Prove that \[ \sin^2\left(\frac{\pi}{8}+\frac{x}{2}\right) – \sin^2\left(\frac{\pi}{8}-\frac{x}{2}\right) = \frac{1}{\sqrt2}\sin x \]

Proof: Using the identity \[ \sin^2A-\sin^2B=(\sin A-\sin B)(\sin A+\sin B) \] let \[ A=\frac{\pi}{8}+\frac{x}{2} \] and \[ B=\frac{\pi}{8}-\frac{x}{2} \] Then, \[ LHS= (\sin A-\sin B)(\sin A+\sin B) \] Using the identities \[ \sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \] and \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] we get \[ LHS= \left( 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \right) \] \[ \times \left( 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \right) \] Now, \[ A+B=\frac{\pi}{4} \] and \[ A-B=x \] Therefore, \[ LHS= 2\cos\frac{\pi}{8}\sin\frac{x}{2} \cdot 2\sin\frac{\pi}{8}\cos\frac{x}{2} \] \[ = 4\sin\frac{\pi}{8}\cos\frac{\pi}{8} \sin\frac{x}{2}\cos\frac{x}{2} \] Using \[ 2\sin\theta\cos\theta=\sin2\theta \] we get \[ 2\sin\frac{\pi}{8}\cos\frac{\pi}{8} = \sin\frac{\pi}{4} = \frac{1}{\sqrt2} \] and \[ 2\sin\frac{x}{2}\cos\frac{x}{2} = \sin x \] Hence, \[ LHS= \frac{1}{\sqrt2}\sin x \] Therefore, \[ \boxed{ \sin^2\left(\frac{\pi}{8}+\frac{x}{2}\right) – \sin^2\left(\frac{\pi}{8}-\frac{x}{2}\right) = \frac{1}{\sqrt2}\sin x } \]