Question

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

• A. $2+4\sin \alpha$
• B. $2-4\sin \alpha$
• C. $2$
• D. None of these

Answer 1

The correct option is C $2$

Given, $\alpha \in (\pi ,\frac{3\pi }{2})$
$\therefore \sin \alpha <0,\cos \alpha <0,\tan \alpha >0$
$\sqrt{4{\sin }^4\alpha +{\sin }^{2}2\alpha }+4{\cos }^2(\frac{\pi }{4}-\frac{\alpha }{2})$
We know $\sin 2\alpha =2\sin \alpha \cos \alpha ,2{\cos }^2=1+\cos 2x$
$⟹\sqrt{4{\sin }^2\alpha ({\sin }^2\alpha +{\cos }^2\alpha )}+4{\cos }^2(\frac{\pi }{4}-\frac{\alpha }{2})$
$=-2\sin \alpha +2(1+\cos (\frac{\pi }{2}-\alpha ))$
$=-2\sin \alpha +2+2\sin \alpha =2$