Question:
Prove that:
$\cos 4 x=1-8 \cos ^{2} x+8 \cos ^{4} x$
Solution:
LHS $=\cos 4 x$
$=\cos (2 \times 2 x)$
$=2 \cos ^{2} \times 2 x-1 \quad\left[\because \cos 2 \theta=2 \cos ^{2} \theta-1\right]$
$=2\left(2 \cos ^{2} x-1\right)^{2}-1\left[\because \cos ^{2} 2 \theta=\left(2 \cos ^{2} \theta-1\right)^{2}\right]$
$=2\left(4 \cos ^{4} x-4 \cos ^{2} x+1\right)-1$
$=8 \cos ^{4} x-8 \cos ^{2} x+1$
$=1-8 \cos ^{2} x+8 \cos ^{4} x=$ RHS
Hence proved.
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