Chapter 5.4, Problem 88E

To determine

To verify: the identity.

Explanation of Solution

Given information:

Given function is

  $\sin \left(n\pi +\theta \right)={\left(-1\right)}^n\sin \theta ,n\text{is}\text{an}\text{integer}$

Calculation:

Consider the following identity

  $\sin \left(n\pi +\theta \right)={\left(-1\right)}^n\sin \theta$

Where $n$ is an integer

Verify the above identity.

Put $n=1$ , an odd number, in above identity.

  $\begin{array}{l}\sin \left(1\times \pi +\theta \right)={\left(-1\right)}^1\sin \theta \\ \sin \left(\pi +\theta \right)=\left(-1\right)\sin \theta \\ -\sin \theta =-\sin \theta \end{array}$

Here both sides are equal.

Put $n=2$ , an even number, in above identity.

  $\begin{array}{c}\sin \left(2\times \pi +\theta \right)={\left(-1\right)}^2\sin \theta \\ \sin \left(2\pi +\theta \right)=1\times \sin \theta \\ \sin \theta =\sin \theta \end{array}$

Here both sides are also equal.

The above identity is correct for both odd and even numbers.

Hence the above identity $\sin \left(n\pi +\theta \right)={\left(-1\right)}^n\sin \theta$ is verified and it is true.