Proving $(b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2}$ for any $\triangle ABC$ I found this problem on Mathematics, Textbook for Class XI by NCERT, ed. January 2021 that uses the sine and cosine formulas along with standard trigonometric identities for proving an identity.
By triangle ABC, the question assumes that the angles of the triangle are A, B and C, and the sides opposite the angles A, B and C, are a, b and c respectively.

For any triangle $ABC$, prove that:
$$(b + c) \cos \frac{B + C}{2} = a \cos \frac{B - C}{2}$$

I've tried to approach this problem by simplifying the
$$\cos \frac{B+C}{2}$$
part into
$$\sin\frac{A}{2}$$
and assuming that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$$
keeping in mind the sine law. However, I haven't been able to solve it yet.
A guidance would be helpful even if the question is not solved in its entirety.
 A: $$\sin(B)+\sin(C)=2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)$$
Multiply $\cos\left(\frac{B+C}{2}\right)$ on both sides, and use double angle formula for sine
$$(\sin(B)+\sin(C))\cos\left(\frac{B+C}{2}\right)=\sin(B+C)\cos\left(\frac{B-C}{2}\right)$$
$\sin(B+C)=\sin(\pi-A)=\sin(A)$
$$(\sin(B)+\sin(C))\cos\left(\frac{B+C}{2}\right)=\sin(A)\cos\left(\frac{B-C}{2}\right)$$
Use sine law: $\sin(A)=\frac{a}k, \sin(B)=\frac{b}k, \sin(C)=\frac{c}k$
$$(b+c)\cos\left(\frac{B+C}{2}\right)=a\cos\left(\frac{B-C}{2}\right)$$
A: First you need to know The sum to product identities:
$$\sin B+\sin C=2\sin(\frac{B+C}{2})\cos(\frac{B-C}{2}) $$
Then the problem is solved:
$$
\begin{equation*}
  \begin{aligned}
    (b+c)\cos(\frac{B+C}{2})-a\cos(\frac{B-C}{2}) &= R[(\sin B+\sin C)\sin(\frac{A}{2})-\sin A\cos(\frac{B-C}{2})] \\
         &= R\sin(\frac{A}{2})[(\sin B+\sin C)-2\cos (\frac{A}{2})\cos(\frac{B-C}{2})] \\
         &= R\sin(\frac{A}{2})[(\sin B+\sin C)-2\sin (\frac{B+C}{2})\cos(\frac{B-C}{2})] \\
         &= 0\\
  \end{aligned}
\end{equation*}
$$
A: For fun, here's a trigonographic proof (for $B>C$):

