Math 3 Quiz: Graphing Sine And Cosine

15 questions · exam conditions

0:00

Graphing Sine And CosineQuestion 1 of 15

A sinusoidal function has amplitude 5, period $\frac{2\pi}{3}$ , and midline $y = -2$ . If the function equals its midline value when $x = 0$ and is increasing at that point, which equation represents this function?

y = 5\sin(3x) - 2

y = 5\cos(3x) - 2

y = 5\sin(\frac{3x}{2}) - 2

y = -5\cos(3x) - 2

← Back to quizzes

Math 3 Quiz

Math 3 Quiz: Graphing Sine And Cosine

Practice Graphing Sine And Cosine in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Graphing Sine And Cosine, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$y = 5\sin(3x) - 2$ (correct answer)
$y = 5\cos(3x) - 2$
$y = 5\sin(\frac{3x}{2}) - 2$
$y = -5\cos(3x) - 2$

Explanation: Period

\frac{2\pi}{3}

means

B = \frac{2\pi}{\frac{2\pi}{3}} = 3

. At

x = 0

, the function equals its midline and is increasing. For sine:

\sin(0) = 0

(midline) and sine is increasing at

x = 0

. For cosine:

\cos(0) = 1

(maximum), not midline. Choice B starts at maximum, not midline. Choice C has wrong B value (

\frac{3}{2}

instead of 3). Choice D starts at minimum due to negative amplitude.

Question 2

A sinusoidal function has the form $y = A\cos(Bx) + C$ where $A > 0$ . If the function has a period of $\frac{4\pi}{3}$ , amplitude of 6, and passes through the point $(0, 2)$ , what is the value of $C$ ?

$C = 8$
$C = 2$
$C = 6$
$C = -4$ (correct answer)

Explanation: When you encounter sinusoidal functions in the form

y = A\cos(Bx) + C

, remember that each parameter has a specific role:

A

controls amplitude,

B

affects period, and

C

represents the vertical shift (midline). Let's work through this systematically. First, we can find

A

and

B

from the given information. Since amplitude = 6 and

A > 0

, we have

A = 6

. The period formula is

\frac{2\pi}{B} = \frac{4\pi}{3}

, so solving for

B

B = \frac{2\pi}{\frac{4\pi}{3}} = \frac{3}{2}

. Now our function is

y = 6\cos\left(\frac{3}{2}x\right) + C

. To find

C

, we use the given point

(0, 2)

2 = 6\cos\left(\frac{3}{2} \cdot 0\right) + C

2 = 6\cos(0) + C

2 = 6(1) + C

2 = 6 + C

C = -4

Looking at the wrong answers: Choice A (

C = 8

) likely comes from adding the amplitude to the y-coordinate instead of recognizing the downward shift. Choice B (

C = 2

) incorrectly assumes the y-intercept equals the vertical shift. Choice C (

C = 6

) confuses the amplitude with the vertical shift parameter. The correct answer is D. Study tip: Always substitute known points into your function to find unknown parameters. Remember that

\cos(0) = 1

, making

(0, y)

points especially useful for finding the vertical shift in cosine functions.

Question 3

A cosine function is vertically stretched by a factor of 3, horizontally compressed by a factor of $\frac{1}{2}$ , and shifted down 4 units. If the original function was $y = \cos(x)$ , what is the midline of the transformed function?

$y = 0$
$y = -4$ (correct answer)
$y = 3$
$y = -1$

Explanation: The transformed function is

y = 3\cos(2x) - 4

. The vertical stretch affects amplitude (becomes 3), horizontal compression affects period (

B = 2

), and the vertical shift down 4 units moves the midline from

y = 0

y = -4

. The midline is determined only by the vertical shift. Choice A gives the original midline before transformation. Choice C confuses the amplitude with the midline. Choice D represents an incorrect partial shift.

Question 4

The graph of $y = \cos(x)$ is transformed to create the function $g(x) = 2\cos(3x - \pi) + 1$ . Which statement correctly describes the relationship between the period of $g(x)$ and the period of the parent cosine function?

The period of $g(x)$ is one-third the period of $\cos(x)$ because the coefficient 3 compresses horizontally (correct answer)
The period of $g(x)$ is three times the period of $\cos(x)$ because the coefficient 3 stretches horizontally
The period of $g(x)$ is twice the period of $\cos(x)$ because the amplitude coefficient 2 affects the period
The period of $g(x)$ equals the period of $\cos(x)$ because the phase shift $-\pi$ cancels the effect of coefficient 3

Explanation: The period of

\cos(x)

2\pi

. For

g(x) = 2\cos(3x - \pi) + 1

, the coefficient 3 multiplying

x

creates a horizontal compression, making the period

\frac{2\pi}{3}

. This is indeed one-third of the original period. The amplitude (2) and vertical shift (+1) do not affect the period, and the phase shift (

-\pi

) only translates the graph horizontally without changing the period.

