Prove that:
$(sin 3 A + sin A) sin A + (cos 3 A - cos A) cos A = 0$

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Solution

$L H S = (sin 3 A + sin A) sin A + cos A (cos 3 A - cos A) = (4 sin A - 4 sin^{3} A) sin A + cos A (4 cos^{3} A - 4 cos A) = 4 sin A (1 - sin^{2} A) sin A + 4 cos A (cos^{2} A - 1) cos A = 4 sin^{2} A (1 - sin^{2} A) - 4 cos^{2} A (1 - cos^{2} A) = 4 sin^{2} A cos^{2} A - 4 sin^{2} A cos^{2} A = 0 = R H S$

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