IB Mathematics: Analysis and Approaches Quiz: Advanced Trig Identities
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Advanced Trig IdentitiesQuestion 1 of 20

In triangle ABC, sin⁡(A)=35\sin(A) = \frac{3}{5} and sin⁡(B)=513\sin(B) = \frac{5}{13}. Given that angle A is acute and angle B is obtuse, find cos⁡(C)\cos(C).

−6365-\frac{63}{65}
−3365-\frac{33}{65}
3365\frac{33}{65}
6365\frac{63}{65}
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Advanced Trig Identities

Practice Advanced Trig Identities in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Advanced Trig Identities, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

In triangle ABC, sin⁡(A)=35\sin(A) = \frac{3}{5} and sin⁡(B)=513\sin(B) = \frac{5}{13}. Given that angle A is acute and angle B is obtuse, find cos⁡(C)\cos(C).

  1. −6365-\frac{63}{65}
  2. −3365-\frac{33}{65}
  3. 3365\frac{33}{65}
  4. 6365\frac{63}{65} (correct answer)
Explanation: Since angle A is acute, cos⁡(A)>0\cos(A) > 0. cos⁡(A)=1−(3/5)2=45\cos(A) = \sqrt{1 - (3/5)^2} = \frac{4}{5}. Since angle B is obtuse, cos⁡(B)<0\cos(B) < 0. cos⁡(B)=−1−(5/13)2=−1213\cos(B) = -\sqrt{1 - (5/13)^2} = -\frac{12}{13}. In a triangle, A+B+C=πA+B+C = \pi, so C=π−(A+B)C = \pi - (A+B). Therefore, cos⁡(C)=cos⁡(π−(A+B))=−cos⁡(A+B)\cos(C) = \cos(\pi - (A+B)) = -\cos(A+B). Using the compound angle identity, cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)=(45)(−1213)−(35)(513)=−4865−1565=−6365\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B) = \left(\frac{4}{5}\right)\left(-\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) = -\frac{48}{65} - \frac{15}{65} = -\frac{63}{65}. Finally, cos⁡(C)=−(−6365)=6365\cos(C) = -(-\frac{63}{65}) = \frac{63}{65}.

Question 2

Find the value of cos⁡(4θ)\cos(4\theta) if cos⁡(θ)=13\cos(\theta) = \frac{1}{3}.

  1. −17981-\frac{179}{81}
  2. −79-\frac{7}{9}
  3. 1781\frac{17}{81} (correct answer)
  4. 4981\frac{49}{81}
Explanation: We can find cos⁡(4θ)\cos(4\theta) by applying the double angle identity for cosine twice. First, find cos⁡(2θ)\cos(2\theta): cos⁡(2θ)=2cos⁡2(θ)−1=2(13)2−1=2(19)−1=29−1=−79\cos(2\theta) = 2\cos^2(\theta) - 1 = 2\left(\frac{1}{3}\right)^2 - 1 = 2\left(\frac{1}{9}\right) - 1 = \frac{2}{9} - 1 = -\frac{7}{9}. Now, use this result to find cos⁡(4θ)=cos⁡(2(2θ))\cos(4\theta) = \cos(2(2\theta)): cos⁡(4θ)=2cos⁡2(2θ)−1=2(−79)2−1=2(4981)−1=9881−8181=1781\cos(4\theta) = 2\cos^2(2\theta) - 1 = 2\left(-\frac{7}{9}\right)^2 - 1 = 2\left(\frac{49}{81}\right) - 1 = \frac{98}{81} - \frac{81}{81} = \frac{17}{81}.

Question 3

If A+B=π4A+B = \frac{\pi}{4}, what is the value of (1+tan⁡A)(1+tan⁡B)(1+\tan A)(1+\tan B)?

