Question

Prove that:

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

Answer 1

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