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$sin A + sin 3 A + sin 5 A + sin 7 A =$

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Solution

$sin A + sin 3 A + sin 5 A + sin 7 A = (sin A + sin 3 A) + (sin 5 A + sin 7 A)$
$= 2 sin (\frac{A + 3 A}{2}) cos (\frac{3 A - A}{2}) + 2 sin (\frac{5 A + 7 A}{2}) cos (\frac{7 A - 5 A}{2})$
$= 2 sin 2 A cos A + 2 sin 6 A cos A = 2 cos A (sin 2 A + sin 6 A)$
$= 2 cos A (2 sin (\frac{2 A + 6 A}{2}) cos (\frac{6 A - 2 A}{2})) = 4 cos A cos 2 A sin 4 A$

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