In triangle ABC, prove the following:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the LHS of the equation $b \cos B+c \cos C=a \cos (B-C)$.

$\mathrm{LHS}=b \cos B+c \cos C$

$=k(\sin B \cos B+\sin C \cos C)$

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

$=\frac{k}{2}(\sin 2 B+\sin 2 C) \quad \ldots(1)$

$\mathrm{RHS}=a \cos (B-C)$

$=k \sin A \cos (B-C)$

$=\frac{k}{2}[2 \sin A \cos (B-C)]$

$=\frac{k}{2}[\sin (A+B-C)+\sin (A-B+C)] \quad[\because 2 \sin A \cos B=\sin (A+B)+\sin (A-B)]$

$=\frac{k}{2}[\sin (\pi-\mathrm{C}-\mathrm{C})+\sin (\pi-\mathrm{B}-\mathrm{B})] \quad[\because \sin (\pi-A)=\sin A, A+B+C=\pi]$

$=\frac{k}{2}(\sin 2 C+\sin 2 B)$

$=\frac{k}{2}(\sin 2 B+\sin 2 C)=$ LHS $\quad[$ from $(1)]$

Hence proved.