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Precalculus : Find the Phase Shift of a Sine or Cosine Function

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Example Questions

Example Question #1 : Find The Phase Shift Of A Sine Or Cosine Function

Find the phase shift of F(t)=3sin(2t+4) .

Possible Answers:

Correct answer:

Explanation:

In the formula,

F(t)=Asin(Bt-C)+D .

$\frac{C}{B}$ represents the phase shift.

Plugging in what we know gives us:

$\frac{C}{B}=\frac{-4}{2}$ .

Simplified, the phase is then .

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Example Question #2 : Find The Phase Shift Of A Sine Or Cosine Function

Please choose the best answer from the following choices.

Describe the phase shift of the following function:

$y=sin(x+\pi )$

Possible Answers:

Vertical stretch by $\pi$ radians.

Shift left by $\pi$ radians.

Shift down by $\pi$ radians.

Shift up by $\pi$ radians.

Correct answer:

Shift left by $\pi$ radians.

Explanation:

Since $\pi$ is being added inside the parentheses, there will be a horizontal shift. The goal is to maintain zero within the parentheses so you will shift left $\pi$ radians.

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Example Question #3 : Find The Phase Shift Of A Sine Or Cosine Function

Which equation would produce this graph?

Phase shift 1

Possible Answers:

$y=sin\left(x-\frac{\pi}{3}\right)$

$y=sin(x-\pi)$

$y=sin\left(x+\frac{\pi}{2}\right)$

$y=sin\left(x-\frac{\pi}{2}\right)$

$y=sin\left(\frac{\pi}{2}x\right)$

Correct answer:

$y=sin\left(x-\frac{\pi}{2}\right)$

Explanation:

This is the graph of sine, but shifted to the right $\frac{\pi}{2}$ units. To reflect this shift, $\frac{\pi}{2}$ should be subtracted from x.

Thus resulting in

$y=sin\left(x-\frac{\pi}{2}\right)$ .

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Example Question #1 : Find The Phase Shift Of A Sine Or Cosine Function

Which equation would produce this sine graph?

Phase shift 2

Possible Answers:

$y=2sin\left(x-\frac{\pi}{2}\right)-1$

$y=sin\left(2x-\frac{\pi}{4}\right)$

y=sin(2x)-3

$y=2sin\left(x-\frac{\pi}{4}\right)-1$

$y=sin\left(x-\frac{3\pi}{4}\right)-1$

Correct answer:

$y=2sin\left(x-\frac{\pi}{4}\right)-1$

Explanation:

The graph has an amplitude of 2 but has been shifted down 1:

Phase shift 2 dots

In terms of the equation, this puts a 2 in front of sin, and -1 at the end.

This makes it easier to see that the graph starts [is at 0] where $x=\frac{\pi}{4}$ .

The phase shift is $\frac{\pi}{4}$ to the right, or $x-\frac{\pi}{4}$ .

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Example Question #5 : Find The Phase Shift Of A Sine Or Cosine Function

Write the equation for a sine graph with a maximum at $\left(\frac{ \pi }{8}, 3\right)$ and a minimum at $\left(\frac{ 9 \pi }{8 } , -1 \right)$ .

Possible Answers:

$y = 2 \sin (x + \frac{ 3 \pi }{ 8 } ) + 1$

$y = 2 \sin ( \frac{1}{2}( x - \frac{ \pi }{8})) + 1$

$y = 3 \sin (\frac{ 1}{2} x )$

$y = 2 \sin (x - \frac{ \pi }{ 8 }) + 1$

$y = 2 \sin (x - \frac{ 3 \pi }{8 } ) - 1$

Correct answer:

$y = 2 \sin (x + \frac{ 3 \pi }{ 8 } ) + 1$

Explanation:

To write this equation, it is helpful to sketch a graph:

Trig graph 1

Indicating the maximum and minimum points, we can see that this graph has been shifted up 1, and it has an amplitude of 2.

The distance from the maximum to the minimum point is half the wavelength. In this case, the wavelength is $\frac{ 9 \pi }{ 8 } - \frac{ \pi }{8 } = \frac{ 8 \pi }{8 } = \pi$ . That means the full wavelength is $2 \pi$ , and the frequency is 1.