  1. −12-\frac{1}{2}
  2. 1
  3. 32\frac{3}{2}
  4. 2 (correct answer)
Explanation: Since A+B=π4A+B = \frac{\pi}{4}, we have tan⁡(A+B)=tan⁡(π4)=1\tan(A+B) = \tan(\frac{\pi}{4}) = 1. Using the compound angle formula for tangent, tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}. So, tan⁡A+tan⁡B1−tan⁡Atan⁡B=1\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1. This implies tan⁡A+tan⁡B=1−tan⁡Atan⁡B\tan A + \tan B = 1 - \tan A \tan B, which can be rearranged to tan⁡A+tan⁡B+tan⁡Atan⁡B=1\tan A + \tan B + \tan A \tan B = 1. Now consider the expression we want to evaluate: (1+tan⁡A)(1+tan⁡B)=1+tan⁡A+tan⁡B+tan⁡Atan⁡B(1+\tan A)(1+\tan B) = 1 + \tan A + \tan B + \tan A \tan B. From our rearrangement, we know that tan⁡A+tan⁡B+tan⁡Atan⁡B=1\tan A + \tan B + \tan A \tan B = 1. Therefore, (1+tan⁡A)(1+tan⁡B)=1+1=2(1+\tan A)(1+\tan B) = 1 + 1 = 2.

Question 4

What is the period of the function f(x)=cos⁡(4x)cos⁡(2x)−sin⁡(4x)sin⁡(2x)f(x) = \cos(4x)\cos(2x) - \sin(4x)\sin(2x)?

  1. π6\frac{\pi}{6}
  2. π3\frac{\pi}{3} (correct answer)
  3. π2\frac{\pi}{2}
  4. π\pi
Explanation: The expression cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\cos(A)\cos(B) - \sin(A)\sin(B) is the compound angle identity for cos⁡(A+B)\cos(A+B). In this case, A=4xA=4x and B=2xB=2x. Therefore, f(x)=cos⁡(4x+2x)=cos⁡(6x)f(x) = \cos(4x+2x) = \cos(6x). The period of a function g(x)=acos⁡(kx)+dg(x) = a\cos(kx)+d is given by 2π∣k∣\frac{2\pi}{|k|}. For f(x)=cos⁡(6x)f(x) = \cos(6x), we have k=6k=6. The period is 2π6=π3\frac{2\pi}{6} = \frac{\pi}{3}.

Question 5

If tan⁡(A)=12\tan(A) = \frac{1}{2} and tan⁡(A+B)=3\tan(A+B) = 3, what is the value of tan⁡(B)\tan(B)?

  1. −5-5
  2. 1 (correct answer)
  3. 52\frac{5}{2}
  4. 5
Explanation: Start with the compound angle identity for tangent: tan⁡(A+B)=tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B)\tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}. Substitute the given values: 3=12+tan⁡(B)1−12tan⁡(B)3 = \frac{\frac{1}{2} + \tan(B)}{1 - \frac{1}{2}\tan(B)}. Let t=tan⁡(B)t = \tan(B). Then 3(1−12t)=12+t3(1 - \frac{1}{2}t) = \frac{1}{2} + t. This simplifies to 3−32t=12+t3 - \frac{3}{2}t = \frac{1}{2} + t. Rearranging to solve for tt: 3−12=t+32t3 - \frac{1}{2} = t + \frac{3}{2}t, which gives 52=52t\frac{5}{2} = \frac{5}{2}t. Therefore, t=tan⁡(B)=1t = \tan(B) = 1.

Question 6

If tan⁡(A)=12\tan(A) = \frac{1}{2} and tan⁡(A+B)=3\tan(A+B) = 3, what is the value of tan⁡(B)\tan(B)?

  1. −5-5
  2. 1 (correct answer)
  3. 52\frac{5}{2}
  4. 5
Explanation: Start with the compound angle identity for tangent: tan⁡(A+B)=tan⁡(A)+tan⁡(B)1−tan⁡(A)tan⁡(B)\tan(A+B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}. Substitute the given values: 3=12+tan⁡(B)1−12tan⁡(B)3 = \frac{\frac{1}{2} + \tan(B)}{1 - \frac{1}{2}\tan(B)}. Let t=tan⁡(B)t = \tan(B). Then 3(1−12t)=12+t3(1 - \frac{1}{2}t) = \frac{1}{2} + t. This simplifies to 3−32t=12+t3 - \frac{3}{2}t = \frac{1}{2} + t. Rearranging to solve for tt: 3−12=t+32t3 - \frac{1}{2} = t + \frac{3}{2}t, which gives 52=52t\frac{5}{2} = \frac{5}{2}t. Therefore, t=tan⁡(B)=1t = \tan(B) = 1.

