Chapter 7.5, Problem 53E

To determine

To find: The solution of the equation $\sin \theta +\sin 3\theta =0$ using a Sum-to-Product formula.

Answer to Problem 53E

The solution of the equation $\sin \theta +\sin 3\theta =0$ is $\theta =\frac{\pi n}{2}$ .

Explanation of Solution

Given information:

The equation $\sin \theta +\sin 3\theta =0$ .

Calculation:

Write the sum-to-product formula

  $\sin c+\sin d=2\sin \frac{c+d}{2}\cos \frac{c-d}{2}$

Simplify the given equation using the above formula

  $\sin \theta +\sin 3\theta =0$

  $\begin{array}{c}2\sin \frac{\theta +3\theta }{2}\cos \frac{\theta -3\theta }{2}=0\\ 2\sin \frac{4\theta }{2}\cos \frac{-2\theta }{2}=0\\ 2\sin 2\theta \cos \left(-\theta \right)=0\end{array}$

The value of $\cos \left(-\theta \right)=\cos \theta$

  $\begin{array}{c}2\sin 2\theta \cos \theta =0\\ \sin 2\theta =0\text{or}\cos \theta =0\end{array}$

For $\sin 2\theta =0$

  $\begin{array}{l}2\theta =0,\pi ,2\pi ,\mathrm{...}\\ 2\theta =\pi n\\ \theta =\frac{\pi n}{2}\end{array}$

For $\cos \theta =0$

  $\begin{array}{l}\theta =\frac{\pi }{2,}\frac{3\pi }{2},\mathrm{...}\\ \theta =\frac{\pi }{2}+n\pi \end{array}$

Here $n$ is any odd integer.

The general solution of $\theta$ is expressed as, $\theta =\frac{\pi n}{2}$

Therefore, the solution of the equation $\sin \theta +\sin 3\theta =0$ is $\theta =\frac{\pi n}{2}$ .