terms to one side. Let me try that.So, sin³θ - cos³θ - cosθ + sinθ ≥ 0. Hmm, that's the left-hand side minus the right-hand side. Now, maybe factor this expression? Let's see. I notice that there's a sin³θ and a sinθ term, and similarly, cos³θ and a cosθ term. Maybe factor sinθ and cosθ from their respective terms. Let's try:sinθ(sin²θ + 1) - cosθ(cos²θ + 1) ≥ 0. Wait, is that right? Let me check:sin³θ - cos³θ - cosθ + sinθ. If I factor sinθ from sin³θ + sinθ, that would be sinθ(sin²θ + 1). Similarly, factoring -cosθ from -cos³θ - cosθ gives -cosθ(cos²θ + 1). So yes, the expression becomes sinθ(sin²θ + 1) - cosθ(cos²θ + 1) ≥ 0. But I'm not sure if that helps. Alternatively, maybe use the identity for a³ - b³? Remember that a³ - b³ = (a - b)(a² + ab + b²). Let's try applying that to sin³θ - cos³θ.So, sin³θ - cos³θ = (sinθ - cosθ)(sin²θ + sinθcosθ + cos²θ). But sin²θ + cos²θ = 1, so this simplifies to (sinθ - cosθ)(1 + sinθcosθ). Therefore, the original inequality becomes:(sinθ - cosθ)(1 + sinθcosθ) ≥ cosθ - sinθ.Hmm, let's rewrite the right-hand side as -(sinθ - cosθ). So, cosθ - sinθ = - (sinθ - cosθ). Therefore, the inequality is:(sinθ - cosθ)(1 + sinθcosθ) + (sinθ - cosθ) ≥ 0.Factor out (sinθ - cosθ):(sinθ - cosθ)[(1 + sinθcosθ) + 1] ≥ 0.Wait, wait. Let me check that again. If the original inequality is:(sinθ - cosθ)(1 + sinθcosθ) ≥ - (sinθ - cosθ).Then, moving all terms to the left:(sinθ - cosθ)(1 + sinθcosθ) + (sinθ - cosθ) ≥ 0.Factor out (sinθ - cosθ):(sinθ - cosθ)[(1 + sinθcosθ) + 1] ≥ 0.Wait, no, that's not quite right. Wait, when we have A ≥ B, then A - B ≥ 0. So here, A is (sinθ - cosθ)(1 + sinθcosθ), and B is - (sinθ - cosθ). Therefore, moving B to the left gives:A - B = (sinθ - cosθ)(1 + sinθcosθ) - (- (sinθ - cosθ)) = (sinθ - cosθ)(1 + sinθcosθ) + (sinθ - cosθ) ≥ 0.So yes, factor out (sinθ - cosθ):(sinθ - cosθ)[(1 + sinθcosθ) + 1] ≥ 0.Simplify the terms inside the brackets:(1 + sinθcosθ) + 1 = 2 + sinθcosθ.Therefore, the inequality becomes:(sinθ - cosθ)(2 + sinθcosθ) ≥ 0.So now we have two factors: (sinθ - cosθ) and (2 + sinθcosθ). Let's analyze each factor.First, 2 + sinθcosθ. Since sinθcosθ is equal to (1/2) sin2θ, which ranges between -1/2 and 1/2. Therefore, the maximum value of sinθcosθ is 1/2 and minimum is -1/2. Therefore, 2 + sinθcosθ ranges from 2 - 1/2 = 3/2 to 2 + 1/2 = 5/2. Therefore, 2 + sinθcosθ is always positive, since 3/2 > 0. Therefore, the sign of the entire expression (sinθ - cosθ)(2 + sinθcosθ) is determined solely by the sign of (sinθ - cosθ). Therefore, the inequality simplifies to (sinθ - cosθ) ≥ 0, since 2 + sinθcosθ is always positive. Therefore, sinθ - cosθ ≥ 0.So, now the problem reduces to solving sinθ - cosθ ≥ 0. Let's solve this inequality.First, sinθ - cosθ ≥ 0. Let's write this as sinθ ≥ cosθ.To find where sinθ ≥ cosθ in the interval [0, 2π). Let's recall when sinθ = cosθ. That occurs when tanθ = 1, so θ = π/4 and θ = 5π/4.So, the solutions to sinθ ≥ cosθ are the intervals where tanθ ≥ 1 (since dividing both sides by cosθ, assuming