Understanding this trigonometric identity $\frac{n}{2} (2 \cos \frac{\alpha_n}{2} \sin \frac{\alpha_n}{2}) = \frac{n}{2} \sin \alpha_n$

Question

I was looking an example that motivates rounding errors using the quadrature of a circle (?). And at a certain point there's this identity:

$$\frac{n}{2} \left( 2 \cos \frac{\alpha_n}{2} \sin \frac{\alpha_n}{2} \right) = \frac{n}{2} \sin \alpha_n$$

which implies that

$$\left( 2 \cos \frac{\alpha_n}{2} \sin \frac{\alpha_n}{2} \right) = \sin \alpha_n$$

I don't remember of having seen this identity, but it's highly possible that my memory was formatted.

Answer 1

Its double angle formula ie $\sin(2x)=2\sin(x)\cos(x)$ which can be proved like this $\sin(2x)=\sin(x+x)=\sin(x)\cos(x)+\sin(x)\cos(x)=2\sin(x)\cos(x)$ as $sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$