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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)

Question 

Where both expressions are defined, which of the following is equivalent to \(\frac{\sec^{2}x-1}{\sec^{2}x}\)?
(A) \(\frac{\tan^{2}x}{\sin^{2}x}\)
(B) \(\frac{\tan^{2}x}{\cos^{2}x}\)
(C) \(\sin^{2}x\)
(D) \(\cos^{2}x\)
▶️ Answer/Explanation
Detailed solution

1. Use Pythagorean Identity:
\(\sec^2 x – 1 = \tan^2 x\).

2. Simplify Fraction:
\(\frac{\tan^2 x}{\sec^2 x} = \tan^2 x \cdot \cos^2 x\)
\(= \frac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x = \sin^2 x\)

✅ Answer: (C)

Question 

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

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

Question 

Where both expressions are defined, which of the following is equivalent to \(\frac{\sec^2 x – 1}{\sec^2 x}\)?
(A) \(\frac{\tan^2 x}{\sin^2 x}\)
(B) \(\frac{\tan^2 x}{\cos^2 x}\)
(C) \(\cos^2 x\)
(D) \(\sin^2 x\)
▶️ Answer/Explanation
Detailed solution

The correct answer is (D).

Step 1: Identify the Pythagorean identity in the numerator. We know that \(\tan^2 x + 1 = \sec^2 x\), therefore \(\sec^2 x – 1 = \tan^2 x\).
Step 2: Substitute this into the expression: \(\frac{\tan^2 x}{\sec^2 x}\).
Step 3: Rewrite in terms of sine and cosine: \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\) and \(\sec^2 x = \frac{1}{\cos^2 x}\).
Step 4: Divide the fractions: \(\frac{\sin^2 x}{\cos^2 x} \div \frac{1}{\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x\).
Step 5: Cancel the \(\cos^2 x\) terms to get the final result: \(\sin^2 x\).

Question 

The function of $h$ is given by $h(x) = \sin \left(\beta + \frac{\pi}{3}\right)$, where $0 < \beta < \frac{\pi}{2}$. Which of the following is an equivalent form for $h(x)$?
(A) $\frac{1}{2} \sin \beta + \frac{\sqrt{3}}{3} \cos \beta$
(B) $\frac{1}{2} \sin \beta – \frac{\sqrt{3}}{2} \cos \beta$
(C) $\frac{\sqrt{3}}{2} \sin \beta + \frac{1}{2} \cos \beta$
(D) $\sin \beta + \frac{\sqrt{3}}{2}$
▶️ Answer/Explanation
Detailed solution
The problem asks for the expansion of the expression $h(x) = \sin \left(\beta + \frac{\pi}{3}\right)$. This requires the sine addition formula:
$\sin(A + B) = \sin A \cos B + \cos A \sin B$
Here, we substitute $A = \beta$ and $B = \frac{\pi}{3}$:
$h(x) = \sin \beta \cos \left(\frac{\pi}{3}\right) + \cos \beta \sin \left(\frac{\pi}{3}\right)$
Next, substitute the exact trigonometric values for the angle $\frac{\pi}{3}$ ($60^\circ$):
$\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
Replacing these values back into the equation:
$h(x) = \sin \beta \left(\frac{1}{2}\right) + \cos \beta \left(\frac{\sqrt{3}}{2}\right)$
Rearranging the terms gives:
$h(x) = \frac{1}{2} \sin \beta + \frac{\sqrt{3}}{2} \cos \beta$
Comparing this result with the options, it corresponds to the form in Option (A). (Note: The option in the image shows $\frac{\sqrt{3}}{3}$, which is likely a typographical error for $\frac{\sqrt{3}}{2}$).
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