Prove that in acute triangle : $\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$ Im going to prove this identity in acute triangle : 
Let $ABC$ acute triangle , $A,B,C$ are angles then : 

$$\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$$

I know that $A,B,C<\frac{π}{2}$ 
$$\sin A\cos (B-C)=\sin A(\cos B\cos C+\sin A\sin B\sin C)$$
And also : 
$$\sin (2B)+\sin (2C)=\sin B\cos B +\sin B\cos B+\sin C\cos C +\sin C\cos C$$ 
But then I don't know how ?
 A: In the standard notation we obtain: $$\sin\alpha\cos(\beta-\gamma)=\frac{1}{2}(\sin(\alpha+\beta-\gamma)+\sin(\alpha-\beta+\gamma))=$$
$$=\frac{1}{2}(\sin(180^{\circ}-2\gamma)+\sin(180^{\circ}-2\beta))=\frac{1}{2}(\sin2\gamma+\sin2\beta).$$
A: Since $A,B,C$ are angles of a triangle, we have that $A+B+C = \pi$. Recall the following trigonometric identities
\begin{align}
\sin(\pi-\theta) & = \sin(\theta)\\
\sin(2\theta) + \sin(2\phi) & = 2 \sin(\theta + \phi) \cos(\theta-\phi)\\
\end{align}
We have that
\begin{align}
\sin(2B) + \sin(2C) & = 2 \sin(B+C) \cos(B-C)\\
& = 2 \sin(\pi-A) \cos(B-C)\\
& = 2 \sin(A) \cos(B-C)
\end{align}
A: First, we need to derive this little identity:
$\sin (x +y) = \sin x\cos y - \cos x\sin y\\
\sin (x - y) = \sin x\cos y + \cos x\sin y\\
\sin (x +y) + \sin(x-y) = 2\sin x\cos y\\
\sin x\cos y = \frac{\sin (x +y) + \sin(x-y)}{2}$
Now we can apply it with $x = A, y = B-C$
And while we are at it, we can derive similar identities to show that
$\cos x\cos y = \frac {\cos (x+y) + \cos (x-y)}{2}\\
\sin x\sin y = \frac {\cos (x-y) - \cos (x+y)}{2}$
