Question

If $si{n}^{3}xsin3x={\sum }_{m=0}^n{C}_{m}cosmx$
where $c_0,c_1,c_2,...........c_n$ are constants and $C_n\ne 0$,then the value of n is

• A. $15$
• B. $6$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$6$

${\sin }^{3}x\sin 3x=\left[\frac{3\sin x-\sin 3x}{4}\right]\sin 3x[\because \sin 3x=3\sin x-4{\sin }^{3}x]$
$=\frac{1}{4}[3\sin x\sin 3x-{\sin }^{2}3x]$
$=\frac{1}{4}\left[\frac{3}{2}\right[\cos 4x-\cos 2x]-{\sin }^{2}3x]$
$=\frac{1}{4}\left[\frac{3}{2}\right[\cos 4x-\cos 2x]+\cos 6x-1]$
$=\frac{1}{8}[3\cos 4x-3\cos 2x+2\cos 6x-2]$
$=\frac{1}{8}[-2-3\cos 2x+3\cos 4x+2\cos 6x]$
$={\sum }_{m=0}^n{C}_m\cos mx$
$n=6$