AP Precalculus Trigonometric and Polar Functions — Worked Answer Explanations

Unit 3 · 12 questions explained

1. Question 1 · Easy

   What is the period of $f(\theta )=\sin (\theta )$?

   • A
     $\pi /2$

     Why not A: That is the period of $\sin (2\theta )/2$ or a similar transformed function.
   • B
     $\pi$

     Why not B: That is the period of $\tan (\theta )$, not $\sin$.
   • C
     $2\pi$Correct
   • D
     $4\pi$

     Why not D: Doubled the standard period; no scaling has been applied.
   Explanation

   The standard sine function repeats every $2\pi$ radians: $\sin (\theta +2\pi )=\sin (\theta )$ for all $\theta$. So the period is $2\pi$.

   Key takeaway

   Standard $\sin$ and $\cos$ both have period $2\pi$; $\tan$ has period $\pi$.

2. Question 2 · Easy

   What is the exact value of $\cos \left(\frac{\pi }{3}\right)$?

   • A
     $\frac{1}{2}$Correct
   • B
     $\frac{\sqrt{2}}{2}$

     Why not B: That is $\cos (\pi /4)$, not $\cos (\pi /3)$.
   • C
     $\frac{\sqrt{3}}{2}$

     Why not C: That is $\sin (\pi /3)$ or $\cos (\pi /6)$ — wrong reference angle output.
   • D
     $0$

     Why not D: $\cos (\theta )=0$ at $\theta =\pi /2$, not $\pi /3$.
   Explanation

   From the unit circle, the point at angle $\pi /3$ (i.e., $60°$) has coordinates $(1/2,\sqrt{3}/2)$. The $x$-coordinate is the cosine: $\cos (\pi /3)=1/2$.

   Key takeaway

   Unit-circle reference values: $\cos(\pi/6) = \sqrt{3}/2$, $\cos(\pi/4) = \sqrt{2}/2$, $\cos(\pi/3) = 1/2$.

3. Question 3 · Easy

   The amplitude of $f(\theta )=3\sin (2\theta )$ is which of the following?

   • A
     $1$

     Why not A: That is the amplitude of the parent $\sin (\theta )$ with no vertical stretch.
   • B
     $2$

     Why not B: Confused the period coefficient with the amplitude.
   • C
     $3$Correct
   • D
     $6$

     Why not D: Multiplied the amplitude coefficient by the period coefficient.
   Explanation

   In $f(\theta )=A\sin (B\theta )$, the amplitude is $|A|$. Here $A=3$, so the amplitude is $3$. The factor of $2$ affects the period, not the amplitude.

   Key takeaway

   Amplitude = $|A|$ in $A\sin(B\theta + C) + D$; $B$ controls period, $C$ phase shift, $D$ vertical shift.

4. Question 4 · Easy

   What is the period of $f(\theta )=\cos (4\theta )$?

   • A
     $\frac{\pi }{2}$Correct
   • B
     $\pi$

     Why not B: Used $\pi /2$ but flipped the formula.
   • C
     $2\pi$

     Why not C: Period of the parent $\cos$; ignored the coefficient of $\theta$.
   • D
     $8\pi$

     Why not D: Multiplied $2\pi$ by $4$ — should have divided.
   Explanation

   For $\cos (B\theta )$, the period is $\frac{2\pi }{|B|}$. With $B=4$, period $=\frac{2\pi }{4}=\frac{\pi }{2}$.

   Key takeaway

   Period of $\sin(B\theta)$ or $\cos(B\theta)$ is $\dfrac{2\pi}{|B|}$ — larger $|B|$ means a more compressed wave.

5. Question 5 · Medium

   If $\sin (\theta )=\frac{3}{5}$ and $\theta$ is in Quadrant II, what is $\cos (\theta )$?

