Question

If $\frac{\pi }{2}<\mathrm{\theta }<\pi ,$ the write the value of $\sqrt{2+\sqrt{2+2\cos 2\mathrm{\theta }}}$ in the simplest form.

Answer 1

We have,

$$\begin{gathered} \sqrt{2+\sqrt{2+2\cos 2\mathrm{\theta }}}=\sqrt{2+\sqrt{2\left(1+\cos 2\mathrm{\theta }\right)}} \\ =\sqrt{2+\sqrt{2.2{\cos }^2\mathrm{\theta }}} \\ =\sqrt{2+2\left|\mathrm{cos\theta }\right|} \\ =\sqrt{2-2\mathrm{cos\theta }}\left(\because \frac{\mathrm{\pi }}{2}<\mathrm{\theta }<\mathrm{\pi }\right) \\ =\sqrt{2\left(1-\mathrm{cos\theta }\right)} \\ =\sqrt{2.2{\sin }^2\frac{\mathrm{\theta }}{2}} \\ =2\left|\sin \frac{\mathrm{\theta }}{2}\right| \\ =2\sin \frac{\mathrm{\theta }}{2}\left(\because \frac{\mathrm{\pi }}{4}<\frac{\mathrm{\theta }}{2}<\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$