Chapter 4.4, Problem 42E

Solution

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

The phase of shift is $y=\sin \left(x-\frac{\pi }{2}\right)$ .

Given information:

The given function is $y=-\cos x$ .

Explanation:

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

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

So the shift of the graph $y=-\cos x$ is $\frac{\pi }{2}$ units right from the graph of $y=\sin x$ .

  $\begin{array}{l}y=-\cos x\\ y=\sin \left(x-\frac{\pi }{2}\right)\end{array}$

Therefore, the phase of shift is $y=\sin \left(x-\frac{\pi }{2}\right)$ .