AP Precalculus Unit 3 Practice Test: Trigonometric & Polar Functions

AP Precalculus Unit 3 Practice Test

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Question 1

The expression $\text{cosec } ⁡A + \cot⁡A $ is equal to,

A
$\dfrac{1 + \cos ⁡A}{\sin ⁡A}$
B
$\dfrac{1 + \tan⁡ A}{1 - \cot⁡ A}$
C
$\dfrac{1}{1 + \tan ⁡A}$
D
$\dfrac{\sin ⁡A}{1 + \cos ⁡A}$
Question 1 Explanation: 
The correct answer is (A).

$\text{cosec}⁡ \ A + \cot⁡ A $ $ =\dfrac{1}{\sin⁡ A} + \dfrac{\cos⁡ A}{\sin ⁡A}$

$=\dfrac{1 + \cos⁡ A}{\sin ⁡A}$
Question 2

The cartesian coordinates $(x,y)$ corresponding to the polar coordinates $(4,\frac{3π}{4})$ are,

A
$(2 \sqrt{2},2 \sqrt{2})$
B
$(-\sqrt{2}, 2 \sqrt{2})$
C
$(-2 \sqrt{2}, 2 \sqrt{2})$
D
$(-2, 2 \sqrt{2})$
Question 2 Explanation: 
The correct answer is (C).

$x=r \cos⁡ θ \ $ and $ \ y=r \ \sin⁡ θ$

Thus, for $(r,θ)=\left(4,\dfrac{3π}{4}\right),$

$x=4 \cos⁡\dfrac{3π}{4} \ $ and $ \ y=4 \sin⁡ \dfrac{3π}{4}$

$x=4×\left(-\dfrac{1}{\sqrt{2}}\right) \ $ and $ \ y=4×\left(\dfrac{1}{\sqrt{2}}\right)$

$x=-2 \sqrt{2} \ $ and $ \ y=2 \sqrt{2}$
Question 3

For some angle $θ$, $\sin⁡ θ = \frac{8}{17}$. If the terminal side is in the 2nd quadrant, what is the value of $\tan⁡ θ + \cot ⁡θ$?

A
$-\dfrac{298}{120}$
B
$-\dfrac{289}{120}$
C
$-\dfrac{256}{120}$
D
$-\dfrac{267}{120}$
Question 3 Explanation: 
The correct answer is (B).

Using $\sin^2 ⁡θ+\cos^2 ⁡θ=1$

$\cos^2⁡ θ=1-\dfrac{64}{289}=\dfrac{225}{289} $

$\cos⁡ θ=-\dfrac{15}{17}$ (as $θ$ is between $\frac{π}{2}$ and $π$)

$\tan⁡ θ+\cot ⁡θ $

$=\dfrac{\sin⁡ θ}{\cos⁡ θ} +\dfrac{\cos⁡ θ}{\sin⁡ θ}$

$=\dfrac{1}{\sin⁡θ \cos⁡θ}$

$=1÷\left(\dfrac{8}{17}×-\dfrac{15}{17}\right)$

$=-\dfrac{289}{120}$
Question 4

What is the amplitude and the period of the below trigonometric function?

A
$\text{Amplitude} = 8 \ \text{ and }$ $\text{Period} = 4$
B
$\text{Amplitude} = 8 \ \text{ and }$ $\text{Period} = 8$
C
$\text{Amplitude} = 4 \ \text{ and }$ $\text{Period} = 6$
D
$\text{Amplitude} = 4 \ \text{ and }$ $\text{Period} = 4$
Question 4 Explanation: 
The correct answer is (D).

Distance from mean value to max value $=4-0=4 \text{ units}$

Amplitude $=4 \text{ units}$

The wave completes 1 oscillation in $6-2=4 \text{ units}$

Period $=4 \text{ units}$
Question 5

If $\tan ⁡α = y$ where $α$ is an angle between $\dfrac{5π}{4}$ and $\dfrac{3π}{2}$, then what is $\cos⁡ α + \sin⁡ α$ equal to?

