Question

In triangle ABC, prove the following:
$b\cos B+c\cos C=a\cos \left(B-C\right)$

Answer 1

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

Then,

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

$$\begin{gathered} \mathrm{LHS}=b\cos B+c\cos C \\ =k\left(\sin B\cos B+\sin C\cos C\right) \\ =\frac{k}{2}\left(2\mathrm{sinBcos}B+2\mathrm{sinCcos}C\right) \\ =\frac{k}{2}\left(\sin 2B+\sin 2C\right)...\left(1\right) \\ \mathrm{RHS}=a\cos \left(B-C\right) \\ =k\sin A\cos \left(B-C\right) \\ =\frac{k}{2}\left[2\sin A\cos \left(B-C\right)\right] \\ =\frac{k}{2}\left[\sin \left(A+B-C\right)+\sin \left(A-B+C\right)\right]\left[\because 2\sin A\cos B=\sin \left(A+B\right)+\sin \left(A-B\right)\right] \\ =\frac{k}{2}\left[\sin \left(\mathrm{\pi }-\mathrm{C}-\mathrm{C}\right)+\sin \left(\mathrm{\pi }-\mathrm{B}-\mathrm{B}\right)\right]\left[\because \sin \left(\mathrm{\pi }-A\right)=\sin A,A+B+C=\mathrm{\pi }\right] \\ =\frac{k}{2}\left(\sin 2C+\sin 2B\right) \\ =\frac{k}{2}\left(\sin 2B+\sin 2C\right)=\mathrm{LHS}\left[\mathrm{from}\left(1\right)\right] \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$