Question

Prove that: $\cos 6x=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$

Answer 1

$\cos 6x=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1(\because \cos 2A=2{\cos }^{2}A-1)$

$L.H.S=\cos 6x=\cos 2\left(3x\right)=2{\cos }^{2}3x-1(\because \cos 3A=4{\cos }^{3}A-3\cos A)$
$=2(4{\cos }^{3}x-3\cos x{)}^2-1$
$=2(16{\cos }^{6}x+9{\cos }^{2}x-24{\cos }^{4}x)-1$
$=32{\cos }^{6}x+18{\cos }^{2}x-48{\cos }^{4}x-1$
$=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$

$\Rightarrow L.H.S=R.H.S.$
Hence proved.