cosθ positive). But we have to be careful because dividing by cosθ when it's negative would reverse the inequality. Maybe a better approach is to consider the equation sinθ - cosθ = 0 and analyze the sign changes.Alternatively, write sinθ - cosθ as √2 sin(θ - π/4). Let's verify that:Using the identity sinθ - cosθ = √2 sin(θ - π/4). Let's check:√2 sin(θ - π/4) = √2 [sinθ cosπ/4 - cosθ sinπ/4] = √2 [sinθ*(√2/2) - cosθ*(√2/2)] = √2*(√2/2)(sinθ - cosθ) = (2/2)(sinθ - cosθ) = sinθ - cosθ. Yes, correct.Therefore, sinθ - cosθ = √2 sin(θ - π/4). Therefore, sinθ - cosθ ≥ 0 is equivalent to √2 sin(θ - π/4) ≥ 0. Since √2 is positive, this simplifies to sin(θ - π/4) ≥ 0.So, sin(θ - π/4) ≥ 0. Let's solve this inequality.The sine function is non-negative in the intervals [0, π] (adjusted for the argument θ - π/4). So, θ - π/4 ∈ [0, π] + 2πk, where k is an integer. But since θ is in [0, 2π), let's find the corresponding θ:θ - π/4 ≥ 0 ⇒ θ ≥ π/4.θ - π/4 ≤ π ⇒ θ ≤ π + π/4 = 5π/4.Therefore, θ ∈ [π/4, 5π/4]. So, the solution is the interval [π/4, 5π/4].Therefore, looking back at the options, option C is [π/4, 5π/4], which matches our solution. Therefore, the answer should be C.Wait, but let me verify this with some test angles to make sure.Take θ = 0: sin0 - cos0 = -1 < 0, which is not in the solution set, which is correct because 0 is not in [π/4, 5π/4].Take θ = π/2: sin(π/2) - cos(π/2) = 1 - 0 = 1 ≥ 0, so π/2 is in the solution. Similarly, θ = π: sinπ - cosπ = 0 - (-1) = 1 ≥ 0, so π is in the solution. θ = 3π/2: sin(3π/2) - cos(3π/2) = -1 - 0 = -1 < 0, which is not in the solution. θ = π/4: sin(π/4) - cos(π/4) = 0, so equality holds, which is included in the inequality. θ = 5π/4: sin(5π/4) - cos(5π/4) = (-√2/2) - (-√2/2) = 0, which is also included. So, between π/4 and 5π/4, the inequality holds, which matches option C. Therefore, the correct answer is C.Final Answerboxed{C}</think>Given the inequality (sin^3 theta - cos^3 theta geq cos theta - sin theta) for (0 leq theta < 2pi), we start by rewriting the inequality:[sin^3 theta - cos^3 theta - cos theta + sin theta geq 0]We use the identity for the difference of cubes: (a^3 - b^3 = (a - b)(a^2 + ab + b^2)). Applying this to (sin^3 theta - cos^3 theta), we get:[(sin theta - cos theta)(sin^2 theta + sin theta cos theta + cos^2 theta)]Since (sin^2 theta + cos^2 theta = 1), this simplifies to:[(sin theta - cos theta)(1 + sin theta cos theta)]The original inequality becomes:[(sin theta - cos theta)(1 + sin theta cos theta) geq -(sin theta - cos theta)]Rearranging terms, we factor out ((sin theta - cos theta)):[(sin theta - cos theta)(1 + sin theta cos theta + 1) geq 0]Simplifying inside the brackets:[(sin theta - cos theta)(2 + sin theta cos theta) geq 0]Since (2 + sin theta cos theta) is always positive (as (sin theta cos theta) ranges between (-frac{1}{2}) and (frac{1}{2})), the inequality reduces to:[sin theta - cos theta geq 0]This can be rewritten using the identity (sin theta - cos theta = sqrt{2} sin left( theta - frac{pi}{4} right)). Thus, we need:[sqrt{2} sin left( theta - frac{pi}{4} right) geq 0]This simplifies to:[sin left( theta - frac{pi}{4} right) geq 0]The solution to this inequality is:[0 leq theta - frac{pi}{4} leq pi implies frac{pi}{4} leq theta leq frac{5pi}{4}]Thus, the range of (theta) is (left[frac{pi}{4}, frac{5pi}{4}right]), which corresponds to option C.