Trigonometry - Phase Shifts

Which of the following is the correct definition of a phase shift?

A measure of the length of a function between vertical asymptotes

The distance a function is shifted diagonally from the general position

The distance a function is shifted horizontally from the general position

The distance a function is shifted vertically from the general position

Explanation

Take the function for example. The graph for is

If we were to change the function to $cos(x + \pi)$ , our phase shift is $-\pi$ . This means we need to shift our entire graph $\pi$ units to the left.

Our new graph $cos(x + \pi)$ is the following

Consider the function $4sec(5(x+\frac{\pi}{3})) +10$ . What is the phase shift of this function?

$\frac{\pi}{3}$

$-\frac{\pi}{3}$

Explanation

The general form for the secant transformation equation is . represents the phase shift of the function. When considering $4sec(5(x+\frac{\pi}{3})) +10$ we see that $C = -\frac{\pi}{3}$ . So our phase shift is $-\frac{\pi}{3}$ and we would shift this function $\frac{\pi}{3}$ units to the left of the original secant function’s graph.

True or False: The function $csc(x + \pi)$ has a phase shift of $\pi$ .

True

False

Explanation

The form of the general cosecant function is . So if we have $csc(x + \pi)$ then , which represents the phase shift, is equal to $-\pi$ . This gives us a phase shift of $-\pi$ .

Which of the following is the phase shift of the function $5 - cot(6x - \pi)$ ?

$\pi$

$-\pi$

$-\frac{\pi}{6}$

$\frac{\pi}{6}$

Explanation

The general form of the cotangent function is . So first we need to get $cot(6x - \pi)$ into the form .

$cot(6x - \pi)$

$cot(6(x - \frac{\pi}{6}))$

From this we see that $C = \frac{\pi}{6}$ giving us our answer.

True or False: If the function has a phase shift of $n2\pi$ , then the graph will not be changed.

True

False

Explanation

This is true because the graph has a period of $2\pi$ , meaning it repeats itself every $2\pi$ units. So if has a phase shift of any multiple of , then it will just overlay the original graph. This is shown below. In orange is the graph of and in purple is the graph of $sin(x + 4\pi)$ .

Which of the following is the graph of with a phase shift of $\frac{\pi}{2}$ ?

Screen shot 2020 08 27 at 2.35.10 pm

Screen shot 2020 08 27 at 2.36.53 pm

Screen shot 2020 08 27 at 2.35.20 pm

Screen shot 2020 08 27 at 2.36.46 pm

Explanation

Start this problem by graphing the function of tangent.

Screen shot 2020 08 27 at 2.35.10 pm

Now we need to shift this graph $\frac{\pi}{2}$ to the right.

Screen shot 2020 08 27 at 2.35.16 pm

This gives us our answer

Screen shot 2020 08 27 at 2.35.20 pm

Identify the phase shift of the following equation.

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

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\frac{1}{2}$

Explanation

If we use the standard form of a sine function

the phase shift can be calculated by $\frac{C}{B}$ . Therefore, in our case, our phase shift is

$\frac\frac{}{}{}{}$ $\frac{\frac{\pi}{4}}{\frac{1}{2}}=\frac{\pi}{4}\cdot2=\frac{\pi}{2}$

Which of the following is equivalent to $2\sin(2x-3\pi)$

$-2\cos(2(x-\frac{\pi}{4}))$

$-2\cos(2x-\frac{\pi}{4})$

$2\sin(2x)$

$-2\cos(\frac{x}{2}-\frac{\pi}{2})$

$2\cos(x)$

Explanation

The first zero can be found by plugging 3π/2 for x, and noting that it is a double period function, the zeros are every π/2, count three back and there is a zero at zero, going down.

A more succinct form for this answer is $-2\sin(2x)$ but that was not one of the options, so a shifted cosine must be the answer.

The first positive peak is at π/4 at -1, so the cosine function will be shifted π/4 to the right and multiplied by -1. The period and amplitude still 2, so the answer becomes $-2\cos(2(x-\frac{\pi}{4}))$ .

To check, plug in π/4 for x and it will come out to -2.

Phase Shifts

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