Question 7

Find the value of cos⁡(4θ)\cos(4\theta) if cos⁡(θ)=13\cos(\theta) = \frac{1}{3}.

  1. −17981-\frac{179}{81}
  2. −79-\frac{7}{9}
  3. 1781\frac{17}{81} (correct answer)
  4. 4981\frac{49}{81}
Explanation: We can find cos⁡(4θ)\cos(4\theta) by applying the double angle identity for cosine twice. First, find cos⁡(2θ)\cos(2\theta): cos⁡(2θ)=2cos⁡2(θ)−1=2(13)2−1=2(19)−1=29−1=−79\cos(2\theta) = 2\cos^2(\theta) - 1 = 2\left(\frac{1}{3}\right)^2 - 1 = 2\left(\frac{1}{9}\right) - 1 = \frac{2}{9} - 1 = -\frac{7}{9}. Now, use this result to find cos⁡(4θ)=cos⁡(2(2θ))\cos(4\theta) = \cos(2(2\theta)): cos⁡(4θ)=2cos⁡2(2θ)−1=2(−79)2−1=2(4981)−1=9881−8181=1781\cos(4\theta) = 2\cos^2(2\theta) - 1 = 2\left(-\frac{7}{9}\right)^2 - 1 = 2\left(\frac{49}{81}\right) - 1 = \frac{98}{81} - \frac{81}{81} = \frac{17}{81}.

Question 8

In triangle ABC, sin⁡(A)=35\sin(A) = \frac{3}{5} and sin⁡(B)=513\sin(B) = \frac{5}{13}. Given that angle A is acute and angle B is obtuse, find cos⁡(C)\cos(C).

  1. −6365-\frac{63}{65}
  2. −3365-\frac{33}{65}
  3. 3365\frac{33}{65}
  4. 6365\frac{63}{65} (correct answer)
Explanation: Since angle A is acute, cos⁡(A)>0\cos(A) > 0. cos⁡(A)=1−(3/5)2=45\cos(A) = \sqrt{1 - (3/5)^2} = \frac{4}{5}. Since angle B is obtuse, cos⁡(B)<0\cos(B) < 0. cos⁡(B)=−1−(5/13)2=−1213\cos(B) = -\sqrt{1 - (5/13)^2} = -\frac{12}{13}. In a triangle, A+B+C=πA+B+C = \pi, so C=π−(A+B)C = \pi - (A+B). Therefore, cos⁡(C)=cos⁡(π−(A+B))=−cos⁡(A+B)\cos(C) = \cos(\pi - (A+B)) = -\cos(A+B). Using the compound angle identity, cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)=(45)(−1213)−(35)(513)=−4865−1565=−6365\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B) = \left(\frac{4}{5}\right)\left(-\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) = -\frac{48}{65} - \frac{15}{65} = -\frac{63}{65}. Finally, cos⁡(C)=−(−6365)=6365\cos(C) = -(-\frac{63}{65}) = \frac{63}{65}.

Question 9

What is the period of the function f(x)=cos⁡(4x)cos⁡(2x)−sin⁡(4x)sin⁡(2x)f(x) = \cos(4x)\cos(2x) - \sin(4x)\sin(2x)?

  1. π6\frac{\pi}{6}
  2. π3\frac{\pi}{3} (correct answer)
  3. π2\frac{\pi}{2}
  4. π\pi
Explanation: The expression cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\cos(A)\cos(B) - \sin(A)\sin(B) is the compound angle identity for cos⁡(A+B)\cos(A+B). In this case, A=4xA=4x and B=2xB=2x. Therefore, f(x)=cos⁡(4x+2x)=cos⁡(6x)f(x) = \cos(4x+2x) = \cos(6x). The period of a function g(x)=acos⁡(kx)+dg(x) = a\cos(kx)+d is given by 2π∣k∣\frac{2\pi}{|k|}. For f(x)=cos⁡(6x)f(x) = \cos(6x), we have k=6k=6. The period is 2π6=π3\frac{2\pi}{6} = \frac{\pi}{3}.

Question 10

Which of the following is an indefinite integral of f(x)=4sin⁡(x)cos⁡(3x)f(x) = 4\sin(x)\cos(3x)?

