Question

Prove that $\frac{\left(\sin 5\theta \right)}{\left(\sin \theta \right)}=16{\cos }^4\theta -12{\cos }^2\theta +1$

Answer 1

L.H.S$=\frac{\sin 5\theta }{\sin \theta }$

$=\frac{\sin (4\theta +\theta )}{\sin \theta }$

$=\frac{\sin 4\theta \cos \theta +\cos 4\theta \sin \theta }{\sin \theta }$

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

$=\frac{4\sin \theta \cos 2\theta {\cos }^2\theta +\cos 4\theta \sin \theta }{\sin \theta }$

$=4\cos 2\theta {\cos }^2\theta +\cos 4\theta$

$=4(2{\cos }^2\theta -1){\cos }^2\theta +(2{\cos }^{2}2\theta -1)$

$=(8{\cos }^2\theta -4){\cos }^2\theta +(2{\cos }^{2}2\theta -1)$

$=8{\cos }^4\theta -4{\cos }^2\theta +2{\cos }^{2}2\theta -1$

$=8{\cos }^4\theta -4{\cos }^2\theta +2{(2{\cos }^2\theta -1)}^2-1$

$=8{\cos }^4\theta -4{\cos }^2\theta +2(4{\cos }^4\theta +1-4{\cos }^2\theta -1$

$=8{\cos }^4\theta -4{\cos }^2\theta +8{\cos }^4\theta +2-8{\cos }^2\theta -1$

$=16{\cos }^4\theta -12{\cos }^2\theta +1=$R.H.S

Hence proved.