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

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Solution

We have $a cos A + b cos B + c cos C$

We knew that sine Role
$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = k$

$a = k sin A, b = k sin B, C = k sin C$

$= a sin A cos A + k sin B cos B + k sin C cos C$

$= \frac{k}{2} [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]$

$= \frac{k}{2} [sin 2 A + 2 sin 2 B + 2 sin 2 C]$

$∴ △ A B C$

$= \frac{k}{2} [4 sin A sin B sin C]$

$= 2 a sin B sin C$

Hence this is the answer

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