Given that $\;\sin^3x\sin3x = \sum^n_{m=0}C_m\cos mx\,,\; C_n \neq 0\;$ is an identity . Find the value of n. Problem : 
Given that $\sin^3 x \sin 3x = \sum^n_{m=0}C_m \cos mx, C_n \neq 0 $ is an identity. Find the value of n.
I tried : $\sin3x = 3\sin x - 4\sin^3 x$ but unable to reach to any point.... Please suggest further ....Thanks..
 A: Using the identity
$$
\sin^3(x)=\frac{3\sin(x)-\sin(3x)}{4}
$$
and the identity
$$
\sin(ax)\sin(bx)=\frac12\big(\cos((a-b)x)-\cos((a+b)x)\big)
$$
we get
$$
\begin{align}
\sin^3(x)\sin(3x)
&=\frac{3\sin(x)-\sin(3x)}{4}\sin(3x)\\
&=\frac38\big(\cos(2x)-\cos(4x)\big)-\frac18\big(\cos(0x)-\cos(6x)\big)\\
&=\frac18\big(\cos(6x)-3\cos(4x)+3\cos(2x)-\cos(0x)\big)
\end{align}
$$
Thus, $n=6$.
A: HINT:
As you have noticed, $4\sin^3x=3\sin x-\sin3x,$
$$\text{So, }\sin^3x\sin3x=\frac{(3\sin x-\sin3x)\sin3x}4=\frac{3(2\sin x\sin3x)-2\sin^23x}8$$
As $2\sin A\sin B=\cos(A-B)-\cos(A+B),$
$\implies  2\sin x\sin3x=\cos(3x-x)-\cos(3x+x)=\cos2x-\cos4x$
and as $\cos2y=1-2\sin^2y\implies 2\sin^2y=1-\cos2y,$
$\implies 2\sin^23x=1-\cos6x$
A: A related problem. Note that, the function $ \sin^3 x \sin 3x  $ is an even function, so its Fourier series will have expansion in terms of the functions $\cos mx$ over the interval $[-\pi,\pi]$, where 

$$ C_m = \frac{1}{\pi}\int_{-\pi}^\pi \sin^3 x \sin 3x   \cos(mx)\, dx=\frac{2}{\pi}\int_{0}^\pi \sin^3 x \sin 3x   \cos(mx)\, dx, \quad m \ge 0.$$

Now, just work out the integral and you will be able to finish the problem. See here for the value of the integral and you will see that the Fourier series has finite number of terms.
Note: This solution is based on the Fourier series knowledge. I just paid attention to the tag. 
A: $n=6$. 
Since $\sin^3x \sin 3 x$ is a degree $6$ polynomial of $\cos x$, and $\cos nx$ is degree $n$ polynomial of $\cos x$, $n$ must be $6$.