   • A
     $\frac{4}{5}$

     Why not A: Magnitude is correct but cosine is negative in Quadrant II.
   • B
     $-\frac{4}{5}$Correct
   • C
     $\frac{3}{5}$

     Why not C: Repeated $\sin$ value; cosine and sine generally differ.
   • D
     $-\frac{3}{5}$

     Why not D: Used $\sin$ magnitude with a sign flip — wrong identity.
   Explanation

   Pythagorean identity: ${\sin }^2\theta +{\cos }^2\theta =1$. So ${\cos }^2\theta =1-9/25=16/25$, giving $\cos \theta =\pm 4/5$. In Quadrant II, cosine is negative, so $\cos \theta =-4/5$.

   Key takeaway

   Use the Pythagorean identity to find the magnitude; let the quadrant fix the sign.

6. Question 6 · Medium

   Find all solutions to $2\sin (\theta )=1$ on the interval $[0,2\pi )$.

   • A
     $\theta =\pi /6$ only

     Why not A: Missed the second solution in Quadrant II where sine is also positive.
   • B
     $\theta =\pi /6$ and $\theta =5\pi /6$Correct
   • C
     $\theta =\pi /3$ and $\theta =2\pi /3$

     Why not C: Confused $\sin (\theta )=1/2$ with $\sin (\theta )=\sqrt{3}/2$.
   • D
     $\theta =\pi /6$ and $\theta =7\pi /6$

     Why not D: Took both Quadrant I and Quadrant III solutions, but Quadrant III has negative sine.
   Explanation

   Solve $\sin (\theta )=1/2$. The reference angle is $\pi /6$. Sine is positive in Quadrants I and II, giving $\theta =\pi /6$ and $\theta =\pi -\pi /6=5\pi /6$.

   Key takeaway

   When solving $\sin\theta = k$, find the reference angle, then include every quadrant where $\sin$ has the right sign.

7. Question 7 · Medium

   What is the range of $g(\theta )=-2\sin (\theta )+1$?

   • A
     $[-1,1]$

     Why not A: Range of parent $\sin$; ignored the amplitude and vertical shift.
   • B
     $[-1,3]$Correct
   • C
     $[-3,1]$

     Why not C: Forgot that the negative amplitude flips min and max around the midline.
   • D
     $[1,3]$

     Why not D: Took only the upper half; missed the negative excursion below the midline.
   Explanation

   $\sin (\theta )$ ranges in $[-1,1]$. Multiplying by $-2$: $-2\sin (\theta )\in [-2,2]$. Add $1$: $-2\sin (\theta )+1\in [-1,3]$.

   Key takeaway

   For $A\sin(\theta) + D$, the range is $[D - |A|, D + |A|]$.

8. Question 8 · Medium

   Convert the rectangular point $(x,y)=(1,\sqrt{3})$ to polar form $(r,\theta )$ with $r>0$ and $\theta \in [0,2\pi )$.

   • A
     $(2,\pi /6)$

     Why not A: Used $\tan \theta =1/\sqrt{3}$ instead of $\sqrt{3}/1$ — flipped the ratio.
   • B
     $(2,\pi /3)$Correct
   • C
     $(4,\pi /3)$

     Why not C: Computed $r^2=4$ but forgot to take the square root.
   • D
     $(2,2\pi /3)$

     Why not D: Confused with Quadrant II — but the point $(1,\sqrt{3})$ is in Quadrant I.
   Explanation

   $r=\sqrt{x^2+y^2}=\sqrt{1+3}=2$. For $\theta$, $\tan \theta =y/x=\sqrt{3}/1=\sqrt{3}$. The point is in Quadrant I, so $\theta =\pi /3$.

   Key takeaway

   Polar conversion: $r = \sqrt{x^2 + y^2}$, $\tan\theta = y/x$. Choose $\theta$ based on the actual quadrant of $(x, y)$.

9. Question 9 · Medium

   What is the graph of the polar equation $r=4$?

   • A
     A horizontal line $y=4$.

     Why not A: A horizontal line in polar form has equation like $r=c/\sin \theta$, not $r=4$.
   • B
     A vertical line $x=4$.

     Why not B: Same idea — that would be $r\cos \theta =4$.
   • C
     A circle of radius $4$ centered at the origin.Correct
   • D
     A spiral.

