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AP Precalculus -3.3 Sine and Cosine Values- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.3 Sine and Cosine Values- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.3 Sine and Cosine Values- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question 

In the \( xy \)-plane, the terminal ray of angle \( \theta \) in standard position intersects a circle of radius \( r \) at the point \( \left( 10, -10\sqrt{3} \right) \). What are the values of \( \theta \) and \( r \)?
(A) \( \theta = \frac{5\pi}{3} \) and \( r = 10 \)
(B) \( \theta = \frac{5\pi}{3} \) and \( r = 20 \)
(C) \( \theta = \frac{11\pi}{6} \) and \( r = 10 \)
(D) \( \theta = \frac{11\pi}{6} \) and \( r = 20 \)
▶️ Answer/Explanation
Detailed solution

We have \( x = 10, y = -10\sqrt{3} \).
The ratio \( \frac{y}{x} = -\sqrt{3} \), so \( \tan\theta = -\sqrt{3} \).
In Quadrant IV, \( \theta = \frac{5\pi}{3} \) (since \(\tan\frac{5\pi}{3} = -\sqrt{3}\)).
Check: \( \cos\frac{5\pi}{3} = \frac12 \), \( \sin\frac{5\pi}{3} = -\frac{\sqrt{3}}{2} \).
Now, \( r\cos\theta = r \cdot \frac12 = 10 \) → \( r = 20 \).
Verify: \( r\sin\theta = 20 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -10\sqrt{3} \), 
Thus \( \theta = \frac{5\pi}{3} \) and \( r = 20 \).
✅ Answer: (B)

Question 

 
 
 
 
 
 
 
 
 
 
The figure shows the terminal ray of angle \(\theta\), in standard position, intersecting the unit circle at point \(P\) in the \(xy\)-plane. The function \(g\) is given by \(g(z) = \cos z\). For the angle \(\omega\) (not shown), \(\theta < \omega < \pi\). Which of the following is true?
(A) \(g(\omega) < g(\theta)\)
(B) \(g(\omega) > g(\theta)\)
(C) \(g(\omega) = g(\theta)\)
(D) Depending on the value of \(\omega\), sometimes \(g(\omega) < g(\theta)\) and sometimes \(g(\omega) > g(\theta)\).
▶️ Answer/Explanation
Detailed solution

On the unit circle, \( \cos z \) gives the \( x \)-coordinate.
From the diagram, \( \theta \) is in Quadrant I (\( 0 < \theta < \frac{\pi}{2} \)) since \( P \) is in Quadrant I.
If \( \theta < \omega < \pi \), then \( \omega \) is in Quadrant II.
Cosine decreases from 1 to -1 as the angle goes from 0 to \( \pi \), so for \( \omega > \theta \), \( \cos\omega < \cos\theta \).
Thus \( g(\omega) < g(\theta) \).
✅ Answer: (A)

Question 

 
 
 
 
 
 
 
 
 
 
 
The figure shows the terminal ray of angle \(\theta\), in standard position, intersecting the unit circle at point \(P\) in the \(xy\)-plane. The function \(f\) is given by \(f(z) = \sin z\). For the angle \(\beta\) (not shown), \(\frac{\pi}{2} < \beta < \theta\). Which of the following is true?
(A) \(f(\beta) < f(\theta)\)
(B) \(f(\beta) > f(\theta)\)
(C) \(f(\beta) = f(\theta)\)
(D) Depending on the value of \(\beta\), sometimes \(f(\beta) < f(\theta)\) and sometimes \(f(\beta) > f(\theta)\).
▶️ Answer/Explanation
Detailed solution

From the diagram, \( \theta \) is in Quadrant I and \( \theta > \frac{\pi}{2} \) ? Actually given \( \frac{\pi}{2} < \beta < \theta \), so \( \theta \) is > \( \frac{\pi}{2} \) as well, but since \( \theta \) is in Quadrant II from the figure (based on answer key reasoning). In Quadrant II, sine decreases as angle increases toward \( \pi \).
If \( \frac{\pi}{2} < \beta < \theta \), then \( \beta \) is closer to \( \frac{\pi}{2} \) where sine = 1, and \( \theta \) is farther from \( \frac{\pi}{2} \) so sine is smaller. Thus \( \sin\beta > \sin\theta \), so \( f(\beta) > f(\theta) \).
✅ Answer: (B)

Question 

 
 
 
 
 
 
 
 
 
 
