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AP Precalculus -3.4 Sine and Cosine Graphs- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.4 Sine and Cosine Graphs- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.4 Sine and Cosine Graphs- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

 
 
 
 
 
 
 
 
 
 
 
 
 
The figure gives the graphs of four functions labeled \( A \), \( B \), \( C \), and \( D \) in the \( xy \)-plane. Which is the graph of \( f(x) = 2 \cos^{-1} x \)?
(A) \( A \)
(B) \( B \)
(C) \( C \)
(D) \( D \)
▶️ Answer/Explanation
Detailed solution

The function \( f(x) = 2\cos^{-1}x \) is a vertical stretch of \( y = \cos^{-1}x \) by factor 2.
– Domain: \( [-1, 1] \)
– Range: \( [0, 2\pi] \) after stretch (since \( \cos^{-1}x \) ranges from \( 0 \) to \( \pi \), multiplying by 2 gives \( 0 \) to \( 2\pi \))
Graph \( C \) matches this vertical dilation.
✅ Answer: (C)

Question 

Which of the following is the graph of \( f(x) = 2 \csc\left(\frac{2\pi}{3}x\right) \) in the \( xy \)-plane?

▶️ Answer/Explanation
Detailed solution

\( f(x) = 2\csc\left(\frac{2\pi}{3}x\right) \)
– Cosecant is reciprocal of sine: \( \csc u = \frac{1}{\sin u} \)
– Vertical asymptotes where \( \sin\left(\frac{2\pi}{3}x\right) = 0 \) → \( \frac{2\pi}{3}x = k\pi \) → \( x = \frac{3k}{2} = 1.5k \)
– Period: \( \frac{2\pi}{2\pi/3} = 3 \)
– Amplitude factor 2 outside → range: \( (-\infty, -2] \cup [2, \infty) \)
Graph (D) matches these features: asymptotes every 1.5 units, period 3, branches outside \( [-2, 2] \).
✅ Answer: (D)

Question 

Which of the following is the graph of \( f(x) = \cot x \) in the \( xy \)-plane?

▶️ Answer/Explanation
Detailed solution

\( f(x) = \cot x = \frac{\cos x}{\sin x} \)
– Vertical asymptotes where \( \sin x = 0 \) → \( x = k\pi \)
– Decreasing on each interval \( (k\pi, (k+1)\pi) \)
– Passes through \( \left(\frac{\pi}{2} + k\pi, 0\right) \)
Graph (B) matches this behavior.
✅ Answer: (B)

Question 

The function \( g \) is given by \( g(\theta) = \cos \theta \). Which of the following describes \( g \) on the interval from \( \theta = \frac{3\pi}{2} \) to \( \theta = 2\pi \)?
(A) \( g \) is decreasing, and the graph of \( g \) is concave down.
(B) \( g \) is decreasing, and the graph of \( g \) is concave up.
(C) \( g \) is increasing, and the graph of \( g \) is concave down.
(D) \( g \) is increasing, and the graph of \( g \) is concave up.
▶️ Answer/Explanation
Detailed solution

On \( \left( \frac{3\pi}{2}, 2\pi \right) \):
– Cosine increases from \( 0 \) to \( 1 \).
– The derivative \( g'(\theta) = -\sin\theta \) is positive (since sine is negative in Quadrant IV), so \( g \) is increasing.
– The second derivative \( g”(\theta) = -\cos\theta \) is negative (since cosine is positive in Quadrant IV), so concave down.
Thus: increasing and concave down.
✅ Answer: (C)

Question 

 
 
 
 
 
 
 
 
 
The figure shows the graph of the function \( f \) in the \( xy \)-plane. The function \( f \) is either a sine function or a cosine function. Consider an angle in standard position in another \( xy \)-plane (not shown). The terminal ray of the angle intersects the unit circle at point \( P \). Which of the following is true about the relationship between \( f \) and \( P \)?
(A) \( f \) gives the vertical distance of \( P \) from the \( x \)-axis for angle measures from 0 to \( 2\pi \).
(B) \( f \) gives the horizontal distance of \( P \) from the \( y \)-axis for angle measures from 0 to \( 2\pi \).
(C) \( f \) gives the vertical displacement of \( P \) from the \( x \)-axis for angle measures from 0 to \( 2\pi \).
(D) \( f \) gives the horizontal displacement of \( P \) from the \( y \)-axis for angle measures from 0 to \( 2\pi \).
▶️ Answer/Explanation
Detailed solution

