Question

In any triangle ABC, prove the following:

$b\cos B+c\cos C=a\cos (B-C)$

Answer 1

Given: ABC is a triangle, $[\therefore A+B+C=\pi ]$

According to sine rule

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k...\left(i\right)$

We have to prove:

$b\cos B+c\cos C=a\cos (B-C)$

$L.H.S=b\cos B+c\cos C$

Putting the values from equation (i)

$L.H.S=k(\sin B\cos B+\sin C\cos C)$

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

$=\frac{k}{2}(\sin 2B+\sin 2C)...\left(ii\right)$

$\text{Now},R.H.S=a\cos (B-C)$

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

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

Using the transformation formula

$[\because 2\sin A\cos B=\sin (A+B)+\sin (A-B\left)\right]$

$R.H.S=\frac{k}{2}\left[\sin \right(A+B-C)+\sin (A-B+C\left)\right]$

$=\frac{k}{2}\left[\sin \right(\pi -C-C)+\sin (\pi -B-B\left)\right]$

$[\because A+B+C=\pi ]$

$R.H.S=\frac{k}{2}(\sin 2C+\sin 2B)$

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

$=L.H.S\left(\text{from}\right(ii\left)\right)$

Hence, proved.