 
The figure shows equilateral triangle \( ABO \) with sides of length 10 in the \( xy \)-plane. Segment \( AB \) is perpendicular to the \( x \)-axis. The terminal ray of an angle \( \theta \) (not shown) in standard position passes through the point \( B \). What is the value of \( 10 \sin \theta \)?
(A) \(-5\sqrt{3}\)
(B) \(-\frac{\sqrt{3}}{2}\)
(C) \(\frac{1}{2}\)
(D) 5
▶️ Answer/Explanation
Detailed solution

Step 1: Geometry of the triangle

Triangle \(ABO\) is equilateral with side length \(10\).
Hence,
\(OA = OB = AB = 10\)
and
\(\angle AOB = 60^\circ.\)

Step 2: Orientation in the coordinate plane

Segment \(AB\) is perpendicular to the \(x\)-axis, so \(A\) and \(B\) have the same
\(x\)-coordinate and are symmetric about the \(x\)-axis.
Thus, the bisector of \(\angle AOB\) lies along the negative \(x\)-axis.

Step 3: Determine the angle \(\theta\)

Since \(\angle AOB = 60^\circ\), the ray \(OB\) makes an angle of
\(\theta = 180^\circ – 30^\circ = 150^\circ\)
with the positive \(x\)-axis.

Step 4: Compute \(10\sin\theta\)

\(\sin 150^\circ = \frac{1}{2}\)

\(10\sin\theta = 10 \cdot \frac{1}{2} = 5\)

\(\boxed{5}\)

\textbf{Correct answer: (D)}

Question 

 
 
 
 
 
 
 
 
 
 
 
 
 
A circle centered at origin with radius 5 is shown. Point \(P = (x, y)\) is on terminal ray of angle \(\theta\), point \(R = (x, -y)\). Which is true about \(\sin\theta\)?
(A) \( \sin\theta = \frac{x}{5} \), ratio of horizontal displacement of \(P\) to distance from origin
(B) \( \sin\theta = \frac{x}{5} \), ratio of horizontal displacement of \(R\) to distance from origin
(C) \( \sin\theta = -\frac{y}{5} \), ratio of vertical displacement of \(R\) to distance from origin
(D) \( \sin\theta = \frac{y}{5} \), ratio of vertical displacement of \(P\) to distance from origin
▶️ Answer/Explanation
Detailed solution

By definition on the unit circle (scaled by radius 5):
\( \sin\theta = \frac{\text{vertical displacement of }P}{ \text{distance from origin}} = \frac{y}{5} \).
✅ Answer: (D)

Question 

The function \( f(x) = \sin(x + \frac{4\pi}{3}) \). What is \( f\left(\frac{3\pi}{4}\right) \)?
(A) \( -\frac{\sqrt{6} – \sqrt{2}}{4} \)
(B) \( \frac{\sqrt{2} + \sqrt{6}}{4} \)
(C) \( \frac{\sqrt{2} – \sqrt{6}}{4} \)
(D) \( \frac{\sqrt{6} – \sqrt{2}}{4} \)
▶️ Answer/Explanation
Detailed solution

Use sum identity: \( \sin\left(\frac{3\pi}{4} + \frac{4\pi}{3}\right) = \sin\frac{3\pi}{4}\cos\frac{4\pi}{3} + \cos\frac{3\pi}{4}\sin\frac{4\pi}{3} \).
\( \sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2} \), \( \cos\frac{4\pi}{3} = -\frac{1}{2} \), \( \cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \), \( \sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2} \).
Compute: \( \frac{\sqrt{2}}{2} \cdot -\frac{1}{2} + \left(-\frac{\sqrt{2}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{6} – \sqrt{2}}{4} \).
✅ Answer: (D)

Question 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
A circle of radius 4.8 centered at origin has angle \( \alpha \) in standard position intersecting the circle at a point. Which is the value of \( \sin\alpha \)?
(A) \( \frac{2.2}{4.8} \)
(B) \( \frac{4.3}{4.8} \)
(C) 2.2
(D) 4.3
▶️ Answer/Explanation
Detailed solution

Sine is vertical displacement / radius.
From the figure (described in PDF), the vertical coordinate is 4.3.
Thus \( \sin\alpha = \frac{4.3}{4.8} \).
✅ Answer: (B)

Question 

 
 
 
 
 
 
 
 
 
 
Unit circle with point \(A(-0.8, 0.6)\) on terminal ray of angle \( \alpha \). What is \( \cos\alpha \)?
(A) \(-0.8\)
(B) \(-0.6\)
(C) \(0.6\)
(D) \(0.8\)
▶️ Answer/Explanation
Detailed solution

On unit circle, \( \cos\alpha \) = \(x\)-coordinate of intersection point = \(-0.8\).
✅ Answer: (A)

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