Question

If $\sin \theta +\sin 2\theta +\sin 3\theta =\sin \alpha$ and $\cos \theta +\cos 2\theta +\cos 3\theta =\cos \alpha$, then $\theta$ is equal to -

• A. $\alpha /2$
• B. $\alpha$
• C. $2\alpha$
• D. $\alpha /6$

Answer 1

The correct option is A $\alpha /2$

Given $\sin \theta +\sin 2\theta +\sin 3\theta =\sin \alpha$ .....(1)
$\cos \theta +\cos 2\theta +\cos 3\theta =\cos \alpha$ .....(2)
Dividing (1) by (2), we get
$\frac{2\sin 2\theta \cos \theta +\sin 2\theta }{2\cos 2\theta \cos \theta +\cos 2\theta }=\frac{\sin \alpha }{\cos \alpha }$
$\Rightarrow \frac{\sin 2\theta [2\cos \theta +1]}{\cos 2\theta [2\cos \theta +1]}=\frac{\sin \alpha }{\cos \alpha }$
$\tan 2\theta =\tan \alpha$
$\Rightarrow 2\theta =\alpha$
$\Rightarrow \theta =\frac{\alpha }{2}$