Question

${\cos }^{3}x\sin 2x=\sum _{r=1}^n{a}_r\sin \left(rx\right)\forall x\in R$, then

• A. $n=5,a_1=\frac{1}{2}$
• B. $n=5,a_1=\frac{1}{4}$
• C. $n=5,a_2=\frac{1}{8}$
• D. $n=5,a_2=\frac{1}{16}$

Answer 1

The correct option is B $n=5,a_1=\frac{1}{4}$

${\cos }^{3}x\sin 2x$
$={\cos }^{2}x\cos x\sin 2x$
$=\left(\frac{1-\cos 2x}{2}\right)\left(\frac{2\sin 2x\cos x}{2}\right)$

$=\frac{1}{4}[\sin 3x+\sin x-\frac{1}{2}(2\sin 3x\cos 2x)-\frac{1}{2}(2\cos 2x\sin x\left)\right]$

$=\frac{1}{4}[\sin 3x+\sin x-\frac{1}{2}(\sin 5x+\sin x)-\frac{1}{2}(\sin 3x-\sin x\left)\right]$

$=\frac{1}{4}[\sin x+\frac{1}{2}\sin 3x-\frac{1}{2}\sin 5x]$

$\Rightarrow a_1=\frac{1}{4},a_3=\frac{1}{8},n=5$