Prove that $a cos A + b cos B + c cos C = 4 R sin A sin B sin C$

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Solution

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

By sine rule

$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = R$

using trigonometric identities and simplifying

$= 2 R (sin A cos A + sin B cos B + sin C cos C)$

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

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

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

$= 4 R sin A sin B sin C$

$=$ R.H.S

Hence proved.

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