  1. cos⁡(2x)−12cos⁡(4x)+C\cos(2x) - \frac{1}{2}\cos(4x) + C (correct answer)
  2. 12sin⁡(4x)+sin⁡(2x)+C\frac{1}{2}\sin(4x) + \sin(2x) + C
  3. 2cos⁡(2x)−2cos⁡(4x)+C2\cos(2x) - 2\cos(4x) + C
  4. −cos⁡(2x)+12cos⁡(4x)+C-\cos(2x) + \frac{1}{2}\cos(4x) + C
Explanation: First, use the product-to-sum identity sin⁡(A)cos⁡(B)=12[sin⁡(A+B)+sin⁡(A−B)]\sin(A)\cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)]. Let A=xA=x and B=3xB=3x. Then 4sin⁡(x)cos⁡(3x)=4⋅12[sin⁡(x+3x)+sin⁡(x−3x)]=2[sin⁡(4x)+sin⁡(−2x)]=2sin⁡(4x)−2sin⁡(2x)4\sin(x)\cos(3x) = 4 \cdot \frac{1}{2}[\sin(x+3x) + \sin(x-3x)] = 2[\sin(4x) + \sin(-2x)] = 2\sin(4x) - 2\sin(2x). Now, integrate this expression: ∫(2sin⁡(4x)−2sin⁡(2x))dx=2(−cos⁡(4x)4)−2(−cos⁡(2x)2)+C=−12cos⁡(4x)+cos⁡(2x)+C\int (2\sin(4x) - 2\sin(2x)) dx = 2\left(-\frac{\cos(4x)}{4}\right) - 2\left(-\frac{\cos(2x)}{2}\right) + C = -\frac{1}{2}\cos(4x) + \cos(2x) + C, which can be written as cos⁡(2x)−12cos⁡(4x)+C\cos(2x) - \frac{1}{2}\cos(4x) + C.

Question 11

Given cos⁡(x)=725\cos(x) = \frac{7}{25} and 3π2<x<2π\frac{3\pi}{2} < x < 2\pi, find the value of cos⁡(x2)\cos(\frac{x}{2}).

  1. −45-\frac{4}{5} (correct answer)
  2. −35-\frac{3}{5}
  3. 35\frac{3}{5}
  4. 45\frac{4}{5}
Explanation: Use the half-angle identity derived from cos⁡(2A)=2cos⁡2(A)−1\cos(2A) = 2\cos^2(A) - 1. Let A=x/2A = x/2, so cos⁡(x)=2cos⁡2(x2)−1\cos(x) = 2\cos^2(\frac{x}{2}) - 1. Rearranging gives cos⁡2(x2)=1+cos⁡(x)2\cos^2(\frac{x}{2}) = \frac{1+\cos(x)}{2}. Substituting the value of cos⁡(x)\cos(x): cos⁡2(x2)=1+7/252=32/252=1625\cos^2(\frac{x}{2}) = \frac{1 + 7/25}{2} = \frac{32/25}{2} = \frac{16}{25}. Thus, cos⁡(x2)=±1625=±45\cos(\frac{x}{2}) = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}. To determine the sign, consider the quadrant of x2\frac{x}{2}. Since 3π2<x<2π\frac{3\pi}{2} < x < 2\pi, dividing by 2 gives 3π4<x2<π\frac{3\pi}{4} < \frac{x}{2} < \pi. This means x2\frac{x}{2} is in Quadrant II, where cosine is negative. Therefore, cos⁡(x2)=−45\cos(\frac{x}{2}) = -\frac{4}{5}.

Question 12

Given cos⁡(x+y)=14\cos(x+y) = \frac{1}{4} and cos⁡(x−y)=34\cos(x-y) = \frac{3}{4}, what is the value of cos⁡(x)cos⁡(y)\cos(x)\cos(y)?

  1. 18\frac{1}{8}
  2. 14\frac{1}{4}
  3. 12\frac{1}{2} (correct answer)
  4. 1
Explanation: We use the product-to-sum identity cos⁡(x)cos⁡(y)=12[cos⁡(x+y)+cos⁡(x−y)]\cos(x)\cos(y) = \frac{1}{2}[\cos(x+y) + \cos(x-y)]. Substituting the given values directly into this formula: cos⁡(x)cos⁡(y)=12[14+34]=12[44]=12(1)=12\cos(x)\cos(y) = \frac{1}{2}\left[\frac{1}{4} + \frac{3}{4}\right] = \frac{1}{2}\left[\frac{4}{4}\right] = \frac{1}{2}(1) = \frac{1}{2}.

