Trigonometry Placement Test Practice with Solutions

The following trigonometry multiple-choice questions guide you in assessing your skills in angles (radians/degrees), coterminal angles, trigonometric ratios, identities, graphs (zeros, max, min, period, amplitude), polar coordinates, sine and cosine laws, and equations.

Question 1

The radian measure of \( \theta = - 220^{\circ} \) is:

  1. \( -220 \pi \)
  2. \( - \frac{220}{ \pi} \)
  3. \( \frac{11}{9} \pi \)
  4. \( - \frac{11}{9} \pi \)
  5. \( \frac{11}{9} \)
Show Detailed Solution

\( 180^{\circ} \) corresponds to \( \pi \). Hence: \( -220^{\circ} \) corresponds to: \( \quad -220 \dfrac{\pi}{180} = - \dfrac{11\pi}{9} \). Answer: D

Question 2

The degree measure of \( \theta = - \dfrac{11 \pi}{4} \) is:

  1. \( -1980^{\circ} \)
  2. \( - 495^{\circ} \)
  3. \( 1980^{\circ} \)
  4. \( -45^{\circ} \)
  5. \( 45^{\circ} \)
Show Detailed Solution

\( \pi \) corresponds to \( 180^{\circ} \). Hence: \( - \dfrac{11 \pi}{4} \) corresponds to: \( \quad - \dfrac{11 \pi}{4} \times \dfrac{180}{\pi} = -495^{\circ} \). Answer: B

Question 3

Which of the following angles is coterminal with the angle \( \theta = - \dfrac{11 \pi}{5} \)?

  1. \( - \frac{4 \pi }{5} \)
  2. \( \frac{9 \pi }{5} \)
  3. \( -\frac{6 \pi }{5} \)
  4. \( \frac{\pi}{5} \)
  5. \( \frac{19 \pi }{5} \)
Show Detailed Solution

Coterminal angles differ by integer multiples of \(2\pi\).

\[ \theta_c = - \frac{11\pi}{5} + k(2\pi) \]

Let \( k = 2 \):

\[ \theta_c = - \frac{11\pi}{5} + 4\pi = - \frac{11\pi}{5} + \frac{20\pi}{5} = \frac{9\pi}{5} \]

Answer: B

Question 4

If \( \theta\) is an acute angle such that \( \cos \theta = 1/4 \), then the exact value of \( \tan \theta \) is:

  1. \( \sqrt {17} \)
  2. \( \frac{1}{\sqrt {17}} \)
  3. \( 4 \)
  4. \( \frac{1}{\sqrt {15}} \)
  5. \( \sqrt {15} \)
Show Detailed Solution

Using a right triangle with \( \cos \theta = \frac{1}{4} \): adjacent = 1, hypotenuse = 4. Opposite = \( \sqrt{4^2 - 1^2} = \sqrt{15} \). \( \tan \theta = \frac{\text{opp}}{\text{adj}} = \sqrt{15} \). Answer: E

Question 5

What is \( \sin \alpha \) in the figure below?

right triangle
  1. \( 3/5 \)
  2. \( 4/5 \)
  3. \( 4 a/5 \)
  4. \( 3 a/5 \)
  5. \( 4/5 a \)
Show Detailed Solution

Pythagorean theorem: hypotenuse \( h = \sqrt{(4a)^2+(3a)^2} = \sqrt{25 a^2} = 5 a \). \( \sin \alpha = \frac{4 a}{5 a} = \frac{4}{5} \). Answer: B

Question 6

Write the height \( h \) in terms of \( \alpha \), \( \beta \), and the distance \( \bar {CD} = x \).

double right triangle
  1. \( h = \dfrac{x}{ \cot \beta - \cot \alpha} \)
  2. \( h = \dfrac{x}{ \cot ( \beta + \alpha) } \)
  3. \( h = \dfrac{x}{ \tan \beta - \tan \alpha} \)
  4. \( h = \dfrac{x}{ \cot \beta} \)
  5. \( h = \dfrac{x}{ \cot \alpha - \cot \beta} \)
Show Detailed Solution

\( \cot \alpha = \frac{\bar{BC}}{h} \), \( \cot \beta = \frac{\bar{BC}+x}{h} \). Solving for \( h \): \( h = \dfrac{x}{ \cot \beta - \cot \alpha} \). Answer: A

Question 7

Find the exact value of \( \tan 75^{\circ} \).