The graph begins at the origin, increases to a maximum at \( \frac{\pi}{2} \), returns to zero at \( \pi \), goes to a minimum at \( \frac{3\pi}{2} \), and returns to zero at \( 2\pi \). This is the graph of \( f(x) = \sin x \).
On the unit circle, for an angle \( \theta \), the sine function gives the \( y \)-coordinate of \( P \), which is the vertical displacement (signed distance) from the \( x \)-axis.
Thus \( f(\theta) = \sin\theta \) gives the vertical displacement of \( P \) from the \( x \)-axis.
✅ Answer: (C)

Question 

The function \( f \) is given by \( f(\theta) = \cos \theta \), and the function \( g \) is given by \( g(\theta) = \sin \theta \). Which of the following describes \( f \) and \( g \) on the interval from \( \theta = \frac{\pi}{2} \) to \( \theta = \pi \)?
(A) Both \( f \) and \( g \) are decreasing.
(B) Both \( f \) and \( g \) are increasing.
(C) \( f \) is decreasing, and \( g \) is increasing.
(D) \( f \) is increasing, and \( g \) is decreasing.
▶️ Answer/Explanation
Detailed solution

On \( \left( \frac{\pi}{2}, \pi \right) \):
– Cosine decreases from 0 to -1.
– Sine decreases from 1 to 0.
Thus both are decreasing.
✅ Answer: (A)

Question 

The function \( f \) is given by \( f(\theta) = \sin \theta \). Which of the following describes \( f \) on the interval from \( \theta = \pi \) to \( \theta = \frac{3\pi}{2} \)?
(A) \( f \) is decreasing, and the graph of \( f \) is concave down.
(B) \( f \) is decreasing, and the graph of \( f \) is concave up.
(C) \( f \) is increasing, and the graph of \( f \) is concave down.
(D) \( f \) is increasing, and the graph of \( f \) is concave up.
▶️ Answer/Explanation
Detailed solution

On \( \left( \pi, \frac{3\pi}{2} \right) \):
– Sine decreases from 0 to -1, so \( f \) is decreasing.
– Derivative \( f'(\theta) = \cos\theta \) is negative in Quadrant III, but becoming less negative (since cosine increases from -1 to 0), so slope is increasing → concave up.
– Second derivative \( f”(\theta) = -\sin\theta \): sine is negative in Quadrant III, so \( -\sin\theta > 0 \) → concave up.
Thus: decreasing and concave up.
✅ Answer: (B)

Question 

The functions \( f \) and \( g \) are given by \( f(x) = \cos x \) and \( g(x) = \sin x \). In the \( xy \)-plane, for how much of the interval \( 0 \leq x \leq 8\pi \) are the graphs of \( f \) and \( g \) both concave up?
(A) \( 0 \) units of \( x \)
(B) \( \frac{\pi}{2} \) units of \( x \)
(C) \( \pi \) units of \( x \)
(D) \( 2\pi \) units of \( x \)
▶️ Answer/Explanation
Detailed solution

Concavity:
– \( f(x)=\cos x \), \( f”(x)=-\cos x \) → concave up when \( -\cos x >0 \) → \( \cos x <0 \).
– \( g(x)=\sin x \), \( g”(x)=-\sin x \) → concave up when \( -\sin x >0 \) → \( \sin x <0 \).
Both concave up when \( \cos x <0 \) and \( \sin x <0 \) → Quadrant III: \( \pi < x < \frac{3\pi}{2} \) per period.
Length per period = \( \frac{\pi}{2} \).
Interval length \( 8\pi \) contains \( \frac{8\pi}{2\pi}=4 \) periods.
Total = \( 4 \times \frac{\pi}{2} = 2\pi \).
✅ Answer: (D)

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