Question

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

Answer 1

Solution:

given

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

multiplying $2$ on both sides

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

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

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

Now for $\cos 2\beta =2{\cos }^2(\frac{\pi }{4}+\alpha )$

We know

$\cos \frac{\pi }{2}=0$

$\sin \frac{\pi }{2}=1$

Taking $RHS$

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

$\Rightarrow 1+\cos (\frac{2\pi }{4}+2\alpha )$

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

$\Rightarrow 1+[\cos \frac{\pi }{2}\cos 2\alpha -\sin \frac{\pi }{2}\sin \alpha ]$

$\Rightarrow 1-\sin 2\alpha$

$\therefore LHD=RHS$ Proved