Question:
Prove that:
$1+\cos ^{2} 2 x=2\left(\cos ^{4} x+\sin ^{4} x\right)$
Solution:
$\mathrm{LHS}=1+\cos ^{2} 2 x$
Using the identity $\cos 2 x=\cos ^{2} x-s \operatorname{in}^{2} x$, we get
LHS $=1+\left(\cos ^{2} x-\sin ^{2} x\right)^{2}$
$=1+\cos ^{4} x+\sin ^{4} x-2 \cos ^{2} x \sin ^{2} x$
$=\left(\cos ^{2} x+\sin ^{2} x\right)^{2}+\cos ^{4} x+\sin ^{4} x-2 \cos ^{2} x \sin ^{2} x \quad\left[\because \cos ^{2} x+\sin ^{2} x=1\right]$
$=\cos ^{4} x+\sin ^{4} x+2 \cos ^{2} x \sin ^{2} x+\cos ^{4} x+\sin ^{4} x-2 \cos ^{2} x \sin ^{2} x$
$=2\left(\cos ^{4} x+\sin ^{4} x\right)=\mathrm{RHS}$
Hence proved.
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