Question

Prove that : $\frac{\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10}{\cos 5\theta +5\cos 3\theta +10\cos \theta }=2\cos \theta$

Answer 1

We have $2\cos \theta (\cos 5\theta +5\cos 3\theta +10\cos \theta )$

$=2\cos \theta \cos 5\theta +10\cos \theta \cos 3\theta +20\cos \theta \cos 5\theta$ using $\cos A\cos B=\frac{1}{2}[\cos (A+B)+\cos (A-B)]$

$=\cos 4\theta +\cos 6\theta +5\cos 2\theta +5\cos 4\theta +10\cos 2\theta +10\cos 0$

$=\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10$

$\Rightarrow 2\cos \theta (\cos 5\theta +5\cos 3\theta +10\cos \theta )=\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10$

$\Rightarrow 2\cos \theta =\frac{\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10}{(\cos 5\theta +5\cos 3\theta +10\cos \theta )}$

or $\frac{\cos 6\theta +6\cos 4\theta +15\cos 2\theta +10}{(\cos 5\theta +5\cos 3\theta +10\cos \theta )}=2\cos \theta$

Hence proved.