Question 5

Consider the function $f(x) = A\sin(Bx)$ where $A > 0$ and $B > 0$ . If doubling the value of $A$ and halving the value of $B$ results in a new function $g(x)$ , how does the graph of $g(x)$ compare to the original graph of $f(x)$ ?

The amplitude doubles and the period doubles, resulting in a vertically and horizontally stretched graph (correct answer)
The amplitude doubles and the period halves, resulting in a vertically stretched and horizontally compressed graph
The amplitude halves and the period doubles, resulting in a vertically compressed and horizontally stretched graph
The amplitude remains the same and the period doubles, because doubling $A$ and halving $B$ have opposite effects

Explanation: Originally:

f(x) = A\sin(Bx)

has amplitude

A

and period

\frac{2\pi}{B}

. After transformation:

g(x) = (2A)\sin(\frac{B}{2}x)

has amplitude

2A

(doubled) and period

\frac{2\pi}{B/2} = \frac{4\pi}{B}

(doubled). So both amplitude and period double, creating vertical and horizontal stretching.

Question 6

A sinusoidal function has the form $y = A\cos(B(x - C)) + D$ . If this function has amplitude 4, period $3\pi$ , phase shift 2 units to the right, and midline $y = -1$ , what is the value of $A + B + C + D$ ?

$\frac{13}{3}$
$\frac{11}{3}$
$\frac{17}{3}$ (correct answer)
$\frac{19}{3}$

Explanation: From the given information: amplitude = 4, so

|A| = 4

, thus

A = 4

(assuming positive). Period =

3\pi

, so

\frac{2\pi}{B} = 3\pi

, giving

B = \frac{2}{3}

. Phase shift = 2 units right, so

C = 2

. Midline = -1, so

D = -1

. Therefore:

A + B + C + D = 4 + \frac{2}{3} + 2 + (-1) = 5 + \frac{2}{3} = \frac{15}{3} + \frac{2}{3} = \frac{17}{3}

Question 7

The function $h(x) = 3\sin(\frac{\pi x}{4}) - 2$ represents the height of a point on a rotating wheel above the ground, where $x$ is time in seconds. At what time $x$ (where $0 \leq x \leq 16$ ) does the point first reach its maximum height?

$x = 8$ seconds
$x = 4$ seconds
$x = 6$ seconds
$x = 2$ seconds (correct answer)

Explanation: When analyzing sinusoidal functions like this one, you need to understand how transformations affect the basic sine curve. The function

h(x) = 3\sin(\frac{\pi x}{4}) - 2

represents a transformed sine wave where the amplitude is 3, there's a vertical shift down by 2 units, and the period is affected by the coefficient

\frac{\pi}{4}

. To find when the sine function reaches its maximum value, recall that

\sin(\theta) = 1

when

\theta = \frac{\pi}{2} + 2\pi k

(where k is any integer). For the first maximum, we need

\frac{\pi x}{4} = \frac{\pi}{2}

. Solving this equation: multiply both sides by

\frac{4}{\pi}

to get

x = 2

. At this point,

h(2) = 3(1) - 2 = 1

, which is indeed the maximum height. Looking at the incorrect answers: Choice A (

x = 8

) occurs when

\frac{\pi x}{4} = 2\pi

, where the sine function completes one full cycle and returns to its starting value, not the maximum. Choice B (

x = 4

) corresponds to

\frac{\pi x}{4} = \pi

, where

\sin(\pi) = 0

, giving a middle value. Choice C (

x = 6

) gives

\frac{\pi x}{4} = \frac{3\pi}{2}

, where

\sin(\frac{3\pi}{2}) = -1

, producing the minimum height. Remember: for sine functions, the first maximum always occurs at

\frac{\pi}{2}

radians. Set the inside expression equal to

\frac{\pi}{2}

and solve for your variable.

Question 8

Two sinusoidal functions are given: $f(x) = 4\cos(2x)$ and $g(x) = 4\sin(2x + \frac{\pi}{2})$ . Which statement about these functions is correct?

