Question

If $\frac{\pi }{2}<\theta <\pi$ the possible answers of $\sqrt{2+\sqrt{2+2\cos \theta }}$ is/are:

• A. $2\sin \theta$
• B. $2\cos \theta$
• C. $-2\sin \theta$
• D. $-2\cos \theta$

Answer 1

The correct option is B $2\cos \theta$

$\sqrt{2+\sqrt{2+2\cos 4\theta }}$ (since $1+\cos 2A=2{\cos }^{2}A$)

$\Rightarrow \sqrt{2+\sqrt{2(1+\cos 4\theta )}}$

$\Rightarrow \sqrt{2+\sqrt{2.2\left({\cos }^{2}2A\right)}}$

$\Rightarrow \sqrt{2+2.\left(\cos 2A\right)}$

$\Rightarrow \sqrt{2(1+\cos 2A)}$

$\Rightarrow \sqrt{2.2{\cos }^{2}A}$

$\Rightarrow \sqrt{{\left(2\cos A\right)}^2}$

$\Rightarrow 2\cos A$

Option $\left(B\right)$