Question

Fill in the blanks so that the following statements are correct.
Suppose that ${\sin }^{3}x\sin 3x=\sum _{m=0}^n{c}_m\cos mx$ is an identity in $x$ where $c_0,c_1,.....c_n$ are constant and $c_n\ne 0$ then the value of $n$ is.......... .

Answer 1

Ans. $n=6$. We have
${\sin }^{3}x\sin 3x=\frac{1}{4}(3\sin x-\sin 3x):\sin 3x$
$=\frac{3}{8}.2\sin x\sin 3x-\frac{1}{8}.2{\sin }^{2}3x$
$=\frac{3}{8}(\cos 2x-\cos 4x)-\frac{1}{8}(1-\cos 6x)$
$=-\frac{1}{8}+\frac{3}{8}\cos 2x-\frac{3}{8}\cos 4x+\frac{1}{8}\cos 6x$
$\therefore -\frac{1}{8}+\frac{3}{8}\cos 2x-\frac{3}{8}\cos 4x+\frac{1}{8}\cos 6x$
$=\stackrel{n}{\underset{m=0}{\Sigma }}{C}_m\cos mx$
$=C_0+C_1\cos x+C_2\cos 2x+C_3\cos 3x+++C_n\cos nx$ $\left(1\right)$
Comparing both sides of $\left(1\right)$, we see that $n=6$.
We also have
$C_0=-1/8,C_1=0,C_2=3/8,C_3=0$,
$C_4=-3/8,C_5=0,C_6=1/8$