The functions have the same amplitude but different periods, so their graphs will intersect at exactly two points per period
The functions have identical graphs because $\sin(2x + \frac{\pi}{2}) = \cos(2x)$ by the cofunction identity (correct answer)
The functions have the same period but different phase shifts, so $g(x)$ leads $f(x)$ by $\frac{\pi}{4}$ units
The functions have different amplitudes because the phase shift in $g(x)$ effectively changes the amplitude from 4 to $4\cos(\frac{\pi}{2}) = 0$

Explanation: Using the cofunction identity

\sin(\theta + \frac{\pi}{2}) = \cos(\theta)

, we have

g(x) = 4\sin(2x + \frac{\pi}{2}) = 4\cos(2x) = f(x)

. Therefore, the functions are identical. Choice A is wrong because they have the same period. Choice C misinterprets phase shift. Choice D incorrectly suggests that phase shift affects amplitude.

Question 9

If $p(x) = 2\sin(3x) + 5$ and $q(x) = -2\cos(3x) + 5$ , which statement about these functions is correct?

Both functions have the same period but different midlines located at different y-values
Function $p(x)$ has a larger amplitude than function $q(x)$ but they have the same range
Both functions have the same amplitude and the same range of values (correct answer)
Function $q(x)$ has the same amplitude as $p(x)$ but a different period due to the negative sign

Explanation: When you encounter trigonometric functions in the form

a\sin(bx) + c

a\cos(bx) + c

, you need to identify three key characteristics: amplitude (

|a|

), period (

\frac{2\pi}{|b|}

), and midline (

y = c

). For

p(x) = 2\sin(3x) + 5

: The amplitude is

|2| = 2

, the period is

\frac{2\pi}{3}

, and the midline is

y = 5

. Since the function oscillates 2 units above and below the midline, the range is

[3, 7]

. For

q(x) = -2\cos(3x) + 5

: The amplitude is

|-2| = 2

(amplitude is always positive), the period is

\frac{2\pi}{3}

, and the midline is

y = 5

. The negative sign flips the cosine function vertically but doesn't change the amplitude. The range is also

[3, 7]

. Both functions have identical amplitude (2) and range ([3, 7]), making C correct. A is wrong because both functions have the same midline at

y = 5

, not different y-values. B incorrectly claims

p(x)

has a larger amplitude than

q(x)

—they're both 2. D makes the common error of thinking the negative sign affects the period, but it only reflects the function across the midline without changing the period. Remember: The coefficient's absolute value determines amplitude, and negative signs only affect vertical reflection, not amplitude or period. Focus on the structure

a\sin(bx) + c

to quickly identify these properties.

Question 10

The graph of $y = \cos(x)$ is transformed to create $y = -2\cos(3x) + 1$ . Which statement correctly describes the transformation?

The amplitude is doubled, period is tripled, graph is reflected over x-axis, then shifted up 1 unit
The amplitude is doubled, period is divided by 3, graph is reflected over x-axis, then shifted up 1 unit (correct answer)
The amplitude is doubled, period is divided by 3, graph is reflected over y-axis, then shifted up 1 unit
The amplitude is halved, period is divided by 3, graph is reflected over x-axis, then shifted up 1 unit

Explanation: From

y = -2\cos(3x) + 1

: amplitude =

|-2| = 2

(doubled), period =

\frac{2\pi}{3}

(divided by 3), the negative sign reflects over the x-axis, and +1 shifts up 1 unit. Choice A incorrectly states period is tripled instead of divided by 3. Choice C confuses x-axis and y-axis reflection. Choice D incorrectly identifies amplitude as halved when

|-2| = 2

means doubled.

Question 11

Consider the function $r(x) = 3\cos(4x) - 1$ . At what value of $x$ in the interval $[0, \frac{\pi}{2}]$ does this function first reach its minimum value?

$x = 0$
$x = \frac{\pi}{8}$
$x = \frac{\pi}{4}$ (correct answer)
$x = \frac{\pi}{2}$

Explanation: The function

r(x) = 3\cos(4x) - 1

reaches its minimum when

\cos(4x) = -1

. This occurs when

4x = \pi + 2\pi k

for integer

k

. For

k = 0

4x = \pi

, so

x = \frac{\pi}{4}

. At

x = 0

r(0) = 3(1) - 1 = 2

(maximum). At

x = \frac{\pi}{4}

r(\frac{\pi}{4}) = 3(-1) - 1 = -4

(minimum). Choice A gives the maximum. Choice B gives

r(\frac{\pi}{8}) = 3\cos(\frac{\pi}{2}) - 1 = -1

(midline). Choice D gives the maximum again since

\cos(2\pi) = 1

Question 12

If $g(x) = 4\cos(\frac{x}{2}) - 3$ , what is the range of function $g$ ?

