Question:
Prove that:
$\cos ^{2}\left(\frac{\pi}{4}-x\right)-\sin ^{2}\left(\frac{\pi}{4}-x\right)=\sin 2 x$
Solution:
LHS $=\cos ^{2}\left(\frac{\pi}{4}-x\right)-\sin ^{2}\left(\frac{\pi}{4}-x\right)$
$=\cos 2\left(\frac{\pi}{4}-x\right)$ $\left[\because \cos ^{2} \alpha-\sin ^{2} \alpha=\cos 2 \alpha\right]$
$=\cos \left(\frac{\pi}{2}-2 x\right)$
$=\sin 2 x=\mathrm{RHS}$ $\left[\because \cos \left(\frac{\pi}{2}-2 \alpha\right)=\sin 2 \alpha\right]$
Hence proved.