Chapter 6.5, Problem 15E

To determine

To find: the phase shift of the function and then graph that function.

Answer to Problem 15E

  $-\frac{\pi }{2}$

Explanation of Solution

Given information:

  $y=\sin \left(2\theta +\pi \right)$

Calculation:

The equation of sine function is as follows:

  $y=A\sin \left(k\theta +c\right)$...........(1)

Here $A$ is an amplitude which is an absolute value of the coefficient and $k=\frac{2\pi }{\text{Period}}$ .

On comparing given equation with equation (1) to get,

  $\begin{array}{l}c=\pi \\ k=2\end{array}$

The phase shift can be evaluated as follows:

  $\begin{array}{c}\text{Phase shift}=-\frac{c}{k}\\ =-\frac{\pi }{2}\left[\text{Put the value of }c\text{ and }k\right]\end{array}$

Thus, the phase shift is $-\frac{\pi }{2}$ .

Since $c>0$ the shift is towards left. The graph of the function is shown below.

  

The blue dots represent the graph of $y=\sin \left(2\theta \right)$ and the red line denotes the graph of $y=\sin \left(2\theta +\pi \right)$ . The graph after phase shift moves towards left to graph of $y=\sin \left(2\theta \right)$ .