Amplitude, Period, Phase Shift of a Trig Function

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Pre-Calculus › Amplitude, Period, Phase Shift of a Trig Function

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Which of the following equations could represent a cosine function with amplitude 3, period $\frac{\pi}{2}$ , and a phase shift of $\frac{\pi}{4}$ ?

$y = 3\cos(x+\pi)$

$y = 3\cos(x-\pi)$

$y = -3\cos(4x+\pi)$

$y=-3\cos(4x-\pi)$

CORRECT

Explanation

The form of the equation will be $y=A\cos(Bx+C)$

First, think about all possible values of A that could give you an amplitude of 3. Either A = -3 or A = 3 could each produce amplitude = 3. Be sure to look for answer choices that satisfy either of these.

Secondly, we know that the period is $\frac{\pi}{2}$ . Normally we know what B is and need to find the period, but this is the other way around. We can still use the same equation and solve:

$\frac{2\pi}{B}=\frac{\pi}{2}$ . You can cross multiply to solve and get B = 4.

Finally, we need to find a value of C that satisfies

$-\frac{C}{B}=\frac{\pi}{4}$ . Cross multiply to get:

$-4C=B\pi$ .

Next, plug in B= 4 to solve for C:

$-4C = 4\pi$

$C=-\pi$

Putting this all together, the equation could either be:

$y = 3\cos(4x-\pi)$ or $y = -3\cos(4x-\pi)$

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

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

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

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

CORRECT

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

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

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

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

Find the phase shift of .

-4

-2

CORRECT

Explanation

In the formula,

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

Plugging in what we know gives us:

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

Simplified, the phase is then .

What is the amplitude of the following function?

$y = -24\sin(x-\pi)+14$

-24

-14

$\pi$

CORRECT

Explanation

When you think of a trigonometric function of the form y=Asin(Bx+C)+D, the amplitude is represented by A, or the coefficient in front of the sine function. While this number is -24, we always represent amplitude as a positive number, by taking the absolute value of it. Therefore, the amplitude of this function is 24.

What is the period of this sine graph?

Trig period 1

$\frac{2}{3}\pi$

CORRECT

$\frac{3}{2}\pi$

$3\pi$

$\pi^2$

Explanation

The graph has 3 waves between 0 and $2\pi$ , meaning that the length of each of the waves is $2\pi$ divided by 3, or $\frac{2}{3}\pi$ .

Which equation would produce this sine graph?

Phase shift 2

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

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

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

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

CORRECT

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 $x-\frac{\pi}{4}$ .

State the amplitude, period, phase shift, and vertical shift of the function $y=-7\sin(6x+\pi)-4$

Amplitude: 7

Period: - $\pi$ /3

Phase Shift: - $\pi$ /6

Vertical Shift: -4

Amplitude: 7

Period: $\pi$ /3

Phase Shift: $\pi$ /6

Vertical Shift: -4

Amplitude: 7

Period: $\pi$ /3

Phase Shift: - $\pi$ /6

Vertical Shift: -4

CORRECT

Amplitude: -7

Period: $\pi$ /3

Phase Shift: - $\pi$ /6

Vertical Shift: 4

Explanation

A common way to make sense of all of the transformations that can happen to a trigonometric function is the following. For the equations y = A sin(Bx + C) + D,

amplitude is |A|
period is 2 $\pi$ /|B|
phase shift is - C/B
vertical shift is D

In our equation, A=-7, B=6, C= $\pi$ , and D=-4. Next, apply the above numbers to find amplitude, period, phase shift, and vertical shift.

To find amplitude, look at the coefficient in front of the sine function. A=-7, so our amplitude is equal to 7.

The period is 2 $\pi$ /B, and in this case B=6. Therefore the period of this function is equal to 2 $\pi$ /6 or $\pi$ /3.

To find the phase shift, take -C/B, or - $\pi$ /6. Another way to find this same value is to set the inside of the parenthesis equal to 0, then solve for x.
6x+ $\pi$ =0
6x=- $\pi$
x=- $\pi$ /6
Either way, our phase shift is equal to - $\pi$ /6.

