Prove that: $\cos 4 x=1-8 \sin ^{2} x \cos ^{2} x$
Prove that: $\cos 4 x=1-8 \sin ^{2} x \cos ^{2} x$
$L.H.S. $ $=\cos 4 x$
$=\cos 2(2 x)$
$=1-2 \sin ^{2} 2 x\left[\cos 2 A=1-2 \sin ^{2} A\right]$
$=1-2(2 \sin x \cos x)^{2}[\sin 2 A=2 \sin A \cos A]$
$=1-8 \sin ^{2} x \cos ^{2} x$
$=$ $R.H.S.$
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