<s>[INST] Prove value of $\sin(15°)$ I'm doing the following exercise: prove that 
$$
\sin(15°)=\frac{1}{2\sqrt{2+\sqrt{3}}}
$$
I'm using this formula: 
$$
\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)
$$
to get: 
$$
\sin(45-30)=\sin(45)\cos(30)-\cos(45)\sin(30) \\
=\frac{1}{\sqrt2}\frac{\sqrt3}{2}-\frac{1}{\sqrt2}\frac{1}{2} \\
=\frac{\sqrt3-1}{2\sqrt2}
$$
However I'm stuck, can't seem to get to the desired result. I tried multiplying for $$\frac{\sqrt{3}+1}{\sqrt{3}+1}$$ but didn't get anything out of it. Any help will be really appreciated. 
 [/INST] They are equivalent.  Since $$(1 + \sqrt{3})^2 = 1 + 2 \sqrt{3} + 3 = 2(2 + \sqrt{3}),$$ it follows that $$\sqrt{2 + \sqrt{3}} = \frac{1 + \sqrt{3}}{\sqrt{2}}.$$  Then $$\frac{1}{2 \sqrt{2 + \sqrt{3}}} = \frac{\sqrt{2}}{2 (1 + \sqrt{3})} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}.$$

It is also worth noting that we can obtain the desired expression more directly via the half-angle identity for $0 \le \theta \le 90^\circ$:  $$\sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} = \frac{\sqrt{1 - \cos^2 \theta}}{\sqrt{2 (1 + \cos \theta)}} = \frac{\sin \theta}{\sqrt{2 (1 + \cos \theta)}},$$ where upon selecting $\theta = 30^\circ$ immediately yields $$\sin 15^\circ = \frac{1}{2 \sqrt{2 + \sqrt{3}}}$$
</s>