Question

Prove that : $\frac{cos3\Theta +2cos5\Theta +cos7\Theta }{cos\Theta +2cos3\Theta +cos5\Theta }=cos2\Theta -sin2\Theta tan3\Theta$

Answer 1

We have,

$L.H.S.$

$\frac{\cos 3\theta +2\cos 5\theta +\cos 7\theta }{\cos \theta +2\cos 3\theta +\cos 5\theta }$

$=\frac{\cos 3\theta +\cos 7\theta +2\cos 5\theta }{\cos \theta +\cos 5\theta +2\cos 3\theta }$

$=\frac{2\cos \frac{3\theta +7\theta }{2}\cos \frac{3\theta -7\theta }{2}+2\cos 5\theta }{2\cos \frac{\theta +5\theta }{2}\cos \frac{\theta -5\theta }{2}+2\cos 3\theta }$

$=\frac{2\cos 5\theta \cos (-2\theta )+2\cos 5\theta }{2\cos 3\theta cos(-2\theta )+2\cos 3\theta }$

$=\frac{2\cos 5\theta (\cos 2\theta +1)}{2\cos 3\theta (\cos 2\theta +1)}\therefore \cos (-\theta )=\cos \theta$

$=\frac{\cos 5\theta }{\cos 3\theta }$

$=\frac{\cos (3\theta +2\theta )}{\cos 3\theta }$

$=\frac{\cos 3\theta \cos 2\theta -\sin 3\theta \sin 2\theta }{\cos 3\theta }$

$=\frac{\cos 3\theta \cos 2\theta }{\cos 3\theta }-\frac{\sin 3\theta \sin 2\theta }{\cos 3\theta }$

$=\cos 2\theta -\sin 2\theta \tan 3\theta$

R.H.S.

Hence, this is the answer.