Chapter 12.8, Problem 17PPS

To determine

Find the amplitude, period and phase shift for each function. Also, draw the graph of the function.

Answer to Problem 17PPS

The amplitude, period and phase shift of the function is $1,2\pi \&\frac{\pi }{2}$ to the left.

Explanation of Solution

Given: It is given in the question that the function is $y=2\sin (\theta +\frac{\pi }{2})$ .

Concept Used:

In this, use the concept of the general equation for sine $y=a\sin b(\theta -h)$ , where a is amplitude, $\frac{2\pi }{\left|b\right|}$ is period and h is the phase shift and if the h is positive , then h units to the right and if h is negative then h units to the left.

Calculation: Here, the function is $y=2\sin (\theta +\frac{\pi }{2})$ .

In this, $a=2$ , so the amplitude is $2$ .

Now, the value of b is $1$ , so the period is: $\frac{2\pi }{\left|b\right|}=\frac{2\pi }{\left|1\right|}=2\pi$

From the equation it is find that h is - $\frac{\pi }{2}$ with is negative sign, so the phase shift is $\frac{\pi }{2}$ to the left.

The graph of the function is given below:

  

Conclusion:

The amplitude, period and phase shift are $1,2\pi \&\frac{\pi }{2}$ to the left..