  1. \( \frac{\sqrt{3}+3}{3} \)
  2. \( \frac{\sqrt{3}}{3}+1 \)
  3. \( \frac{3-\sqrt{3}}{3} \)
  4. \( 2+\sqrt{3} \)
  5. \( \frac{1}{2+\sqrt{3}} \)
Show Detailed Solution

\( \tan(30^{\circ} + 45^{\circ}) = \frac{\tan 30 + \tan 45}{1 - \tan 30 \tan 45} = \frac{\sqrt{3}/3 + 1}{1 - \sqrt{3}/3} = 2+\sqrt{3} \). Answer: D

Question 8

Find \( y \) such that \( \sin \alpha = 1/5 \) and \( \alpha \) is an angle in standard position whose terminal side passes through \( (-2, y) \).

  1. \( \frac{\sqrt{6}}{6} \)
  2. \( -\frac{\sqrt{6}}{6} \)
  3. \( \frac{5}{2} \)
  4. \( -\frac{1}{2} \)
  5. \( -\frac{2}{5} \)
Show Detailed Solution

Since the terminal side passes through \( (-2,y) \),

\[ r = \sqrt{(-2)^2 + y^2} = \sqrt{4+y^2} \]

Using the definition of sine:

\[ \sin \alpha = \frac{y}{r} \]

Since \( \sin \alpha = \frac15 \),

\[ \frac{y}{\sqrt{4+y^2}} = \frac15 \]

Square both sides:

\[ \frac{y^2}{4+y^2} = \frac1{25} \]

\[ 25y^2 = 4 + y^2 \]

\[ 24y^2 = 4 \]

\[ y^2 = \frac16 \]

\[ y = \pm \frac1{\sqrt6} \]

Since the point \((-2,y)\) lies in Quadrant II and \( \sin \alpha >0 \), we must have \( y>0 \).

\[ y = \frac1{\sqrt6} = \frac{\sqrt6}{6} \]

Answer: A

Question 9

Find the exact value of \( \cot \alpha \) if \( \sin \alpha = - 11/16 \) and \( \sec \alpha > 0 \).

  1. \( \frac{3\sqrt{15}}{11} \)
  2. \( -\frac{3\sqrt{15}}{11} \)
  3. \( -\frac{11\sqrt{15}}{45} \)
  4. \( \frac{11\sqrt{15}}{45} \)
  5. \( -1/2 \)
Show Detailed Solution

\( \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \frac{3\sqrt{15}}{16} \). \( \cot \alpha = \frac{\cos \alpha}{\sin \alpha} = \frac{3\sqrt{15}/16}{-11/16} = -\frac{3\sqrt{15}}{11} \). Answer: B

Question 10

Angle \( \beta \) passes through \( (2,k) \), \( k \neq 0 \). Which is true?

  1. \( \tan \beta < 0 \)
  2. \( \sin \beta > 0 \)
  3. \( \cos \beta = \sqrt{1-\sin^2 \beta} \)
  4. \( \sec \beta < 0 \)
  5. \( \sin \beta + \cos \beta = 1 \)
Show Detailed Solution

Since \( x=2 > 0 \), \( \cos \beta > 0 \). The identity \( \cos^2 \beta + \sin^2 \beta = 1 \) implies \( \cos \beta = \sqrt{1-\sin^2 \beta} \) since the cosine is positive. Answer: C

Question 11

If \( \cos \theta = 1/5 \), find the exact value of \( \cos (\theta + 3\pi) \).

  1. \( -1/5 \)
  2. \( 1/5 \)
  3. \( - \frac{2\sqrt{6}}{5} \)
  4. \( 1 \)
  5. \( \frac{2\sqrt{6}}{5} \)
Show Detailed Solution

\( \cos (\theta + 3\pi) = -\cos \theta = -1/5 \). Answer: A

Question 12

What is the range of \( f(x) = - \dfrac{1}{3} \sin(2x-\pi/4) - 3 \)?

  1. \( [ -1/3 , 1/3 ] \)
  2. \( [ -3 , -3 ] \)
  3. \( [ -10/3 , -8/3 ] \)
  4. \( [ -1 , 1 ] \)
  5. \( [ -3 , 0 ] \)
Show Detailed Solution

Sine ranges \([-1, 1]\). Multiplying by \(-1/3\) gives \([-1/3, 1/3]\). Shifting by \(-3\) gives \([-10/3, -8/3]\). Answer: C

Question 13

Find zeros of \( f(x) = 2 \cos (2x - \pi/3) + 1 \) in \( [0,\pi) \).