$[-7, 1]$ (correct answer)
$[-1, 7]$
$[-4, 4]$
$[-3, 3]$

Explanation: The cosine function has range

[-1, 1]

. With amplitude 4,

4\cos(\frac{x}{2})

has range

[-4, 4]

. Subtracting 3 shifts everything down 3 units, giving range

[-4-3, 4-3] = [-7, 1]

. Choice B incorrectly adds 3 instead of subtracting. Choice C ignores the vertical shift. Choice D uses the vertical shift value as the amplitude.

Question 13

The function $y = A\cos(Bx) + C$ passes through the points $(0, 7)$ , $(\frac{\pi}{4}, 2)$ , and $(\frac{\pi}{2}, -3)$ . What is the value of the amplitude $A$ ?

$A = 4$
$A = 5$ (correct answer)
$A = 7$
$A = 10$

Explanation: From

(0, 7)

7 = A\cos(0) + C = A + C

. From

(\frac{\pi}{2}, -3)

-3 = A\cos(\frac{B\pi}{2}) + C

. If

B = 2

, then

\cos(\pi) = -1

, so

-3 = -A + C

. Solving:

A + C = 7

and

-A + C = -3

gives

2A = 10

, so

A = 5

and

C = 2

. Checking with middle point:

2 = 5\cos(\frac{\pi}{2}) + 2 = 5(0) + 2 = 2

✓. Choice A results from calculation error. Choice C confuses amplitude with the y-value at x = 0. Choice D uses

2A

instead of

A

Question 14

Two sinusoidal functions $f(x) = \sin(2x)$ and $g(x) = 2\sin(x)$ are compared on the interval $[0, 2\pi]$ . Which statement correctly describes their relationship?

Both functions have the same period length, but $g(x)$ oscillates with greater amplitude than $f(x)$
Function $g(x)$ has twice the amplitude of $f(x)$ and both functions complete the same number of cycles
Function $f(x)$ has twice the amplitude and twice the frequency compared to function $g(x)$ over the given interval
Function $f(x)$ completes twice as many cycles as $g(x)$ and has the same amplitude throughout the interval (correct answer)

Explanation: When comparing sinusoidal functions, you need to analyze two key characteristics: amplitude (how high and low the function oscillates) and frequency (how often it completes a full cycle). For

f(x) = \sin(2x)

, the coefficient 2 inside the sine function affects the period. The standard sine function

\sin(x)

has period

2\pi

, but

\sin(2x)

has period

\frac{2\pi}{2} = \pi

. This means

f(x)

completes one full cycle every

\pi

units, so over the interval

[0, 2\pi]

, it completes 2 full cycles. The amplitude remains 1 since there's no coefficient in front of the sine. For

g(x) = 2\sin(x)

, the coefficient 2 is outside the sine function, affecting amplitude but not period. This function has amplitude 2 and keeps the standard period of

2\pi

, completing just 1 cycle over

[0, 2\pi]

. Answer choice A is incorrect because the functions have different periods (

\pi

2\pi

). Choice B is wrong because while

g(x)

does have twice the amplitude, the functions don't complete the same number of cycles. Choice C incorrectly states that

f(x)

has twice the amplitude when both functions actually have the same amplitude of 1 and 2 respectively, and mischaracterizes the amplitude relationship. Choice D correctly identifies that

f(x)

completes twice as many cycles (2 vs 1) and recognizes that both functions have consistent amplitudes throughout the interval. Study tip: Remember that coefficients inside the function

\sin(bx)

affect frequency/period, while coefficients outside

a\sin(x)

affect amplitude.

Question 15

A function $f(x) = A\sin(Bx) + C$ has a maximum value of 7 and a minimum value of -1. If the function completes exactly 3 full cycles over the interval $[0, 2\pi]$ , what is the value of $B$ ?

$B = 3$ (correct answer)
$B = 6$
$B = \frac{3}{2\pi}$
$B = \frac{2\pi}{3}$

Explanation: First, find the amplitude and midline. The amplitude is

A = \frac{7-(-1)}{2} = 4

and midline is

C = \frac{7+(-1)}{2} = 3

. For 3 complete cycles over

[0, 2\pi]

, the period must be

\frac{2\pi}{3}

. Since period =

\frac{2\pi}{B}

, we have

\frac{2\pi}{B} = \frac{2\pi}{3}

, so

B = 3

. Choice B confuses cycles with the coefficient. Choice C inverts the relationship. Choice D uses the period value as B.