Prove that \[ \cos^2\left(\frac{\pi}{4}-x\right) – \sin^2\left(\frac{\pi}{4}-x\right) = \sin2x \]
Proof:
Using the identity
\[
\cos^2A-\sin^2A=\cos2A
\]
let
\[
A=\frac{\pi}{4}-x
\]
Then,
\[
LHS
=
\cos\left[
2\left(
\frac{\pi}{4}-x
\right)
\right]
\]
\[
=
\cos\left(
\frac{\pi}{2}-2x
\right)
\]
Using the identity
\[
\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta
\]
we get
\[
LHS=\sin2x
\]
Hence proved,
\[
\boxed{
\cos^2\left(\frac{\pi}{4}-x\right)
–
\sin^2\left(\frac{\pi}{4}-x\right)
=
\sin2x
}
\]