Chapter 4.4, Problem 41E

Solution

To state: The function $y=-\sin x$ as a phase of shift of $y=\sin x$ .

The phase of shift is $y=\sin \left(x+\pi \right)$ .

Given information:

The given function is $y=-2\sin x$ .

Explanation:

Consider the given function: $y=-\sin x$

Compare the function $y=-\sin x$ with the standard function $y=a\sin bx$ .

Then the values are $a=-1\text{ and }b=1$ .

Here $a$ is the amplitude of the function and its value is 1.

Now find the period of the function.

  $\text{Period}=\frac{2\pi }{\left|b\right|}$

Substitute the values $b=1$ in the formula:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{\left|1\right|}\\ =2\pi \end{array}$

The minus sign in the given function $y=-\sin x$ shows that its graph is reflected across x- axis, so its maximum value is at $\frac{3\pi }{2}$ on the other hand the maximum of the function $y=\sin x$ is at $\frac{\pi }{2}$ .

So the shift of the graph $y=-\sin x$ is $\pi$ units left from the graph of $y=\sin x$ .

  $\begin{array}{l}y=-\sin x\\ y=\sin \left(x+\pi \right)\end{array}$

Therefore, the phase of shift is $y=\sin \left(x+\pi \right)$ .