[boxed{C}],<think>Okay, so I need to solve this inequality: sin³θ - cos³θ ≥ cosθ - sinθ, where θ is between 0 and 2π. The answer choices are given, so I need to figure out which interval θ falls into. Let's start by understanding the equation step by step.First, I remember that a³ - b³ can be factored as (a - b)(a² + ab + b²). Maybe that can help simplify the left side of the inequality. Let me try that:sin³θ - cos³θ = (sinθ - cosθ)(sin²θ + sinθcosθ + cos²θ)Hmm, sin²θ + cos²θ is equal to 1, right? So that simplifies the expression:(sinθ - cosθ)(1 + sinθcosθ)So the left side becomes (sinθ - cosθ)(1 + sinθcosθ). The right side of the inequality is cosθ - sinθ. Let me write the inequality again:(sinθ - cosθ)(1 + sinθcosθ) ≥ cosθ - sinθWait, the right side is cosθ - sinθ, which is the same as -(sinθ - cosθ). So maybe I can factor that out. Let me rearrange the inequality:(sinθ - cosθ)(1 + sinθcosθ) + (sinθ - cosθ) ≥ 0Wait, no. Let me bring everything to the left side:(sinθ - cosθ)(1 + sinθcosθ) - (cosθ - sinθ) ≥ 0But since cosθ - sinθ = -(sinθ - cosθ), this becomes:(sinθ - cosθ)(1 + sinθcosθ) + (sinθ - cosθ) ≥ 0Now factor out (sinθ - cosθ):(sinθ - cosθ)[(1 + sinθcosθ) + 1] ≥ 0Wait, that would be:(sinθ - cosθ)(1 + sinθcosθ + 1) ≥ 0So that's:(sinθ - cosθ)(2 + sinθcosθ) ≥ 0Hmm, is that right? Let me check the steps again.Original inequality:sin³θ - cos³θ ≥ cosθ - sinθLeft side factored: (sinθ - cosθ)(1 + sinθcosθ)Right side: cosθ - sinθ = -(sinθ - cosθ)So bringing everything to the left:(sinθ - cosθ)(1 + sinθcosθ) + (sinθ - cosθ) ≥ 0Yes, that's correct. Then factor out (sinθ - cosθ):(sinθ - cosθ)[(1 + sinθcosθ) + 1] ≥ 0Wait, 1 + sinθcosθ + 1 is 2 + sinθcosθ. So the inequality simplifies to:(sinθ - cosθ)(2 + sinθcosθ) ≥ 0Okay, so now the inequality is (sinθ - cosθ)(2 + sinθcosθ) ≥ 0. Let's analyze this.First, note that 2 + sinθcosθ is always positive, right? Because sinθcosθ has a maximum value of 0.5 (since sin2θ = 2sinθcosθ, so sinθcosθ = (sin2θ)/2, which has a maximum of 1/2). Therefore, 2 + sinθcosθ ≥ 2 - 0.5 = 1.5, which is always positive. Therefore, the sign of the entire expression depends solely on (sinθ - cosθ). Since 2 + sinθcosθ is always positive, the inequality (sinθ - cosθ)(2 + sinθcosθ) ≥ 0 is equivalent to sinθ - cosθ ≥ 0.Because multiplying by a positive number doesn't change the inequality direction. So essentially, the original inequality simplifies to sinθ - cosθ ≥ 0. Therefore, the solution is all θ where sinθ ≥ cosθ.So now the problem reduces to finding θ in [0, 2π) where sinθ ≥ cosθ. Let's solve that.When is sinθ ≥ cosθ? Let's recall that sinθ = cosθ at θ = π/4 and θ = 5π/4. Let's analyze the intervals between these points.From