Question

Given that ${\sin }^2\beta =\sin \alpha .\cos \alpha ;$ show that $\cos 2\beta =2{\cos }^2\left(\frac{\pi }{4}+\alpha \right)$.

Answer 1

Given,

$\Rightarrow {\sin }^2\beta =\sin \alpha \cos \alpha$ (multiplying by $2$ both side)

$\Rightarrow 2{\sin }^2\beta =2\sin \alpha \cos \alpha$

$\Rightarrow 1-\cos 2\beta =\sin 2\alpha$

$\Rightarrow \cos 2\beta =1-\sin 2\alpha$

Taking RHs

$\Rightarrow 2{\cos }^2(\frac{\pi }{4}+\alpha )$

$\Rightarrow 1+\cos (\frac{\pi }{2}+2\alpha )$

$\Rightarrow 1+(\cos \frac{\pi }{2}\cos 2\alpha -\sin \frac{\pi }{2}\sin 2\alpha )$

$\Rightarrow 1-\sin 2\alpha$

$\therefore$ LHS$=$RHS proved.