$\sin ^{2}\left(\frac{\pi}{8}+\frac{\theta}{2}\right)- \sin ^{2}\left(\frac{\pi}{8}-\frac{\theta}{2}\right)= ?$

Solve:$\begin{array}{l} =\sin ^{2}\left(\frac{\pi}{8}+\frac{\theta}{2}\right)-\sin ^{2}\left(\frac{\pi}{8}-\frac{\theta}{2}\right) \\ =\frac{1}{2}\left\{1-\cos 2\left(\frac{\pi}{8}+\frac{\theta}{2}\right)\right\} \\ -\frac{1}{2}\left\{1-\cos 2\left(\frac{\pi}{8}-\frac{\theta}{2}\right)\right\} \\ =\frac{1}{2}\left\{1-\cos \left(\frac{\pi}{4}+\theta\right)-1+\cos \left(\frac{\pi}{4}-\theta\right)\right\} \\ =\frac{1}{2}\left\{\cos \left(\frac{\pi}{4}-\theta\right)-\cos \left(\frac{\pi}{4}+\theta\right)\right\} \\ =\frac{1}{2} .2 \sin \frac{\pi}{4} \sin \theta=\frac{1}{\sqrt{2}} \sin \theta \\ \end{array}$