Trigonometric identity. How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$?

Question

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 $\cos\frac02 = \cos0 = 1$

Answer 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 $$\sin x= 2\sin (x/2)\cos (x/2)$$

Answer 2

Hint:

Your first step is wrong because $\sin(x)$ is not a linear function so: $$\sin x \ne \sin(\frac{x}{2})+\sin(\frac{x}{2})$$

Answer 3

In your work, you used the sum-to-product formula, $$\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big),$$

which is correct. The mistake is that $$\sin A\ne\sin\frac{A}2 + \sin\frac{A}2$$

You can still use the sum-to-product formula.

$$\sin A + \sin 0 = 2\sin\Big(\frac{A+0}2\Big)\cos\Big(\frac{A-0}2\Big),$$