Question

Compare the graphs of $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$.

Short answer

From the graph of the function $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$ it can be interpreted that the graph of the functions $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$ overlap each other.

Step-by-step solution

Step 1. Write down the given information.

The given functions are $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$.

Step 2. Concept used.

A function of the form:

$y=a\sin b\left(\theta -h\right)+k,y=a\cos b\left(\theta -h\right)+k$has vertical shift $\left(k\right)$. And, a period of $\frac{360°}{\left|b\right|}\text{or}\frac{2\pi }{\left|b\right|}$ for sine, cosecant, secant and cosine functions and a period of $\frac{180°}{\left|b\right|}\text{or}\frac{\pi }{\left|b\right|}$ for tangent and cotangent function. The phase shift for the functions is $\left(h\right)$.

The amplitude of secant, cosecant, tangent and cotangent functions is not defined.

The equation for the midline is written as, $y=k$. A midline is a new reference line when the parent graph is stretched vertically up or down and then the graph oscillates about new reference line called the midline.

Step 3. Explanation.

The given functions are $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$.

The parent function $y=\sin \left(\theta \right)$ is shifted towards right by a phase shift of $\frac{\pi }{2}$ to get the function $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]$.

The parent function $y=\cos \left(\theta \right)$ is shifted towards left by a phase shift of $-\frac{3\pi }{2}$ to get the function $y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$.

It can also be seen from the graph drawn of the functions $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$.

Step 4. Sketch the graph of given function.

The graph of the functions $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$ are shown below.

Step 5. Interpretation from the graph of the function.

From the graph of the function $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$ it can be interpreted that the graph of the functions $y=-\sin \left[\frac{1}{4}\left(\theta -\frac{\pi }{2}\right)\right]\text{and}y=\cos \left[\frac{1}{4}\left(\theta +\frac{3\pi }{2}\right)\right]$ overlap each other.