Question

Prove the identity
$\cos h^{2}x{\cos }^{2}x-\sin h^{2}x{\sin }^{2}x\equiv \frac{1}{2}(1+\cos h2x\cos 2x)$

Answer 1

Lets take the RHS

$\cos h^{2}x.{\cos }^{2}x-\sin h^{2}x.{\sin }^{2}x$.

We know

$\cos h^{2}x=\frac{\cos h2x+1}{2}$ ; $\sin h^{2}x=\frac{\cos h2x-1}{2}$

on substituting it,

$\left(\frac{\cos h2x+1}{2}\right).{\cos }^{2}x-\left(\frac{\cos h2x-1}{2}\right){\sin }^{2}x.$

$\frac{1}{2}(\cos h2x{\cos }^{2}x+{\cos }^{2}x-\cos h2x.{\sin }^{2}x+{\sin }^{2}x)$

$\frac{1}{2}\left(\cos h2x\right({\cos }^{2}x-{\sin }^{2}x)+{\cos }^{2}x+{\sin }^{2}x)$

$=\frac{1}{2}(\cos h2x.\cos 2x+{\cos }^{2}x+1-{\cos }^{2}x)$

$=\frac{1}{2}(\cos 2x.\cos 2x+1)$

hence RHS=LHS proved.