If $\sin^{3}{xsin\ 3x = \sum_{m = 0}^{n}C_{m}\cos{mx}}$ is an identity in $x$, where $C_{0},\ C_{1},\ldots,C_{n}$ are constants and $C_{n}
eq 0$, then the value of $n$ equals

Question

If $\sin^{3}{xsin\ 3x = \sum_{m = 0}^{n}C_{m}\cos{mx}}$ is an identity in $x$, where $C_{0},\ C_{1},\ldots,C_{n}$ are constants and $C_{n} 
eq 0$, then the value of $n$ equals

Dr. Abubacker · Accepted Answer

(c)
Given, $\sin^{3}{x\sin{3x = \sum_{m = 0}^{n}C_{m}}}\cos{mx}$
$\Rightarrow \ \ \ \left( \frac{3\sin{x - \sin{3x}}}{4} ight)\sin{3x = \sum_{m = 0}^{n}C_{m}\cos{mx}}$
$\Rightarrow \ \frac{3}{8}\ \left( 2\sin{3x\sin x} ight) - \frac{1}{8}\left( 2\sin^{2}{3x} ight) = \sum_{m = 0}^{n}C_{m}\cos{mx}$
$\Rightarrow \ \frac{3}{8}\ \left( \cos{2x - \cos{4x}} ight) - \frac{1}{8}\left( 1 - \cos{6x} ight) = \sum_{m = 0}^{n}{C_{m}\cos{mx}}$
$\Rightarrow \ \frac{1}{8}\cos{6x + \frac{3}{8}\left( \cos{2x - \cos{4x}} ight) - \frac{1}{8}}$
$= C_{0}\cos{0 + C_{1}\cos{x + C_{2}\cos{2x + ... + C_{n}\cos{nx}}}}$
$	herefore\ \ \ n = 6$

Q. If $\sin^{3}{xsin\ 3x = \sum_{m = 0}^{n}C_{m}\cos{mx}}$ is an identity in $x$, where $C_{0},\ C_{1},\ldots,C_{n}$ are constants and $C_{n} \neq 0$, then the value of $n$ equals

Explanation

Reviewed by Dr. Abubacker

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