<s>[INST] 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 ?
 [/INST] 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}
</s>