Question

Prove that : $\cos 4A=1-8{\cos }^{2}A+8{\cos }^{4}A$

Answer 1

L.H.S$=\cos 4A=\cos 2\left(2A\right)$

$=2{\cos }^{2}2A-1$

$=2{(2{\cos }^{2}A-1)}^2-1$

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

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

$=8{\cos }^{4}A-8{\cos }^{2}A-1=$R.H.S

Hence proved.