Question

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

Answer 1

Consider the L.H.S.

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

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

We know that

$\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cdot \cos \left(\frac{C-D}{2}\right)$

Therefore,

$=\frac{2\cos \left(\frac{5\theta +3\theta }{2}\right)\cdot \cos \left(\frac{5\theta -3\theta }{2}\right)+2\cos \left(\frac{7\theta +5\theta }{2}\right)\cdot \cos \left(\frac{7\theta -5\theta }{2}\right)}{2\cos \left(\frac{5\theta +3\theta }{2}\right)\cdot \cos \left(\frac{5\theta -3\theta }{2}\right)+2\cos \left(\frac{3\theta +\theta }{2}\right)\cdot \cos \left(\frac{3\theta -\theta }{2}\right)}$

$=\frac{\cos \left(\frac{8\theta }{2}\right)\cdot \cos \left(\frac{2\theta }{2}\right)+\cos \left(\frac{12\theta }{2}\right)\cdot \cos \left(\frac{2\theta }{2}\right)}{\cos \left(\frac{8\theta }{2}\right)\cdot \cos \left(\frac{2\theta }{2}\right)+\cos \left(\frac{4\theta }{2}\right)\cdot \cos \left(\frac{2\theta }{2}\right)}$

$=\frac{\cos 4\theta \cdot \cos \theta +\cos 6\theta \cdot \cos \theta }{\cos 4\theta \cdot \cos \theta +\cos 2\theta \cdot \cos \theta }$

$=\frac{\cos 6\theta +\cos 4\theta }{\cos 4\theta +\cos 2\theta }$

$=\frac{2\cos \left(\frac{6\theta +4\theta }{2}\right)\cdot \cos \left(\frac{6\theta -4\theta }{2}\right)}{2\cos \left(\frac{4\theta +2\theta }{2}\right)\cdot \cos \left(\frac{4\theta -2\theta }{2}\right)}$

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

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

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

We know that

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

Therefore,

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

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

R.H.S

Hence, proved.