A
$\dfrac{(1+y)}{\sqrt{1-y^2}}$
B
$\dfrac{(1-y)}{\sqrt{1+y^2}}$
C
$ - \dfrac{(1-y)}{\sqrt{1+y^2}}$
D
$ - \dfrac{(1+y)}{\sqrt{1+y^2}}$
Question 5 Explanation: 
The correct answer is (D).

$α$ is in the 3rd Quadrant

$→\cos⁡ α$ and $\sin⁡ α$ are negative

$1+\tan^2⁡ α=sec^2 ⁡α$

So, $\sec⁡ α=-\sqrt{1+y^2 } $ $ →\cos⁡ α=- \dfrac{1}{\sqrt{1+y^2}}$

And, $\sin ⁡α=-\sqrt{1-( \dfrac{1}{1+y^2 )} } $ $ =-\dfrac{y}{\sqrt{1+y^2}}$

$\sin⁡ α + \cos α=- \dfrac{1+y}{\sqrt{1+y^2}}$
Question 6

For some angle $x, \sin ⁡x = -\dfrac{\sqrt{3}}{2}$ and $\sec x = 2$. Which option shows the size of angle $x$?

A
$\dfrac{11π}{6}$
B
$\dfrac{4π}{3}$
C
$\dfrac{5π}{3}$
D
$\dfrac{2π}{3}$
Question 6 Explanation: 
The correct answer is (C).

$\sin⁡ x$ is $-ive$ and $\cos⁡ x$ is $+ive$

Thus, angle $x$ is in the 4th Quadrant

Using $\sin ⁡x=-\dfrac{\sqrt{3}}{2} \ $ and $ \ \cos ⁡x=\dfrac{1}{2}$ on a unit circle,

$→x$ must be equal to $\dfrac{5π}{3}$
Question 7

$\tan \left(\dfrac{πx}{4}\right)$ is plotted on a graph paper for $x∈ \left(-\dfrac{5}{3},\dfrac{5}{3}\right)$. Which option shows the correct graph for the function?

A
B
C
D
Question 7 Explanation: 
The correct answer is (A).

$\left(\dfrac{πx}{4}\right)$ is an odd function and is positive in the 1st Quadrant.

So, options (B), (C) and (D) are ruled out.
Question 8

Two functions, $f(θ)$ and $g(θ)$, are defined as $\sqrt{2} \sin ⁡θ$ and $\sqrt{6} \cos ⁡θ$, respectively. If $ π < θ < 2π$, find $θ$ for which $f(θ)=g(θ)$.

A
$\dfrac{5π}{4}$
B
$\dfrac{4π}{3}$
C
$\dfrac{5π}{6}$
D
$\dfrac{6π}{5}$
Question 8 Explanation: 
The correct answer is (B).

$\sqrt{2} \sin ⁡θ=√6 \cos ⁡θ$

$ → \tan ⁡θ=\sqrt{3}$

Using a unit circle,

$θ=\dfrac{π}{3}$ or $\dfrac{4π}{3}$

For $π < θ < 2π$, answer should be $\dfrac{4π}{3}$.
Question 9

The polar coordinates $(r,θ)$ corresponding to the cartesian coordinates $(1,-\sqrt{3})$ are,

A
$\left(2, \dfrac{π}{3}\right)$
B
$\left(2, \dfrac{2π}{3}\right)$
C
$\left(2, \dfrac{5π}{3}\right)$
D
$\left(2, \dfrac{4π}{3}\right)$
Question 9 Explanation: 
The correct answer is (C).

$r=\sqrt{x^2+y^2} $ $ = \sqrt{1+3} =2$

$\cos⁡θ= \dfrac{x}{r}=\dfrac{1}{2}$

$\sin⁡θ=\dfrac{y}{r}=-\dfrac{\sqrt{3}}{2} $

Using a unit circle,

$θ = \dfrac{5π}{3}$

Polar coordinates $=\left(2, \dfrac{5π}{3}\right)$
Question 10

Look at the graph of $g(x)$.

