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AP Precalculus -3.12 Equivalent Trigonometric Forms- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.12 Equivalent Trigonometric Forms- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.12 Equivalent Trigonometric Forms- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

The function \( f \) is defined by \( f(\theta) = \cos(2\theta) \). Which is an equivalent expression for \( f(\theta) \)?
(A) 1
(B) \( 1 – 2\cos^2\theta \)
(C) \( 1 – 2\sin^2\theta \)
(D) \( 2\cos\theta\sin\theta \)
▶️ Answer/Explanation
Detailed solution

Using the double-angle identity:
\( \cos(2\theta) = \cos^2\theta – \sin^2\theta \).
Replace \( \cos^2\theta = 1 – \sin^2\theta \):
\( \cos(2\theta) = (1 – \sin^2\theta) – \sin^2\theta = 1 – 2\sin^2\theta \).
✅ Answer: (C)

Question

On the unit circle, first angle \( \theta \) gives point \((a, b)\). Second angle \( 2\theta \) gives \((c, d)\). Express \( c, d \) in terms of \( a, b \).
(A) \( c = -b, d = -a \)
(B) \( c = 2a, d = 2b \)
(C) \( c = a^2 + b^2, d = 2ab \)
(D) \( c = a^2 – b^2, d = 2ab \)
▶️ Answer/Explanation
Detailed solution

On unit circle: \( a = \cos\theta, b = \sin\theta \).
Then \( c = \cos(2\theta) = \cos^2\theta – \sin^2\theta = a^2 – b^2 \).
\( d = \sin(2\theta) = 2\sin\theta\cos\theta = 2ab \).
✅ Answer: (D)

Question

The function \( f(x) = \cos x (1 + \tan^2 x) \). Which expression is equivalent to \( f(x) \)?
(A) \( \sec x \)
(B) \( \cos^3 x \)
(C) \( \tan x \sec x \)
(D) \( \cot x \csc x \)
▶️ Answer/Explanation
Detailed solution

Using \( 1 + \tan^2 x = \sec^2 x \):
\( f(x) = \cos x \cdot \sec^2 x = \cos x \cdot \frac{1}{\cos^2 x} = \frac{1}{\cos x} = \sec x \).
✅ Answer: (A)

Question 

 
 
 
 
 
 
 
 
 
 
Right triangle: hypotenuse = 1, side opposite angle \( \frac{5\pi}{12} \) has length \( x \). Given \( \frac{5\pi}{12} = \frac{3\pi}{12} + \frac{2\pi}{12} \), find \( x \).
(A) \( \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{2}}{2} \right) + \left( \frac{1}{2} \right) \left( \frac{\sqrt{3}}{2} \right) \)
(B) \( \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) \)
(C) \( \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) – \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) \)
(D) \( \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) – \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) \)
▶️ Answer/Explanation
Detailed solution

\( x = \sin\frac{5\pi}{12} \).
Let \( \frac{3\pi}{12} = \frac{\pi}{4} \), \( \frac{2\pi}{12} = \frac{\pi}{6} \).
Using sum identity: \( \sin\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \sin\frac{\pi}{4}\cos\frac{\pi}{6} + \cos\frac{\pi}{4}\sin\frac{\pi}{6} \).
\( \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \), \( \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}, \sin\frac{\pi}{6} = \frac{1}{2} \).
Thus \( x = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) \).
✅ Answer: (B)

Question 

The identity \( \sin(x + \frac{\pi}{2}) = \cos x \) shows cosine is a horizontal translation of sine. Which identity verifies this directly?
(A) The sum identity for sine
(B) The sum identity for cosine
(C) The double-angle identity for sine
(D) The double-angle identity for cosine
▶️ Answer/Explanation
Detailed solution

The sum identity for sine: \( \sin(A + B) = \sin A \cos B + \cos A \sin B \).
Let \( A = x, B = \frac{\pi}{2} \):
\( \sin(x + \frac{\pi}{2}) = \sin x \cos\frac{\pi}{2} + \cos x \sin\frac{\pi}{2} \).
Since \( \cos\frac{\pi}{2} = 0 \) and \( \sin\frac{\pi}{2} = 1 \), we get \( \cos x \).
✅ Answer: (A)

Question 

The function \( g(x) = 7\sin(2x) \). Which is an equivalent form?
(A) \( 14\cos x \sin x \)
(B) \( (7\cos x)(7\sin x) \)
(C) \( 7\cos^2 x – 7\sin^2 x \)
(D) \( 7 – 14\sin^2 x \)
▶️ Answer/Explanation
Detailed solution

Using the double-angle identity: \( \sin(2x) = 2\sin x \cos x \).
Thus \( g(x) = 7 \cdot 2\sin x \cos x = 14\sin x \cos x \).
✅ Answer: (A)

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