Question

A value of $\theta$ satisfying $\sin \left(5\theta \right)-\sin \left(3\theta \right)+\sin \left(\theta \right)=0,$ such that $0<\theta <\frac{\pi }{2}$ is

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$\frac{\pi }{6}$

Explanation for correct answer:

Finding the value.

Step-1: Simplify the given data.

Given, $\sin \left(5\theta \right)-\sin \left(3\theta \right)+\sin \left(\theta \right)=0,$

$\Rightarrow$ $\sin \left(5\theta \right)+\sin \left(\theta \right)=\sin \left(3\theta \right)$

Step-2: Apply formula $\mathit{s}\mathit{i}\mathit{n}\left(c\right)\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\left(d\right)\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\left(\frac{c+d}{2}\right)\mathit{c}\mathit{o}\mathit{s}\left(\frac{c-d}{2}\right)$

$\Rightarrow$ $2\sin \left(\frac{5\theta +\theta }{2}\right)\cos \left(\frac{5\theta -\theta }{2}\right)=\sin \left(3\theta \right)$

$\Rightarrow$ $2\sin \left(3\theta \right)\cos \left(2\theta \right)=\sin \left(3\theta \right)$

$\Rightarrow$ $2\sin \left(3\theta \right)\cos \left(2\theta \right)-\sin \left(3\theta \right)=0$

$\Rightarrow$ $\sin \left(3\theta \right)\left(2\cos \left(2\theta \right)-1\right)=0$

$\Rightarrow$ $\sin \left(3\theta \right)=0,\cos \left(2\theta \right)=\frac{1}{2}$

$\Rightarrow$ $$\begin{gathered} \theta =n\pi ,2\theta =\frac{\pi }{3} \\ \theta =\mathrm{\pi }/6 \end{gathered}$$

Hence, option $\left(B\right)$ is correct.