If the function $f(x)=\dfrac{x}{2}$ is drawn, for how many values of $x$ is $f(x)=g(x)$?

A
$4$
B
$1$
C
$2$
D
$6$
Question 10 Explanation: 
The correct answer is (A).

Plotting $\dfrac{x}{2}$ on the graph gives,



The number of points of intersection of the line and the graph is 4.

Thus, $f(x)$ and $g(x)$ are equal 4 times.
Question 11

What is the value of $\sin⁡\left(\cos^{-1}⁡\dfrac{1}{2}\right) + \sin ⁡\left(\cos^{-1}⁡\dfrac{\sqrt{3}}{2}\right)$?

You may use a calculator.

A
$1.40$
B
$1.37$
C
$1.20$
D
$1.11$
Question 11 Explanation: 
The correct answer is (B).

$\sin⁡\left(\cos^{-1} \dfrac{1}{2}\right)+\sin\left(\cos^{-1}⁡\dfrac{\sqrt{3}}{2}\right)$

$=\sin⁡ \dfrac{π}{3}+\sin⁡ \dfrac{π}{6}$

$=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}$

$=0.87+0.5$

$=1.37$
Question 12

Which of the following options shows the range of $2(1 − 2 \cos ⁡x)$?

A
$[0,8]$
B
$[-2,4]$
C
$[-2,6]$
D
$[0,6]$
Question 12 Explanation: 
The correct answer is (C).

$-1 ≤ - \cos⁡x≤1$

$-2 ≤ -2 \cos ⁡x≤2$

$-1 ≤ 1-2 \cos⁡ x≤3$

$-2 ≤ 2(1-2 \cos ⁡x )≤6$

Thus, the range of the function is $[-2,6]$.
Question 13

Which of the below graphs shows the polar function $r = 1 + \cos ⁡θ \ ?$

A
B
C
D
Question 13 Explanation: 
The correct answer is (D).

$r=1+\cos⁡ θ$

Check values of $r$ for multiple values of $θ$ :

$r(0)=1+1=2 $

$r(π)=1-1=0$

$\left[\text{Similarly, for } \dfrac{π}{2} \text{ and } \dfrac{3π}{2} \right]$

Which implies option (D) is true.
Question 14

A polar function $r=1-\sin⁡ θ$ is plotted for $\dfrac{3π}{2} < θ < 2π$. Which of the below statements is true about the graph as we move from $\dfrac{3π}{2}$ to $2π$?

A
The points get closer to the origin
B
The points are below the $x$-axis and are at equal distance from the origin
C
The points are above the $x$-axis and keep moving towards the origin
D
The points get closer to the origin and then move away
Question 14 Explanation: 
The correct answer is (A).

For $r=1-\sin ⁡θ,$

$θ$ $\dfrac{3π}{2}$ $\dfrac{5π}{3}$ $\dfrac{7π}{4}$ $\dfrac{11π}{6}$ $2π$
$r(θ)$ $2$ $1.87$ $1.71$ $1.5$ $1$
Thus, the points are below the $x$-axis and get closer to the origin.
Question 15

Which function could represent the below graph?

A
$r=1-2θ$
B
$r=1-θ$
C
$r=2+θ$
D
$r=1+θ$
Question 15 Explanation: 
The correct answer is (D).

Option (A):
$θ=0→ r=1$ & $θ=π →r=1-2π $ $ =-5.28$
(Incorrect function)

Option (B):
$θ=0→ r=1$ & $θ=π →r=1-π $ $ =-2.14$
(Incorrect function)

Option (C):
$θ=0→ r=2$
(Incorrect function)

Option (D):
$θ=0→ r=1 $ & $ θ=π →r=1+π $ $ =4.14$
(Correct function)
Question 16

For what values of $θ$ is $\cos⁡ θ < -\dfrac{1}{2}$ and $\sin⁡θ > - \dfrac{1}{\sqrt{2}} \ ?$
(Note: $π < θ < 2π$)

A
$π < θ < \dfrac{7π}{6}$
B
$\dfrac{5π}{4} < θ < \dfrac{3π}{2}$
C
$π < θ < \dfrac{5π}{4}$
D
$π < θ < \dfrac{3π}{2}$
Question 16 Explanation: 
The correct answer is (C).

