Question

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

Answer 1

b'

L.H.S$=\cos \theta +\cos 7\theta +cos3\theta +\cos 5\theta$

$=(\cos \theta +\cos 7\theta )+(\cos 3\theta +\cos 5\theta )$

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

$=2\cos 4\theta \cos 3\theta +2\cos 4\theta \cos \theta$

$=2\cos 4\theta (\cos 3\theta +\cos \theta )$

$=2\cos 4\theta \left(2\cos \left(\frac{3\theta +\theta }{2}\right)\cos \left(\frac{3\theta -\theta }{2}\right)\right)$

$=4\cos 4\theta \cos 2\theta \cos \theta =$R.H.S

Hence proved