Question

Prove that:
${\sin }^2\left(\frac{\pi }{8}+\frac{A}{2}\right)-{\sin }^2\left(\frac{\pi }{8}-\frac{A}{2}\right)=\frac{1}{\sqrt{2}}\sin A$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^2\left(\frac{\mathrm{\pi }}{8}+\frac{A}{2}\right)-{\sin }^2\left(\frac{\mathrm{\pi }}{8}-\frac{A}{2}\right) \\ =\frac{1}{2}\left\{1-\cos 2\left(\frac{\mathrm{\pi }}{8}+\frac{A}{2}\right)\right\}-\frac{1}{2}\left\{1-\cos 2\left(\frac{\mathrm{\pi }}{8}-\frac{A}{2}\right)\right\} \\ =\frac{1}{2}\left\{\cos \left(\frac{\mathrm{\pi }}{4}-A\right)-\cos \left(\frac{\mathrm{\pi }}{4}+A\right)\right\} \end{gathered}$$

Using the identity $\cos C-\cos D=-2\sin \frac{C+D}{2}\sin \frac{C-D}{2}$, we get

$$\begin{gathered} =\frac{1}{2}\left\{-2\sin \left(\frac{\left(\frac{\mathrm{\pi }}{4}-A\right)+\left(\frac{\mathrm{\pi }}{4}+A\right)}{2}\right)\sin \left(\frac{\left(\frac{\mathrm{\pi }}{4}-A\right)-\left(\frac{\mathrm{\pi }}{4}+A\right)}{2}\right)\right\} \\ =-\sin \frac{\mathrm{\pi }}{4}\sin \left(-A\right) \\ =\sin \frac{\mathrm{\pi }}{4}\sin A\left[\because \sin \left(-\mathrm{A}\right)=-\sin A\right] \\ =\frac{1}{\sqrt{2}}\sin A=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$