Question 13

Given that sin⁡(A)=35\sin(A) = \frac{3}{5} for π2<A<π\frac{\pi}{2} < A < \pi and cos⁡(B)=−1213\cos(B) = -\frac{12}{13} for π<B<3π2\pi < B < \frac{3\pi}{2}, find the value of cos⁡(A+B)\cos(A+B).

  1. −6365-\frac{63}{65}
  2. −3365-\frac{33}{65}
  3. 3365\frac{33}{65}
  4. 6365\frac{63}{65} (correct answer)
Explanation: First, determine the values of cos⁡(A)\cos(A) and sin⁡(B)\sin(B). Since AA is in Quadrant II, cos⁡(A)\cos(A) is negative. Using sin⁡2(A)+cos⁡2(A)=1\sin^2(A) + \cos^2(A) = 1, we have cos⁡(A)=−1−(3/5)2=−1−9/25=−16/25=−45\cos(A) = -\sqrt{1 - (3/5)^2} = -\sqrt{1 - 9/25} = -\sqrt{16/25} = -\frac{4}{5}. Since BB is in Quadrant III, sin⁡(B)\sin(B) is negative. Using sin⁡2(B)+cos⁡2(B)=1\sin^2(B) + \cos^2(B) = 1, we have sin⁡(B)=−1−(−12/13)2=−1−144/169=−25/169=−513\sin(B) = -\sqrt{1 - (-12/13)^2} = -\sqrt{1 - 144/169} = -\sqrt{25/169} = -\frac{5}{13}. Now, use the compound angle identity for cosine: cos⁡(A+B)=cos⁡(A)cos⁡(B)−sin⁡(A)sin⁡(B)\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B). Substituting the values: cos⁡(A+B)=(−45)(−1213)−(35)(−513)=4865+1565=6365\cos(A+B) = \left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(-\frac{5}{13}\right) = \frac{48}{65} + \frac{15}{65} = \frac{63}{65}.

Question 14

Simplify the expression 2sin⁡(5x)cos⁡(3x)−sin⁡(8x)2\sin(5x)\cos(3x) - \sin(8x).

  1. −sin⁡(2x)-\sin(2x)
  2. sin⁡(2x)\sin(2x) (correct answer)
  3. cos⁡(2x)\cos(2x)
  4. sin⁡(8x)+sin⁡(2x)\sin(8x) + \sin(2x)
Explanation: Use the product-to-sum identity 2sin⁡(A)cos⁡(B)=sin⁡(A+B)+sin⁡(A−B)2\sin(A)\cos(B) = \sin(A+B) + \sin(A-B). Let A=5xA=5x and B=3xB=3x. Then, 2sin⁡(5x)cos⁡(3x)=sin⁡(5x+3x)+sin⁡(5x−3x)=sin⁡(8x)+sin⁡(2x)2\sin(5x)\cos(3x) = \sin(5x+3x) + \sin(5x-3x) = \sin(8x) + \sin(2x). The original expression becomes (sin⁡(8x)+sin⁡(2x))−sin⁡(8x)=sin⁡(2x)(\sin(8x) + \sin(2x)) - \sin(8x) = \sin(2x).

Question 15

The expression sin⁡(3x)\sin(3x) is equivalent to which of the following?

  1. 3sin⁡(x)−4sin⁡3(x)3\sin(x) - 4\sin^3(x) (correct answer)
  2. 4sin⁡3(x)−3sin⁡(x)4\sin^3(x) - 3\sin(x)
  3. 4cos⁡3(x)−3cos⁡(x)4\cos^3(x) - 3\cos(x)
  4. 3sin⁡(x)−2sin⁡3(x)3\sin(x) - 2\sin^3(x)
Explanation: To derive the identity for sin⁡(3x)\sin(3x), we use the compound angle and double angle formulas: sin⁡(3x)=sin⁡(2x+x)=sin⁡(2x)cos⁡(x)+cos⁡(2x)sin⁡(x)=(2sin⁡(x)cos⁡(x))cos⁡(x)+(1−2sin⁡2(x))sin⁡(x)=2sin⁡(x)cos⁡2(x)+sin⁡(x)−2sin⁡3(x)=2sin⁡(x)(1−sin⁡2(x))+sin⁡(x)−2sin⁡3(x)=2sin⁡(x)−2sin⁡3(x)+sin⁡(x)−2sin⁡3(x)=3sin⁡(x)−4sin⁡3(x)\sin(3x) = \sin(2x+x) = \sin(2x)\cos(x) + \cos(2x)\sin(x) = (2\sin(x)\cos(x))\cos(x) + (1-2\sin^2(x))\sin(x) = 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^3(x) = 2\sin(x)(1-\sin^2(x)) + \sin(x) - 2\sin^3(x) = 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) = 3\sin(x) - 4\sin^3(x).