     Why not D: Spirals require $r$ to depend on $\theta$, e.g. $r=\theta$.
   Explanation

   $r=4$ means every point is exactly $4$ units from the origin — that's the definition of a circle of radius $4$ centered at the origin.

   Key takeaway

   $r = c$ is a circle of radius $|c|$ about the origin; $\theta = c$ is a line through the origin at angle $c$.

10. Question 10 · Hard

    A Ferris wheel of radius $10$ m has its center $12$ m above the ground and completes one revolution every $30$ seconds. Starting from the bottom at $t=0$, which function models the rider's height $h(t)$ in meters?

    • A
      $h(t)=10\sin \left(\frac{2\pi }{30}t\right)+12$

      Why not A: Starts at the midline ($12$) at $t=0$ rather than at the bottom ($2$).
    • B
      $h(t)=-10\cos \left(\frac{2\pi }{30}t\right)+12$Correct
    • C
      $h(t)=10\cos \left(\frac{2\pi }{30}t\right)+12$

      Why not C: Starts at the top ($22$) at $t=0$ instead of the bottom — wrong sign on amplitude.
    • D
      $h(t)=10\cos \left(\frac{\pi }{30}t\right)+12$

      Why not D: Period would be $60$ seconds — twice the given period.
    Explanation

    Amplitude is the radius: $10$. Midline is the center height: $12$. Period $30$ gives angular frequency $\frac{2\pi }{30}$. At $t=0$ the rider is at the bottom ($h=2$), so we need $-10\cos (0)+12=-10+12=2$. ✓

    Key takeaway

    Sinusoidal modeling: amplitude = radius, midline = center, period = revolution time. Use $-\cos$ to start at the minimum.

11. Question 11 · Hard

    Simplify $\frac{\sin (2\theta )}{\sin (\theta )}$ for $\sin (\theta )\ne 0$.

    • A
      $2$

      Why not A: Cancelled $\sin (\theta )$ without first applying the double-angle identity correctly.
    • B
      $\cos (\theta )$

      Why not B: Off by a factor of $2$ — forgot the coefficient in $\sin (2\theta )=2\sin \theta \cos \theta$.
    • C
      $2\cos (\theta )$Correct
    • D
      $\sin (\theta )$

      Why not D: Did not use the double-angle identity.
    Explanation

    Double-angle identity: $\sin (2\theta )=2\sin (\theta )\cos (\theta )$. Therefore $\frac{\sin (2\theta )}{\sin (\theta )}=\frac{2\sin \theta \cos \theta }{\sin \theta }=2\cos \theta$.

    Key takeaway

    Double-angle identity: $\sin(2\theta) = 2\sin\theta\cos\theta$.

12. Question 12 · Hard

    Find all $\theta \in [0,2\pi )$ that satisfy $2{\cos }^2(\theta )-\cos (\theta )-1=0$.

    • A
      $\theta =0,\pi /2,\pi$

      Why not A: Solved $\cos \theta =1$ correctly but solved $\cos \theta =-1/2$ as $\cos \theta =0$.
    • B
      $\theta =0,2\pi /3,4\pi /3$Correct
    • C
      $\theta =\pi /3,5\pi /3$

      Why not C: Used $\cos \theta =1/2$ — sign error when factoring.
    • D
      $\theta =\pi$, $2\pi /3$, $4\pi /3$

      Why not D: Took $\cos \theta =-1$ instead of $\cos \theta =1$.
    Explanation

    Let $u=\cos \theta$. Then $2u^2-u-1=0$. Factor: $(2u+1)(u-1)=0$, so $u=-1/2$ or $u=1$. Solve: $\cos \theta =1$ gives $\theta =0$. $\cos \theta =-1/2$ gives $\theta =2\pi /3$ or $\theta =4\pi /3$.

    Key takeaway

    Trig equations that are quadratic in $\cos\theta$ (or $\sin\theta$): substitute $u$, factor, then solve each piece on the given interval.