Question

If $\pi <\alpha <\frac{3\pi }{2}$, then find the value of expression $\sqrt{4{\sin }^4\alpha +{\sin }^{2}2\alpha }+4{\cos }^2(\frac{\pi }{4}-\frac{\alpha }{2})$.

Answer 1

As we know that

$\pi <\alpha <\frac{3\pi }{2},\sin \alpha =-ve$

Given that:

$\sqrt{4{\sin }^4\alpha +{\sin }^{2}2\alpha }+4co{s}^2(\frac{\pi }{4}-\frac{\alpha }{2})$

$=\sqrt{4{\sin }^4\alpha +4{\sin }^2\alpha {\cos }^2\alpha }+4co{s}^2(\frac{\pi }{4}-\frac{\alpha }{2})$

$=\sqrt{4{\sin }^2\alpha ({\sin }^2\alpha +{\cos }^2\alpha )}+4co{s}^2(\frac{\pi }{4}-\frac{\alpha }{2})$

$=\sqrt{4{\sin }^2\alpha }+4co{s}^2\frac{\alpha }{2}$

$=2\sin \alpha +2[1+\cos (\frac{\pi }{2}-\alpha )]$

$=2\sin \alpha +2[1-\sin \alpha ]$

$=2\sin \alpha +2-2\sin \alpha$

$=2.$