Prove that

Solution:

To Prove: $\cos ^{4} x+\sin ^{4} x=\frac{1}{2}\left(2-\sin ^{2} 2 x\right)$

Taking LHS,

$=\cos ^{4} x+\sin ^{4} x$

Adding and subtracting $2 \sin ^{2} x \cos ^{2} x$, we get

$=\cos ^{4} x+\sin ^{4} x+2 \sin ^{2} x \cos ^{2} x-2 \sin ^{2} x \cos ^{2} x$

We know that,

$a^{2}+b^{2}+2 a b=(a+b)^{2}$

$=\left(\cos ^{2} x+\sin ^{2} x\right)-2 \sin ^{2} x \cos ^{2} x$

$=(1)-2 \sin ^{2} x \cos ^{2} x\left[\because \cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$=1-2 \sin ^{2} x \cos ^{2} x$

Multiply and divide by 2, we get

$=\frac{1}{2}\left[2 \times\left(1-2 \sin ^{2} x \cos ^{2} x\right)\right]$

$=\frac{1}{2}\left[2-4 \sin ^{2} x \cos ^{2} x\right]$

$=\frac{1}{2}\left[2-(2 \sin x \cos x)^{2}\right]$

$=\frac{1}{2}\left[2-(\sin 2 x)^{2}\right][\because \sin 2 x=2 \sin x \cos x]$

$=\frac{1}{2}\left(2-\sin ^{2} 2 x\right)$

= RHS

∴ LHS = RHS

Hence Proved