Trigonometric identity. How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$? How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$ ?
My failed take on this matter is:
$$
\sin A = 
\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big)=
2\sin\Big(\frac{\frac{2A}2}2\Big)\cos\frac02 =
2\sin\frac{A}2
$$
where $\cos\frac02 = \cos0 = 1$

First of all $$\sin x\neq \sin (x/2)+\sin (x/2)$$
In fact $$\sin (a+b)=\sin a\cos b+\cos a\sin b$$
Substitute $a=b=x/2$  to get $$\sin x= 2\sin (x/2)\cos (x/2)$$
