In any $△ A B C$ < prove that
$a cos A + b cos B + c cos C = 2 a sin B sin C$ .

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Solution

$L H S - k sin A cos A + k = s i n B cos B + k sin C cos C$
$= \frac{k}{2} [sin 2 A + sin 2 B + sin 2 C]$
$= \frac{k}{2} [2 sin (A + B) cos (A - B) + 2 sin C cos C]$
$= k [sin (π - C) cos (A - B) + sin C cos (π - (A + B))]$
$= k [sin C cos (A - B) - sin C cos (A + B)]$
$= k sin C [- 2 sin A . sin (- B)]$
$= 2 k sin C sin A sin B$
$= 2 a sin B sin C$
$= R H S$

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