θ = 0 to θ = π/4: Let's pick θ = 0. sin0 = 0, cos0 = 1. So sinθ < cosθ here. At θ = π/4, sinθ = cosθ = √2/2.From θ = π/4 to θ = 5π/4: Let's pick θ = π/2. sin(π/2) = 1, cos(π/2) = 0. So sinθ > cosθ here. Similarly, θ = π: sinπ = 0, cosπ = -1. So sinθ = 0 > cosπ = -1. At θ = 3π/2: sin3π/2 = -1, cos3π/2 = 0. So sinθ = -1 < cosθ = 0. Wait, but 5π/4 is in the third quadrant. Let me check θ = 5π/4: sin5π/4 = -√2/2, cos5π/4 = -√2/2. So sinθ = cosθ here.So the interval between π/4 and 5π/4, sinθ ≥ cosθ. Let me confirm with θ = 3π/4: sin3π/4 = √2/2, cos3π/4 = -√2/2. So sinθ = √2/2 > cosθ = -√2/2. Yes, sinθ > cosθ here. Similarly, θ = π: sinθ = 0 > cosθ = -1.At θ = 5π/4: sinθ = cosθ. Then from 5π/4 to 2π, let's check θ = 3π/2: sinθ = -1, cosθ = 0. So sinθ < cosθ. θ = 7π/4: sinθ = -√2/2, cosθ = √2/2. So sinθ < cosθ. So the interval where sinθ ≥ cosθ is [π/4, 5π/4].Therefore, the answer should be option C: [π/4, 5π/4]. Let me cross-check with the answer choices:(A) [0, π/4](B) [π/4, π](C) [π/4, 5π/4](D) [π/2, 3π/2)Yes, option C is [π/4, 5π/4], which matches our solution. Let me just make sure that we didn't make a mistake in simplifying the original inequality.Original inequality: sin³θ - cos³θ ≥ cosθ - sinθFactored left side: (sinθ - cosθ)(1 + sinθcosθ)Right side: -(sinθ - cosθ)Bring everything to left: (sinθ - cosθ)(1 + sinθcosθ) + (sinθ - cosθ) ≥ 0Factor: (sinθ - cosθ)(1 + sinθcosθ + 1) = (sinθ - cosθ)(2 + sinθcosθ) ≥ 0Since 2 + sinθcosθ is always positive, the inequality is equivalent to sinθ - cosθ ≥ 0. Then, solving sinθ ≥ cosθ gives θ ∈ [π/4, 5π/4]. So yes, option C is correct.Final Answerboxed{C}</think>Given the inequality (sin^3 theta - cos^3 theta geq cos theta - sin theta) for (0 leq theta < 2pi), we start by factoring the left-hand side using the difference of cubes formula:[sin^3 theta - cos^3 theta = (sin theta - cos theta)(sin^2 theta + sin theta cos theta + cos^2 theta)]Since (sin^2 theta + cos^2 theta = 1), this simplifies to:[(sin theta - cos theta)(1 + sin theta cos theta)]The right-hand side of the inequality is (cos theta - sin theta), which can be rewritten as (-(sin theta - cos theta)). Substituting this into the inequality, we get:[(sin theta - cos theta)(1 + sin theta cos theta) geq -(sin theta - cos theta)]Bringing all terms to the left-hand side, we factor out ((sin theta - cos theta)):[(sin theta - cos theta)(1 + sin theta cos theta + 1) geq 0]This simplifies to:[(sin theta - cos theta)(2 + sin theta cos theta) geq 0]Since (2 + sin theta cos theta) is always positive (as the maximum value of (sin theta cos theta) is (0.5)), the inequality reduces to:[sin theta - cos theta geq 0]This means (sin theta geq cos theta). Solving this inequality, we find the values of (theta) where (sin theta geq cos theta). The points where (sin theta = cos theta) are (theta = frac{pi}{4}) and (theta = frac{5pi}{4}). Analyzing the intervals, we find that (sin theta geq cos theta) in the interval (left[frac{pi}{4}, frac{5pi}{4}right]).Thus, the range of (theta) is (boxed{C}).]

