Question

In any triangle, prove that $b\cos B+c\cos C=a\cos (B-C)$

Answer 1

We have;

According to Sine law;

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Now;

$b\cos B=c.\cos C$

$=2R\sin B.\cos B+2R\sin C.\cos C$

$=R(\sin 2B+\sin 2C)$

$=R\times 2\times \sin \left(\frac{2B+2C}{2}\right).\cos \left(\frac{2B-2C}{2}\right)$

$[\because \sin C+\sin D=2\sin (\frac{C+D}{2}\left)\cos \right(\frac{C-D}{2}\left)\right]$

$=2R\sin (B+C).\cos (B-C)$

$=2R\sin (\pi -A).\cos (B-C)$ $[\because A+B+C=\pi ]$

$=2R\sin A.\cos (B-C)$

$=a.\cos (B-C)$

Hence;

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