  1. \( \{ \frac{2\pi}{3}, \frac{4\pi}{3} \} \)
  2. \( \{ \frac{\pi}{2} \} \)
  3. \( \{ \frac{4\pi}{3} \} \)
  4. \( \{ \frac{5\pi}{6} \} \)
  5. \( \{ \frac{\pi}{2}, \frac{5\pi}{6} \} \)
Show Detailed Solution

\( \cos(2x - \pi/3) = -1/2 \implies 2x - \pi/3 = 2\pi/3 \text{ or } 4\pi/3 \implies x = \pi/2, 5\pi/6 \). Answer: E

Question 14

Values of \( x \) in \( [0,2\pi) \) where \( f(x) = - \sin (x + \pi/6) + 1 \) has a maximum.

  1. \( 3\pi/2 \)
  2. \( \pi/2 \)
  3. \( 5\pi/3 \)
  4. \( 4\pi/3 \)
  5. \( 2\pi/3 \)
Show Detailed Solution

\( -\sin(x+\pi/6) + 1 = 2 \implies \sin(x+\pi/6) = -1 \implies x+\pi/6 = 3\pi/2 \implies x = 4\pi/3 \). Answer: D

Question 15

Period of \( f(x) = - \cos (-\frac{x}{4} - 2 \pi) + 10 \).

  1. \( 8\pi \)
  2. \( 4\pi \)
  3. \( 2\pi \)
  4. \( \pi/4 \)
  5. \( - 8\pi \)
Show Detailed Solution

Period formula: \( 2\pi / |B| \). \( B = 1/4 \), Period = \( 8\pi \). Answer: A

Question 16

Amplitude of \( f(x) = - 2 \cos (-5x - \pi) + 5 \)?

  1. 7
  2. 2
  3. 3
  4. 1
  5. -2
Show Detailed Solution

Amplitude = \( |A| = |-2| = 2 \). Answer: B

Question 17

Equation of \( f(x) = A \sin(B x + C) + D \), period \( \pi/2 \), amplitude 2, max 3, max at \( x = \pi/4 \).

  1. \( 2 \sin((\pi/2) x - \pi/2 ) \)
  2. \( 2 \sin(4 x + \pi/2 ) + 1 \)
  3. \( - 2 \sin(4 x + \pi/2 ) + 3 \)
  4. \( 2 \sin(4 x - \pi/2 ) + 1 \)
  5. \( 2 \sin(4 x + \pi/2 ) + 3 \)
Show Detailed Solution

B=4, A=2, D=1. Max at \( \pi/4 \implies C = -\pi/2 \). Result: \( 2 \sin(4x - \pi/2) + 1 \). Answer: D

Question 18

Equation of \( f(x) = A \cos(B x + C) \) based on graph?

  1. \( 2.5 \cos(3 x) \)
  2. \( 2.5 \cos(x + \pi/2) \)
  3. \( 2.5 \cos(3 x + \pi/2) \)
  4. \( 2.5 \cos(3 \pi x + \pi/2) \)
  5. \( 2.5 \cos(3 x + \pi) \)
Show Detailed Solution

Amplitude = 2.5, Period = \( 2\pi/3 \), so B=3. Matches \( 2.5 \cos(3x + \pi/2) \). Answer: C

Question 19

\( \frac{\cos x - \sin x}{1 - \tan x} \) equals:

  1. \( \frac{\cos x - \sin x}{1 + \tan x} \)
  2. \( \frac{\cos x + \sin x}{1 - \tan x} \)
  3. \( \cos x \)
  4. \( \cos x - \sin x \)
  5. \( \frac{\cos x + \sin x}{1 + \tan x} \)
Show Detailed Solution

Replace \( \tan x \) with \( \dfrac{\sin x}{\cos x} \):

\[ \frac{\cos x - \sin x}{1-\tan x} = \frac{\cos x - \sin x} {1-\frac{\sin x}{\cos x}} \]

Simplify the denominator:

\[ 1-\frac{\sin x}{\cos x} = \frac{\cos x-\sin x}{\cos x} \]

Therefore,

\[ \frac{\cos x-\sin x} {\frac{\cos x-\sin x}{\cos x}} = \cos x \]

Answer: C

Question 20

Find \( \cos (\alpha - \beta) \) given \( \alpha \in Q2, \beta \in Q3, \sin \alpha = 9/15, \tan \beta = 28/45 \).

  1. \( 265/276 \)
  2. \( -84/265 \)
  3. \( 84/265 \)
  4. \( 36/53 \)
  5. \( 85/267 \)
Show Detailed Solution

Calculations: \( \cos \alpha = -4/5 \). \( \sin \beta = -28/53, \cos \beta = -45/53 \). \( \cos(\alpha - \beta) = (-4/5)(-45/53) + (3/5)(-28/53) = 84/265 \). Answer: C