Using a unit circle,

$\sin⁡θ>-\dfrac{1}{\sqrt{2}} → θ < \dfrac{5π}{4} \ $ OR $ \ θ > \dfrac{7π}{4}$

$\cos ⁡θ< - \dfrac{1}{2} → θ < \dfrac{4π}{3} \ $ AND $ \ θ > \dfrac{2π}{3}$

Thus, in the interval $π < θ < 2π,$

$π < θ < \dfrac{5π}{4} $
Question 17

If $g(y)=\sin⁡(2y+\dfrac{π}{12})$, then which of the below domains guarantees the existence of an inverse?

A
$-\dfrac{13π}{24} < y < \dfrac{11π}{24}$
B
$-\dfrac{13π}{12} < y < \dfrac{11π}{12}$
C
$0 < y < 2π$
D
$-\dfrac{π}{2} < y < \dfrac{3π}{4}$
Question 17 Explanation: 
The correct answer is (A). Inverse exists only for one period.

$2y + \dfrac{π}{12} ∈ (-π,π)$

$-π < 2y + \dfrac{π}{12} < π$

$-\dfrac{13π}{12} < 2y < \dfrac{11π}{12}$

$-\dfrac{13π}{24} < y < \dfrac{11π}{24}$
Question 18

The temperature of a city during winters can be modeled using a function of the form $f(t)=300-100 \sin \left(\dfrac{πt}{45}\right)$, where $t$ is a specific winter day (typically between 1 and 90).

What is the trend in the city temperature on the 22nd day?

You may use a calculator.

A
The temperature is falling down at an increasing rate
B
The temperature is falling down at a decreasing rate
C
The temperature is rising up at a decreasing rate
D
The temperature is rising up at an increasing rate
Question 18 Explanation: 
The correct answer is (B). A rough sketch of the temperature function for $t$ from 0 to 90 is:



As seen from the sketch,

For $t$ = 22, the temperature is falling down at a decreasing rate.
Question 19

A rose curve is defined by the equation $r = \cos⁡(2θ)$. Which option shows the graph of the function in the interval $(0,2π)$?

A
B
C
D
Question 19 Explanation: 
The correct answer is (D). Check values of $\cos⁡(2θ)$ at different points on the graph:

For $θ=0, r(0)=1$

For $θ=\dfrac{π}{2}, r(\dfrac{π}{2})=-1$ (Which means $r=1$ in the opposite direction)

→ Option (A) and (B) are ruled out

For $θ=\dfrac{π}{4}, r(\dfrac{π}{4})=0$

→ Option (C) is ruled out
Question 20

Which of the below expressions is equal to $\dfrac{\sin ⁡x}{1 - \cos ⁡x} \ ?$

A
$\dfrac{1 + \sin⁡ x}{\cos ⁡x} $
B
$\dfrac{1 + \cos ⁡x}{\sin⁡ x} $
C
$\dfrac{1 - \sin⁡ x}{\cos⁡ x} $
D
$\dfrac{1 + 2 \sin⁡ x}{\cos ⁡x} $
Question 20 Explanation: 
The correct answer is (B).

$\sin^2⁡ x=1-\cos^2 ⁡x$

$\sin^2⁡ x $ $ =(1-\cos⁡ x )(1+\cos⁡ x)$

Divide by $\sin ⁡x (1- \cos⁡ x)$ on both sides,

$\dfrac{\sin⁡ x}{1 - \cos⁡ x} = \dfrac{1 + \cos⁡ x}{\sin⁡ x} $
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