Question 16

Given cos⁡(x)=725\cos(x) = \frac{7}{25} and 3π2<x<2π\frac{3\pi}{2} < x < 2\pi, find the value of cos⁡(x2)\cos(\frac{x}{2}).

  1. −45-\frac{4}{5} (correct answer)
  2. −35-\frac{3}{5}
  3. 35\frac{3}{5}
  4. 45\frac{4}{5}
Explanation: Use the half-angle identity derived from cos⁡(2A)=2cos⁡2(A)−1\cos(2A) = 2\cos^2(A) - 1. Let A=x/2A = x/2, so cos⁡(x)=2cos⁡2(x2)−1\cos(x) = 2\cos^2(\frac{x}{2}) - 1. Rearranging gives cos⁡2(x2)=1+cos⁡(x)2\cos^2(\frac{x}{2}) = \frac{1+\cos(x)}{2}. Substituting the value of cos⁡(x)\cos(x): cos⁡2(x2)=1+7/252=32/252=1625\cos^2(\frac{x}{2}) = \frac{1 + 7/25}{2} = \frac{32/25}{2} = \frac{16}{25}. Thus, cos⁡(x2)=±1625=±45\cos(\frac{x}{2}) = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}. To determine the sign, consider the quadrant of x2\frac{x}{2}. Since 3π2<x<2π\frac{3\pi}{2} < x < 2\pi, dividing by 2 gives 3π4<x2<π\frac{3\pi}{4} < \frac{x}{2} < \pi. This means x2\frac{x}{2} is in Quadrant II, where cosine is negative. Therefore, cos⁡(x2)=−45\cos(\frac{x}{2}) = -\frac{4}{5}.

Question 17

What is the maximum value of the function f(x)=sin⁡(x)cos⁡(x)+cos⁡2(x)−sin⁡2(x)f(x) = \sin(x)\cos(x) + \cos^2(x) - \sin^2(x)?

  1. 1
  2. 52\frac{\sqrt{5}}{2} (correct answer)
  3. 32\frac{3}{2}
  4. 2\sqrt{2}
Explanation: The function can be rewritten using double angle identities. We know that sin⁡(x)cos⁡(x)=12sin⁡(2x)\sin(x)\cos(x) = \frac{1}{2}\sin(2x) and cos⁡2(x)−sin⁡2(x)=cos⁡(2x)\cos^2(x) - \sin^2(x) = \cos(2x). So, f(x)=12sin⁡(2x)+cos⁡(2x)f(x) = \frac{1}{2}\sin(2x) + \cos(2x). This is in the form asin⁡(θ)+bcos⁡(θ)a\sin(\theta) + b\cos(\theta) where θ=2x\theta = 2x, a=1/2a = 1/2, and b=1b = 1. The maximum value of this expression is R=a2+b2R = \sqrt{a^2 + b^2}. In this case, R=(12)2+12=14+1=54=52R = \sqrt{(\frac{1}{2})^2 + 1^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}.

Question 18

Given cos⁡(x+y)=14\cos(x+y) = \frac{1}{4} and cos⁡(x−y)=34\cos(x-y) = \frac{3}{4}, what is the value of cos⁡(x)cos⁡(y)\cos(x)\cos(y)?

