Question

$\sin \theta +sin2\theta +sin3\theta =\sin \alpha$ and $\cos \theta +cos2\theta +cos3\theta =\cos \alpha$, then $\theta =$

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

Answer 1

The correct option is A $\alpha /2$

$\sin \theta +\sin 2\theta +\sin 3\theta =\sin \alpha$

$2\sin 2\theta \cos \theta +\sin 2\theta =\sin \alpha$

$\sin 2\theta (2\cos \theta +1)=\sin \alpha$ .....(1)

$\cos \theta +\cos 2\theta +\cos 3\theta =\cos \alpha$

$2\cos 2\theta \cos \theta +\cos 2\theta =\cos \alpha$

$\cos 2\theta (2\cos \theta +1)=\cos \alpha$ ......(2)

From equation 1 and 2, we get,

$\tan 2\theta =\tan \alpha$

$\alpha =2\theta$

$\theta =\frac{\alpha }{2}$