🔑:6.C.Solution 1: From sin ^{3} theta-cos ^{3} theta geqslant cos theta-sin theta, we get (sin theta-cos theta)left(2+frac{1}{2} sin 2 thetaright) geqslant 0.Since 2+frac{1}{2} sin 2 theta>0, we have sin theta-cos theta geqslant 0.Solving this, we get frac{pi}{4} leqslant theta leqslant frac{5 pi}{4}.Solution 2: The original inequality can be transformed intosin ^{3} theta+sin theta geqslant cos ^{3} theta+cos theta.Construct the function f(x)=x^{3}+x, then the original inequality becomes f(sin theta) geqslant f(cos theta). It is easy to see that f(x) is an increasing function on mathbf{R}, therefore, sin theta geqslant cos theta. Noting that 0 leqslant theta leqslant 2 pi, solving this, we get frac{pi}{4} leqslant theta leqslant frac{5 pi}{4}.

🧠:[<think>Alright, let's try to solve this problem step by step. So, we have two equations here:1. (7^{2x} = 36)2. (7^{-x} = 6^{-frac{a}{2}})We need to find the value of (a). Hmm, okay. Let me start by looking at the first equation. It's an exponential equation where the base is 7, and the exponent is (2x), resulting in 36. Maybe I can take the logarithm of both sides to solve for (x)? But before jumping into logarithms, let me see if I can manipulate the equations to relate them somehow.First, equation 1: (7^{2x} = 36). If I can express (7^{2x}) in terms of (7^x), that might help. Since (7^{2x}) is the same as ((7^x)^2), right? So, let me rewrite equation 1 as:((7^x)^2 = 36)Taking the square root of both sides gives me:(7^x = 6) or (7^x = -6)But wait, (7^x) is always positive because any positive number raised to a real power is positive. So, (7^x = -6) doesn't make sense. Therefore, we discard that and have:(7^x = 6)Okay, so that gives me (7^x = 6). Now, maybe I can use this in equation 2. Let me check equation 2 again: (7^{-x} = 6^{-frac{a}{2}}). Hmm, (7^{-x}) is the reciprocal of (7^x). Since we know (7^x = 6), then (7^{-x} = frac{1}{6}). Let me substitute that into equation 2:(frac{1}{6} = 6^{-frac{a}{2}})Now, this equation is (6^{-1} = 6^{-frac{a}{2}}). Since the bases are the same and the equation holds, the exponents must be equal. Therefore:(-1 = -frac{a}{2})Solving for (a), multiply both sides by -1:(1 = frac{a}{2})Then multiply both sides by 2:(a = 2)Wait, that seems straightforward. Let me verify the steps again to make sure I didn't make any mistakes.Starting with (7^{2x} = 36), which is ((7^x)^2 = 36). Taking the square root gives (7^x = 6), since the negative solution is extraneous. Then, using the second equation (7^{-x} = 6^{-frac{a}{2}}), substitute (7^{-x}) with (1/6) because (7^{-x} = 1/7^x = 1/6). Therefore, (1/6 = 6^{-a/2}). Since (1/6 = 6^{-1}), equate the exponents: (-1 = -a/2), leading to (a = 2). Yeah, that seems right.Alternatively, maybe I can approach it using logarithms. Let's try that method to confirm.Starting with equation 1: (7^{2x} = 36). Taking the natural logarithm of both sides:(ln(7^{2x}) = ln(36))Using the logarithm power rule, (2x ln(7) = ln(36))Solving for (x):(x = frac{ln(36)}{2 ln(7)})Simplify (ln(36)) as (ln(6^2) = 2 ln(6)), so:(x = frac{2 ln(6)}{2 ln(7)} = frac{ln(6)}{ln(7)})So, (x = log_7 6). That makes sense because (7^{log_7 6} = 6), which matches our previous result.Now, plug this value of (x) into equation 2: (7^{-x} = 6^{-frac{a}{2}})Substitute (x = log_7 6):(7^{-log_7 6} = 6^{-frac{a}{2}})Left side: (7^{-log_7 6} = (7^{log_7 6})^{-1} = 6^{-1})So, (6^{-1} = 6^{-frac{a}{2}})Again, since the bases are equal, set the exponents equal:(-1 = -frac{a}{2})Multiply both sides by -2:(2 = a)Same result. So both methods lead to (a = 2). Therefore, the answer must be 2.Just to be thorough, let me check if there's another way to approach this. Maybe express both equations in terms of exponents with base 6 or base 7 and compare?Starting with equation 1: (7^{2x} = 6^2). Taking both sides to the power of something. If I can write equation 2 in terms of 6's exponent, which it already is. So, equation 2 is (7^{-x} = 6^{-a/2}). If I can relate (7^{-x}) to (6^{-a/2}). From equation 1, (7^{2x} = 6^2), so (7^{x} = 6). Therefore, (7^{-x} = 6^{-1}). Therefore, equation 2 becomes (6^{-1} = 6^{-a/2}), leading to ( -1 = -a/2 implies a=2). Yep, same answer again.I think that's solid. All methods converge to (a=2). So I can confidently say the answer is 2.Final AnswerThe value of (a) is boxed{2}.</think>Given the equations:1. (7^{2x} = 36)2. (7^{-x} = 6^{-frac{a}{2}})We start by solving the first equation. Rewriting (7^{2x}) as ((7^x)^2), we have:[(7^x)^2 = 36]Taking the square root of both sides, we get:[7^x = 6]Next, we use this result in the second equation. Since (7^{-x}) is the reciprocal of (7^x), we have:[7^{-x} = frac{1}{6}]Substituting this into the second equation:[frac{1}{6} = 6^{-frac{a}{2}}]Recognizing that (frac{1}{6} = 6^{-1}), we equate the exponents:[-1 = -frac{a}{2}]Solving for (a), we multiply both sides by -2:[a = 2]Thus, the value of (a) is (boxed{2}).,<think>Okay, so I need to find the value of 'a' given these two equations: 7^(2x) = 36 and 7^(-x) = 6^(-a/2). Hmm, let's see. Let me start by understanding what each equation represents and how they might relate to each other.First, the equation 7^(2x) = 36. I know that exponents can sometimes be manipulated by taking logarithms, but maybe there's a way to express x in terms of logarithms here. Let me recall that if I have an equation like a^b = c, then taking the logarithm of both sides gives b * log(a) = log(c), so b = log(c) / log(a). Maybe that can help here.Alternatively, since both equations involve 7^x and 6, maybe I can express one equation in terms of the other. Let me see. The first equation is 7^(2x) = 36. That can be rewritten as (7^x)^2 = 36, right? Because when you raise a power to a power, you multiply the exponents. So 7^(2x) is the same as (7^x)^2. So that gives me (7^x)^2 = 36. Taking the square root of both sides, I get 7^x = 6 or 7^x = -6. But since 7^x is always positive (because any positive number to a real exponent is positive), we can discard the negative solution. So 7^x = 6.Wait, that's helpful! So from the first equation, we can directly find that 7^x = 6. Then, perhaps I can use this in the second equation. Let me check the second equation: 7^(-x) = 6^(-a/2). Hmm. If I know 7^x = 6, then 7^(-x) is 1/(7^x) = 1/6. So substituting that into the second equation, 1/6 = 6^(-a/2).So now we have 1/6 = 6^(-a/2). Let me recall that 1/6 is the same as 6^(-1). Therefore, 6^(-1) = 6^(-a/2). Since the bases are equal (both are 6), the exponents must be equal. Therefore, -1 = -a/2. Solving for a, multiply both sides by -1: 1 = a/2. Then multiply both sides by 2: a = 2. So is that the answer? Let me verify each step to make sure I didn't make any mistakes.Starting with the first equation: 7^(2x) = 36. Rewriting as (7^x)^2 = 36. Taking the square root, 7^x = 6. That's correct because sqrt(36) is 6, and since 7^x is positive, we ignore the negative root. Then, in the second equation: 7^(-x) = 6^(-a/2). Since 7^x = 