Question

If ${\sin }^{3}x\sin 3x=\sum _{m=0}^n{C}_{m}cosmx$ where $c_0,c_1,c_2,\dots \dots \dots ..c_n$ are constants and $C_n\ne 0$, then the value of $n$ is:

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C

$6$

Explanation for the correct option.

Given that, ${\sin }^{3}x\sin 3x=\sum _{m=0}^n{C}_{m}cosmx....\left(1\right)$

Using $\sin 3x=3\sin x-4{\sin }^{3}x$ & substituting in equation (1) we get:

$\begin{array}{rcl}\therefore {\sin }^{3}x\sin 3x& =& \left[\frac{3\sin x-\sin 3x}{4}\right]\sin 3x\\ & =& \frac{1}{4}\left[3\sin x\sin 3x-{\sin }^{2}3x\right]\\ & =& \frac{1}{4}\left[\frac{3}{2}[\cos 4x-\cos 2x]-{\sin }^{2}3x\right]\left[\sin a+\sin b=\cos \left(a+b\right)-\cos (a-b)\right]\\ & =& \frac{1}{4}\left[\frac{3}{2}[\cos 4x-\cos 2x]+\cos 6x-1\right]\\ & =& \frac{1}{8}[3\cos 4x-3\cos 2x+2\cos 6x-2]\\ & =& \frac{1}{8}[-2-3\cos 2x+3\cos 4x+2\cos 6x]\\ & =& \sum _{m=0}^n{C}_m\cos mx\end{array}$

On comparing the LHS & RHS we get :

$n=6$

Hence, option (C) is correct.