Question

Prove that:

${\cos }^2\left(\frac{\pi }{4}-x\right)-{\sin }^2\left(\frac{\pi }{4}-x\right)=\sin 2x$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\cos }^2\left(\frac{\mathrm{\pi }}{4}-x\right)-{\sin }^2\left(\frac{\mathrm{\pi }}{4}-x\right) \\ =\cos 2\left(\frac{\mathrm{\pi }}{4}-x\right)\left[\because {\cos }^2\mathrm{\alpha }-{\sin }^2\mathrm{\alpha }=\cos 2\mathrm{\alpha }\right] \\ =\cos \left(\frac{\mathrm{\pi }}{2}-2x\right) \\ =\sin 2x=\mathrm{RHS}\left[\because \cos \left(\frac{\mathrm{\pi }}{2}-2\mathrm{\alpha }\right)=\sin 2\mathrm{\alpha }\right] \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$