Chapter 13.7, Problem 2BCYP

To determine

To determine the amplitude and period of the given function.

Answer to Problem 2BCYP

Amplitude $=\frac{1}{2}$ , Period $=\pi$

Explanation of Solution

Given:

  $y=\frac{1}{2}\sin 2\theta$

Concept used:

Amplitude is the distance between the center line of the function and the top or bottom of the function.

Period is the distance between two peaks of the graph.

The generalized equation for a sine function is:

  $y=A\sin \left(Bx\right)$

Amplitude $=A$

Period $=\frac{2\pi }{B}$

  

Calculation:

For $y=\frac{1}{2}\sin 2\theta$ ,

Comparing with the general sine equation $y=A\sin \left(Bx\right)$ ,

  $A=\frac{1}{2},B=2$

So, the amplitude of the given function is $\frac{1}{2}$ .

Period $=\frac{2\pi }{B}$

  $\begin{array}{l}=\frac{2\pi }{2}\\ =\pi \end{array}$

The period of the given function is $0.4\pi$ .

Conclusion:

Therefore, the amplitude of the given function is $\frac{1}{2}$ and the period is $\pi$ .