Chapter 1.6, Problem 59E

To determine

To evaluate: Whether the function is periodic or not. Also, find its period.

Answer to Problem 59E

It is proved that the function is periodic and the period of the function is $2\pi$ .

Explanation of Solution

Given information:

The given function is $y={\sin }^{3}x$ .

Calculation:

Consider the function $y={\sin }^{3}x$ . It is known that $\sin x$ has period $2\pi$ . So,

  $\sin \left(x+2\pi \right)=\sin x$

Check the periodicity of the function for $2\pi$ .

  $\begin{array}{c}{\sin }^3\left(x+2\pi \right)={\left[\sin \left(x+2\pi \right)\right]}^3\\ ={\left[\sin x\right]}^3\left(\therefore \sin \left(x+2\pi \right)=\sin x\right)\\ ={\sin }^{3}x\end{array}$

The graph for the given function $y={\sin }^{3}x$ is shown below.

  

Figure (1)

Interpretation: From the above graph it can be observed that the function is periodic and the period of the function is $2\pi$ .