<s>[INST] How to express the equations as the Square root Like? $$2\sin \frac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$$
What law is need to be applied here? Do I have to make the $\frac{\pi}{16}$ in a form  that will be give us $\sqrt{2}$ like  sin 45 degree? 
 [/INST] Since, $\sin{\dfrac{\theta}{2}}=\sqrt{\dfrac{1-\cos\theta}{2}}$ and $\cos{\dfrac{\theta}
{2}}=\sqrt{\dfrac{1+\cos\theta}{2}}$
$\sin{\dfrac{\theta}{4}}=\sqrt{\dfrac{1-\cos{\dfrac{\theta}{2}}}{2}}$
$\sin{\dfrac{\theta}{4}}=\sqrt{\dfrac{1-\sqrt{\dfrac{1+\cos{\theta}}{2}}}{2}}$
Putting, $\theta=\dfrac{\pi}{4}$
$\sin{\dfrac{\pi}{16}}=\sqrt{\dfrac{1-\sqrt{\dfrac{1+\dfrac{1}{\sqrt2}}{2}}}{2}}=\dfrac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2}$
Now, multiply both sides by $2$ to get the required equation.
$2\sin \dfrac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$
</s>