The vertical shift is equal to D, which is -4.

y=-7\sin(6x+\pi)-4

State the amplitude, period, phase shift, and vertical shift of the function $y=-\cos(x-\pi)+3$

Amplitude: 1

Period: $2\pi$

Phase Shift: $\pi$

Vertical Shift: 3

CORRECT

Amplitude: 1

Period: $\pi$

Phase Shift: $2\pi$

Vertical Shift: 3

Amplitude: 1

Period: $2\pi$

Phase Shift: $\pi$

Vertical Shift: 0

Amplitude: 1

Period: $2\pi$

Phase Shift: $2\pi$

Vertical Shift: 3

Amplitude: 1

Period: $\pi$

Phase Shift: $\pi$

Vertical Shift: 3

Explanation

A common way to make sense of all of the transformations that can happen to a trigonometric function is the following. For the equations y = A sin(Bx + C) + D,

amplitude is |A|
period is 2 $\pi$ /|B|
phase shift is - C/B
vertical shift is D

In our equation, A=-1, B=1, C=- $\pi$ , and D=3. Next, apply the above numbers to find amplitude, period, phase shift, and vertical shift.

To find amplitude, look at the coefficient in front of the sine function. A=-1, so our amplitude is equal to 1.

The period is 2 $\pi$ /B, and in this case B=1. Therefore the period of this function is equal to 2 $\pi$ .

To find the phase shift, take -C/B, or $\pi$ . Another way to find this same value is to set the inside of the parenthesis equal to 0, then solve for x.
x- $\pi$ =0
x= $\pi$
Either way, our phase shift is equal to $\pi$ .

The vertical shift is equal to D, which is 3.

State the amplitude, period, phase shift, and vertical shift of the function $y=\sin(2x-3)+2$

Amplitude: 1

Period: $\pi$

Phase Shift: -3/2

Vertical Shift: -2

Amplitude: 1

Period: $\pi$

Phase Shift: -3/2

Vertical Shift: 2

Amplitude: 1

Period: 3/2

Phase Shift: $\pi$

Vertical Shift: 2

Amplitude: 1

Period: $\pi$

Phase Shift: 3/2

Vertical Shift: 2

CORRECT

Explanation

A common way to make sense of all of the transformations that can happen to a trigonometric function is the following. For the equations y = A sin(Bx + C) + D,

amplitude is |A|
period is 2 $\pi$ /|B|
phase shift is - C/B
vertical shift is D

In our equation, A=1, B=2, C=-3, and D=2. Next, apply the above numbers to find amplitude, period, phase shift, and vertical shift.

To find amplitude, look at the coefficient in front of the sine function. A=1, so our amplitude is equal to 1.

The period is 2 $\pi$ /B, and in this case B=2. Therefore the period of this function is equal to $\pi$ .

To find the phase shift, take -C/B, or 3/2. Another way to find this same value is to set the inside of the parenthesis equal to 0, then solve for x.
2x-3=0
2x=3
x=3/2
Either way, our phase shift is equal to 3/2.

The vertical shift is equal to D, which is 2.

Select the answer choice that correctly matches each function to its period.

$\sin x \rightarrow \pi$

$\cos x \rightarrow \pi$

$\tan x \rightarrow \pi$

$\cot x \rightarrow \pi$

$\sec x \rightarrow \pi$

$\csc x \rightarrow \pi$

$\sin x \rightarrow \pi$

$\cos x \rightarrow \pi$

$\tan x \rightarrow 2\pi$

$\cot x \rightarrow 2\pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow 2\pi$

$\cot x \rightarrow 2\pi$

$\sec x \rightarrow \pi$

$\csc x \rightarrow \pi$

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow \pi$

$\cot x \rightarrow \pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

CORRECT

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow 2\pi$

$\cot x \rightarrow 2\pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

Explanation

The following matches the correct period with its corresponding trig function:

$\sin x \rightarrow 2\pi$

$\cos x \rightarrow 2\pi$

$\tan x \rightarrow \pi$

$\cot x \rightarrow \pi$

$\sec x \rightarrow 2\pi$

$\csc x \rightarrow 2\pi$

In other words, sin x, cos x, sec x, and csc x all repeat themselves every $2\pi$ units. However, tan x and cot x repeat themselves more frequently, every $\pi$ units.

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