  1. 18\frac{1}{8}
  2. 14\frac{1}{4}
  3. 12\frac{1}{2} (correct answer)
  4. 1
Explanation: We use the product-to-sum identity cos⁡(x)cos⁡(y)=12[cos⁡(x+y)+cos⁡(x−y)]\cos(x)\cos(y) = \frac{1}{2}[\cos(x+y) + \cos(x-y)]. Substituting the given values directly into this formula: cos⁡(x)cos⁡(y)=12[14+34]=12[44]=12(1)=12\cos(x)\cos(y) = \frac{1}{2}\left[\frac{1}{4} + \frac{3}{4}\right] = \frac{1}{2}\left[\frac{4}{4}\right] = \frac{1}{2}(1) = \frac{1}{2}.

Question 19

Which of the following is an indefinite integral of f(x)=4sin⁡(x)cos⁡(3x)f(x) = 4\sin(x)\cos(3x)?

  1. cos⁡(2x)−12cos⁡(4x)+C\cos(2x) - \frac{1}{2}\cos(4x) + C (correct answer)
  2. 12sin⁡(4x)+sin⁡(2x)+C\frac{1}{2}\sin(4x) + \sin(2x) + C
  3. 2cos⁡(2x)−2cos⁡(4x)+C2\cos(2x) - 2\cos(4x) + C
  4. −cos⁡(2x)+12cos⁡(4x)+C-\cos(2x) + \frac{1}{2}\cos(4x) + C
Explanation: First, use the product-to-sum identity sin⁡(A)cos⁡(B)=12[sin⁡(A+B)+sin⁡(A−B)]\sin(A)\cos(B) = \frac{1}{2}[\sin(A+B) + \sin(A-B)]. Let A=xA=x and B=3xB=3x. Then 4sin⁡(x)cos⁡(3x)=4⋅12[sin⁡(x+3x)+sin⁡(x−3x)]=2[sin⁡(4x)+sin⁡(−2x)]=2sin⁡(4x)−2sin⁡(2x)4\sin(x)\cos(3x) = 4 \cdot \frac{1}{2}[\sin(x+3x) + \sin(x-3x)] = 2[\sin(4x) + \sin(-2x)] = 2\sin(4x) - 2\sin(2x). Now, integrate this expression: ∫(2sin⁡(4x)−2sin⁡(2x))dx=2(−cos⁡(4x)4)−2(−cos⁡(2x)2)+C=−12cos⁡(4x)+cos⁡(2x)+C\int (2\sin(4x) - 2\sin(2x)) dx = 2\left(-\frac{\cos(4x)}{4}\right) - 2\left(-\frac{\cos(2x)}{2}\right) + C = -\frac{1}{2}\cos(4x) + \cos(2x) + C, which can be written as cos⁡(2x)−12cos⁡(4x)+C\cos(2x) - \frac{1}{2}\cos(4x) + C.

Question 20

Which expression is equivalent to sin⁡(x)+sin⁡(5x)cos⁡(x)+cos⁡(5x)\frac{\sin(x) + \sin(5x)}{\cos(x) + \cos(5x)}?

  1. tan⁡(2x)\tan(2x)
  2. tan⁡(3x)\tan(3x) (correct answer)
  3. cot⁡(2x)\cot(2x)
  4. cot⁡(3x)\cot(3x)
Explanation: Apply the sum-to-product identities. Numerator: sin⁡(x)+sin⁡(5x)=2sin⁡(x+5x2)cos⁡(x−5x2)=2sin⁡(3x)cos⁡(−2x)=2sin⁡(3x)cos⁡(2x)\sin(x) + \sin(5x) = 2\sin\left(\frac{x+5x}{2}\right)\cos\left(\frac{x-5x}{2}\right) = 2\sin(3x)\cos(-2x) = 2\sin(3x)\cos(2x). Denominator: cos⁡(x)+cos⁡(5x)=2cos⁡(x+5x2)cos⁡(x−5x2)=2cos⁡(3x)cos⁡(−2x)=2cos⁡(3x)cos⁡(2x)\cos(x) + \cos(5x) = 2\cos\left(\frac{x+5x}{2}\right)\cos\left(\frac{x-5x}{2}\right) = 2\cos(3x)\cos(-2x) = 2\cos(3x)\cos(2x). The fraction becomes 2sin⁡(3x)cos⁡(2x)2cos⁡(3x)cos⁡(2x)=sin⁡(3x)cos⁡(3x)=tan⁡(3x)\frac{2\sin(3x)\cos(2x)}{2\cos(3x)\cos(2x)} = \frac{\sin(3x)}{\cos(3x)} = \tan(3x).