This sketch shows that the graph starts to the left of the y-axis. To figure out exactly where, subtract $\frac{ \pi }{2}$ from the maximum x-coordinate, $\frac{ \pi }{ 8 }$ :

$\frac{ \pi }{ 8 } - \frac{ \pi }{2} = \frac{ \pi }{8 } - \frac{ 4 \pi }{ 8 } = \frac{ - 3 \pi }{ 8 }$ .

Our equation will be in the form $y = A \cdot \sin ( f (x - h )) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This graph has an equation of

$y = 2 \sin ( x + \frac{ 3 \pi }{ 8 } ) + 1$ .

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Example Question #6 : Find The Phase Shift Of A Sine Or Cosine Function

Write the equation for a cosine graph with a maximum at $\left( \frac{ \pi }{3} , 3 \right)$ and a minimum at $\left(\frac{ 4 \pi }{3} , -3 \right)$ .

Possible Answers:

$y = 3 \cos (\frac{ 1}{2} ( x - \frac{ \pi }{3} ))$

$y = 2 \cos (x + \frac{ \pi }{3} )+ 1$

$y = -3 \cos (x - \frac{ \pi } {3 })$

$y = 3 \cos (x - \frac{ \pi }{3} )$

$y = 2 \cos (\frac{ 1}{2} ( x + \frac{ \pi }{3} )) + 1$

Correct answer:

$y = 3 \cos (x - \frac{ \pi }{3} )$

Explanation:

In order to write this equation, it is helpful to sketch a graph:

Trig graph 2

The dotted line is at $x = \frac{ \pi }{3}$ , where the maximum occurs and therefore where the graph starts. This means that the graph is shifted to the right $\frac{ \pi }{3}$ .

The distance from the maximum to the minimum is half the entire wavelength. Here it is $\frac{ 4 \pi }{3} - \frac{ \pi }{3} = \frac{ 3 \pi }{3} = \pi$ .

Since half the wavelength is $\pi$ , that means the full wavelength is $2 \pi$ so the frequency is just 1.

The amplitude is 3 because the graph goes symmetrically from -3 to 3.

The equation will be in the form $y = A \cdot \cos (f (x - h)) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

This equation is

$y = 3 \cos (x - \frac{ \pi } {3 })$ .

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Example Question #7 : Find The Phase Shift Of A Sine Or Cosine Function

Write the equation for a sine function with a maximum at $\left(\frac{ 7 \pi }{10} , 7 \right)$ and a minimum at $\left(\frac{ 17 \pi }{10 } , 1\right)$ .

Possible Answers:

$y = 7 \sin (x + \frac{ \pi }{5} ) - 3$

$y = 3 \sin (\frac{ 1}{2} (x - \frac{ 7 \pi }{10 })) + 4$

$y = \sin (x - \frac{ \pi}{10}) + 7$

$y = 3 \sin (x - \frac{ \pi }{5 }) + 4$

$y = 7 \sin (x - \frac{\pi }{10}) + 1$

Correct answer:

$y = 3 \sin (x - \frac{ \pi }{5 }) + 4$

Explanation:

The equation will be in the form $y = A \sin (f (x - h)) + k$ where A is the amplitude, f is the frequency, h is the horizontal shift, and k is the vertical shift.

To write the equation, it is helpful to sketch a graph:

Trig graph 3

From plotting the maximum and minimum, we can see that the graph is centered on y = 4 with an amplitude of 3.

The distance from the maximum to the minimum is half the wavelength. For this graph, this distance is $\frac{ 17 \pi }{10 } - \frac{ 7 \pi }{10 } = \frac{ 10 \pi }{10 } = \pi$ .

This means that the total wavelength is $2 \pi$ and the frequency is 1.

The graph starts $\frac{ \pi }{2}$ behind the maximum point. To determine this x value, subtract $\frac{ \pi }{2 }$ from the x-coordinate of the maximum:

$\frac{ 7 \pi }{10 } - \frac{ \pi }{2} = \frac{ 7 \pi }{10 } - \frac{ 5 \pi }{10 } = \frac{ 2 \pi }{10 } = \frac{ \pi }{5}$

Our equation is:

$y = 3 \sin (x - \frac{ \pi }{5} ) + 4$ .

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Precalculus : Find the Phase Shift of a Sine or Cosine Function

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #1 : Find The Phase Shift Of A Sine Or Cosine Function

Example Question #2 : Find The Phase Shift Of A Sine Or Cosine Function

Example Question #3 : Find The Phase Shift Of A Sine Or Cosine Function

Example Question #1 : Find The Phase Shift Of A Sine Or Cosine Function

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