If {sin ^3}x sin 3x = sumlimits_{m = 0}^n {{c | vedclass

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(B) We know that $\sin 3x = 3 \sin x - 4 \sin^3 x$,so $\sin^3 x = \frac{1}{4}(3 \sin x - \sin 3x)$.Substituting this into the expression:${\sin ^3}x \

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$6$

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(B) We know that $\sin 3x = 3 \sin x - 4 \sin^3 x$,so $\sin^3 x = \frac{1}{4}(3 \sin x - \sin 3x)$.Substituting this into the expression:${\sin ^3}x \sin 3x = \frac{1}{4}(3 \sin x - \sin 3x) \sin 3x$$= \frac{3}{4} \sin x \sin 3x - \frac{1}{4} \sin^2 3x$Using the product-to-sum formulas $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$:$= \frac{3}{8}(2 \sin x \sin 3x) - \frac{1}{8}(2 \sin^2 3x)$$= \frac{3}{8}(\cos 2x - \cos 4x) - \frac{1}{8}(1 - \cos 6x)$$= -\frac{1}{8} + \frac{3}{8} \cos 2x - \frac{3}{8} \cos 4x + \frac{1}{8} \cos 6x$Comparing this with $\sum_{m=0}^n c_m \cos mx = c_0 + c_1 \cos x + c_2 \cos 2x + \dots + c_n \cos nx$,we see that the highest frequency term is $\cos 6x$,which implies $n = 6$.

If ${\sin ^3}x \sin 3x = \sum\limits_{m = 0}^n {{c_m} \cos mx} $ where ${c_0}, {c_1}, {c_2}, \dots, {c_n}$ are constants and ${c_n} \ne 0$,then the value of $n$ is

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