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If $\sin^{3} x \sin 3 x = \sum_{m = 0}^{6} c_{m} \cos^{m} x$ where $c_{0}, c_{1}, ..... c_{6}$ are constant then______

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$2 c_{2} + 3 c_{6} = 0$

$c_{1} + c_{3} + c_{5} = 6$

$c_{4} + 2 c_{6} = 0$

$c_{0} + c_{2} + c_{4} + c_{6} = 0$

answer is A.

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Detailed Solution

$\sin^{3} x (3 \sin x - 4 \sin^{3} x) = \sin^{4} x (3 - 4 \sin^{2} x) = {(1 - \cos^{2} x)}^{2} [3 - 4 (1 - \cos^{2} x)] = 4 \cos^{6} x - 9 \cos^{4} x + 6 \cos^{2} x - 1$

$C_{0} = - 1, C_{1} = 0, C_{2} = 6, C_{3} = 0, C_{4} = - 9, C_{5} = 0, C_{6} = 4$

$c_{0} + c_{2} + c_{4} + c_{6} = 0$

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