Question

If ${\cos }^{3}x\sin 2x=a_1\sin x+a_2\sin 2x+\cdots +a_n\sin nx$ $\forall x\in R$, where $a_n\ne 0$, then which of the following is/are correct ?

• A. $a_5=\frac{1}{8}$
• B. $a_2+a_3=\frac{3}{8}$
• C. $a_4=0$
• D. $n=5$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $a_5=\frac{1}{8}$
B $a_2+a_3=\frac{3}{8}$
C $a_4=0$
D $n=5$

${\cos }^{3}x\sin 2x=\left({\cos }^{2}x\right)\left(\cos x\sin 2x\right)=\left(\frac{1+\cos 2x}{2}\right)\left(\frac{2\sin 2x\cos x}{2}\right)$
$=\frac{1}{4}(1+\cos 2x)(\sin 3x+\sin x)$
$=\frac{1}{4}[\sin 3x+\sin x+\frac{1}{2}(2\sin 3x\cos 2x)+\frac{1}{2}(2\cos 2x\sin x\left)\right]$
$=\frac{1}{4}[\sin 3x+\sin x+\frac{1}{2}(\sin 5x+\sin x)+\frac{1}{2}(\sin 3x-\sin x\left)\right]$
$=\frac{1}{4}[\sin x+\frac{3}{2}\sin 3x+\frac{1}{2}\sin 5x]\therefore a_1\sin x+a_2\sin 2x+\cdots +a_n\sin nx=\frac{1}{4}[\sin x+\frac{3}{2}\sin 3x+\frac{1}{2}\sin 5x]$

So,
$n=5a_5=\frac{1}{8}{a}_2+a_3=0+\frac{3}{8}=\frac{3}{8}{a}_4=0$