6, then 7^(-x) = 1/6. So replacing 7^(-x) with 1/6 gives 1/6 = 6^(-a/2). Since 1/6 is 6^(-1), we set -1 = -a/2. Solving for a gives a = 2. All steps check out. So the value of 'a' should be 2.Alternatively, another way to approach this is by using logarithms. Let's try that method to confirm. Starting with 7^(2x) = 36. Take the natural logarithm of both sides: ln(7^(2x)) = ln(36). Using the power rule, 2x * ln(7) = ln(36). Then, solving for x: x = ln(36)/(2 ln(7)). Similarly, from the second equation: 7^(-x) = 6^(-a/2). Take natural logs again: -x * ln(7) = (-a/2) * ln(6). Multiply both sides by -1: x * ln(7) = (a/2) * ln(6). Then, solving for a: a = (2 x ln(7))/ln(6). But from the first equation, we know that x = ln(36)/(2 ln(7)). Substituting that into the expression for a: a = (2 * [ln(36)/(2 ln(7))] * ln(7))/ln(6). The 2 and 2 cancel, ln(7) cancels with the denominator, so we have a = ln(36)/ln(6). Now, ln(36) can be written as ln(6^2) which is 2 ln(6). Therefore, a = (2 ln(6))/ln(6) = 2. So again, a = 2. This method also confirms that the answer is 2.Alternatively, maybe using logarithm base 7. Let's try that. From 7^(2x) = 36. Taking log base 7 of both sides: log_7(7^(2x)) = log_7(36). Simplifies to 2x = log_7(36). Therefore, x = (log_7(36))/2. Then, 7^(-x) = 6^(-a/2). Take log base 7 of both sides: log_7(7^(-x)) = log_7(6^(-a/2)). Which simplifies to -x = (-a/2) * log_7(6). Multiply both sides by -1: x = (a/2) * log_7(6). Substitute the expression for x from the first equation: (log_7(36))/2 = (a/2) * log_7(6). Multiply both sides by 2: log_7(36) = a * log_7(6). Then, log_7(36) is log_7(6^2) = 2 log_7(6). So 2 log_7(6) = a log_7(6). Divide both sides by log_7(6): 2 = a. Therefore, again, a = 2. So regardless of the method used, logarithm base 7, natural logarithm, or manipulating exponents directly, all methods lead to a = 2. Therefore, I can be confident that the answer is 2.Another way to check: Let's plug a = 2 back into the original equations and see if they hold. First equation: 7^(2x) = 36. From the second equation, if a = 2, then 7^(-x) = 6^(-2/2) = 6^(-1) = 1/6. Therefore, 7^x must be 6. Then, 7^(2x) is (7^x)^2 = 6^2 = 36, which matches the first equation. So yes, it's consistent. Therefore, the solution is correct.I think that's thorough enough. I considered multiple methods, checked consistency, and each time arrived at a = 2. So the answer is 2.Final AnswerThe value of ( a ) is boxed{2}.</think>Given the equations (7^{2x} = 36) and (7^{-x} = 6^{-frac{a}{2}}), we need to find the value of (a).Starting with the first equation:[7^{2x} = 36]We can rewrite this as:[(7^x)^2 = 36]Taking the square root of both sides, we get:[7^x = 6]Since (7^x) must be positive, we discard the negative solution.Next, consider the second equation:[7^{-x} = 6^{-frac{a}{2}}]From (7^x = 6), we find:[7^{-x} = frac{1}{6}]Substituting this into the second equation, we have:[frac{1}{6} = 6^{-frac{a}{2}}]Since (frac{1}{6} = 6^{-1}), we equate the exponents:[-1 = -frac{a}{2}]Solving for (a), we multiply both sides by (-1):[1 = frac{a}{2}]Then, multiplying both sides by 2:[a = 2]To verify, we substitute (a = 2) back into the original equations and confirm they hold true. The first equation (7^{2x} = 36) is satisfied because (7^x = 6) implies (7^{2x} = 6^2 = 36). The second equation (7^{-x} = 6^{-1}) is also satisfied.Thus, the value of (a) is (boxed{2}).]

🔑:begin{array}{l}7^{x}=6 Rightarrow 7^{-x}=(6)^{-frac{a}{2}}=6^{-1} Rightarrow a=2end{array}