Question

that $\cos 7\theta +\cos 5\theta +\cos 3\theta +\cos \theta =4\cos \theta \cos 2\theta \cos 4\theta$

Answer 1

$LHS=\cos 7\theta +\cos 5\theta +\cos 3\theta +\cos \theta$

$=2\cos 6\theta \cos \theta +2\cos 2\theta \cos \theta$

$=2\cos \theta (\cos 6\theta +\cos 2\theta )$

$=2\cos \theta \left(2\cos 4\theta \cos 2\theta \right)$

$=4\cos \theta \cos 2\theta \cos 4\theta =RHS$

$LHS=RHS$

Hence proved