AP Precalculus Unit 3 Practice Test

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Question 1 |

### The expression $\text{cosec } A + \cotA $ is equal to,

$\dfrac{1 + \cos A}{\sin A}$ | |

$\dfrac{1 + \tan A}{1 - \cot A}$ | |

$\dfrac{1}{1 + \tan A}$ | |

$\dfrac{\sin A}{1 + \cos A}$ |

Question 1 Explanation:

The correct answer is (A).

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

$=\dfrac{1 + \cos A}{\sin A}$

$\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,

$(2 \sqrt{2},2 \sqrt{2})$ | |

$(-\sqrt{2}, 2 \sqrt{2})$ | |

$(-2 \sqrt{2}, 2 \sqrt{2})$ | |

$(-2, 2 \sqrt{2})$ |

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}$

$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 2^{nd} quadrant, what is the value of $\tan θ + \cot θ$?

$-\dfrac{298}{120}$ | |

$-\dfrac{289}{120}$ | |

$-\dfrac{256}{120}$ | |

$-\dfrac{267}{120}$ |

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 $

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

$=\dfrac{1}{\sinθ \cosθ}$

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

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

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 $

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

$=\dfrac{1}{\sinθ \cosθ}$

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

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

Question 4 |

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

$\text{Amplitude} = 8 \ \text{ and }$ $\text{Period} = 4$ | |

$\text{Amplitude} = 8 \ \text{ and }$ $\text{Period} = 8$ | |

$\text{Amplitude} = 4 \ \text{ and }$ $\text{Period} = 6$ | |

$\text{Amplitude} = 4 \ \text{ and }$ $\text{Period} = 4$ |

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 $=4 \text{ units}$

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?

$\dfrac{(1+y)}{\sqrt{1-y^2}}$ | |

$\dfrac{(1-y)}{\sqrt{1+y^2}}$ | |

$ - \dfrac{(1-y)}{\sqrt{1+y^2}}$ | |

$ - \dfrac{(1+y)}{\sqrt{1+y^2}}$ |

Question 5 Explanation:

The correct answer is (D).

$α$ is in the 3

$→\cos α$ 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}}$

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

$α$ is in the 3

^{rd}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 $x$?

$\dfrac{11π}{6}$ | |

$\dfrac{4π}{3}$ | |

$\dfrac{5π}{3}$ | |

$\dfrac{2π}{3}$ |

Question 6 Explanation:

The correct answer is (C).

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

Thus, angle $x$ is in the 4

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

$→x$ must be equal to $\dfrac{5π}{3}$

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

Thus, angle $x$ is in the 4

^{th}QuadrantUsing $\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?

Question 7 Explanation:

The correct answer is (A).

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

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

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

^{st}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 $f(θ)=g(θ)$.

$\dfrac{5π}{4}$ | |

$\dfrac{4π}{3}$ | |

$\dfrac{5π}{6}$ | |

$\dfrac{6π}{5}$ |

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 $\dfrac{4π}{3}$.

$\sqrt{2} \sin θ=√6 \cos θ$

$ → \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,

$\left(2, \dfrac{π}{3}\right)$ | |

$\left(2, \dfrac{2π}{3}\right)$ | |

$\left(2, \dfrac{5π}{3}\right)$ | |

$\left(2, \dfrac{4π}{3}\right)$ |

Question 9 Explanation:

The correct answer is (C).

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

$\cosθ= \dfrac{x}{r}=\dfrac{1}{2}$

$\sinθ=\dfrac{y}{r}=-\dfrac{\sqrt{3}}{2} $

Using a unit circle,

$θ = \dfrac{5π}{3}$

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

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

$\cosθ= \dfrac{x}{r}=\dfrac{1}{2}$

$\sinθ=\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 $f(x)=g(x)$?

$4$ | |

$1$ | |

$2$ | |

$6$ |

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.

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.

$1.40$ | |

$1.37$ | |

$1.20$ | |

$1.11$ |

Question 11 Explanation:

The correct answer is (B).

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

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

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

$=0.87+0.5$

$=1.37$

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

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

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

$=0.87+0.5$

$=1.37$

Question 12 |

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

$[0,8]$ | |

$[-2,4]$ | |

$[-2,6]$ | |

$[0,6]$ |

Question 12 Explanation:

The correct answer is (C).

$-1 ≤ - \cosx≤1$

$-2 ≤ -2 \cos x≤2$

$-1 ≤ 1-2 \cos x≤3$

$-2 ≤ 2(1-2 \cos x )≤6$

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

$-1 ≤ - \cosx≤1$

$-2 ≤ -2 \cos x≤2$

$-1 ≤ 1-2 \cos x≤3$

$-2 ≤ 2(1-2 \cos x )≤6$

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

Question 13 |

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

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.

$r=1+\cos θ$

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 $2π$?

The points get closer to the origin | |

The points are below the $x$-axis and are at equal distance from the origin | |

The points are above the $x$-axis and keep moving towards the origin | |

The points get closer to the origin and then move away |

Question 14 Explanation:

The correct answer is (A).

For $r=1-\sin θ,$

Thus, the points are below the $x$-axis and get closer to the origin.

For $r=1-\sin θ,$

$θ$ | $\dfrac{3π}{2}$ | $\dfrac{5π}{3}$ | $\dfrac{7π}{4}$ | $\dfrac{11π}{6}$ | $2π$ |

$r(θ)$ | $2$ | $1.87$ | $1.71$ | $1.5$ | $1$ |

Question 15 |

### Which function could represent the below graph?

$r=1-2θ$ | |

$r=1-θ$ | |

$r=2+θ$ | |

$r=1+θ$ |

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)

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 $\sinθ > - \dfrac{1}{\sqrt{2}} \ ?$

(Note: $π < θ < 2π$)

$π < θ < \dfrac{7π}{6}$ | |

$\dfrac{5π}{4} < θ < \dfrac{3π}{2}$ | |

$π < θ < \dfrac{5π}{4}$ | |

$π < θ < \dfrac{3π}{2}$ |

Question 16 Explanation:

The correct answer is (C).

Using a unit circle,

$\sinθ>-\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} $

Using a unit circle,

$\sinθ>-\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?

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

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

$0 < y < 2π$ | |

$-\dfrac{π}{2} < y < \dfrac{3π}{4}$ |

Question 17 Explanation:

The correct answer is (A). Inverse exists only for one period.

$2y + \dfrac{π}{12} ∈ (-π,π)$

$-π < 2y + \dfrac{π}{12} < π$

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

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

$2y + \dfrac{π}{12} ∈ (-π,π)$

$-π < 2y + \dfrac{π}{12} < π$

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

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

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 22^{nd} day?

### You may use a calculator.

The temperature is falling down at an increasing rate | |

The temperature is falling down at a decreasing rate | |

The temperature is rising up at a decreasing rate | |

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.

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 $(0,2π)$?

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

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 $\dfrac{\sin x}{1 - \cos x} \ ?$

$\dfrac{1 + \sin x}{\cos x} $ | |

$\dfrac{1 + \cos x}{\sin x} $ | |

$\dfrac{1 - \sin x}{\cos x} $ | |

$\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} $

$\sin^2 x=1-\cos^2 x$

$\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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