In any triangle, prove that:

Question

In any triangle, prove that: $a\cos A+b\cos B+c\cos C= 4R\sin A.\sin B.\sin C$

My Attempt: $$L.H.S=a\cos A+ b\sin A+ c\sin C$$ $$=2R.\sin A. \cos A+ 2R\sin B\cos B+2R\sin C\cos C$$ $$=R.\sin (2A)+R.\sin (2B)+R.\sin (2C)$$ $$=R(\sin (2A)+\sin (2B)+\sin (2C)$$

Answer 1

$$\cfrac{a}{\sin A} = \cfrac{b}{\sin B} = \cfrac{c}{\sin C} = 2R$$
L.H.S.
$$R(\sin {2A} + \sin {2B} + \sin {2C}) = R(2\sin (A+B) \cos(A-B) + 2 \sin C \cos C) \\= 2R\sin C ( \cos(A-B) + \cos C)\\ = 2R\sin C(\cos(A-B) + \cos(\pi - A - B))\\ = 2R\sin C(\cos(A-B) - \cos(A + B)) \\ =4R \sin A \sin B \sin C$$