Question

Prove that: A cosA+b cos B+c cos C=2 a sin B sinC

Answer 1

= a cos A + b cos B + c cos C
Let, $k=\frac{a}{sinA}$
So, the LHS reduces to-
$=\left(\frac{k}{2}\right)$ [ 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ]
$=\left(\frac{k}{2}\right)$ [ ( sin 2A + sin 2B ) + sin 2C ]
$=\left(\frac{k}{2}\right)$ [ 2 sin (A+B)· cos(A-B) + sin 2C ]
Using the identity:
$sinA\pm sinB=2sin\left(\frac{A\pm B}{2}\right)cos\left(\frac{A\mp B}{2}\right)$
$=\left(\frac{k}{2}\right)$ [ 2 sin C. cos(A-B) + 2 sin C cos C ]
= k sin C [ cos(A-B) + cos C ] Substituting, cos \(C =cos [\pi-(A+B)]=-cos(A+B)\)
= k sin C [ cos(A-B) - cos(A+B) ]
= k sin C [ 2 sin A sin B ]
= 2 ( k sin A ) sin B sin C
= 2 a sin B sin C = RHS