If ${\cos }^{3}x.\sin 2x={\displaystyle \sum _{m=1}^n{a}_m}$ $\sin mx$ is an identity $x$ . Then

We know that

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta \Rightarrow \cos 3\theta +3\cos \theta =4{\cos }^3\theta$

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

$=\frac{\cos 3x.\sin 2x+3\cos x.\sin 2x}{4}$

$=\frac{2\cos 3x.\sin 2x+3\times 2\cos x.\sin 2x}{8}$

$=\frac{\sin \left[3x+2x\right]-\sin \left[3x-2x\right]+3\left[\sin 3x+\sin x\right]}{8}$

$=\frac{\sin 5x-\sin x+3\sin 3x+3\sin x}{8}$

$=\frac{\sin 5x+3\sin x+2\sin x}{8}$

$=\frac{2\sin x}{8}+\frac{3\sin 3x}{8}+\frac{\sin 5x}{8}$

$=\frac{1}{4}\sin x+\frac{3}{8}\sin 3x+\frac{1}{8}\sin 5x$

$a_1=\frac{1}{4},a_2=0,a_3=\frac{3}{8},a_4=0$

$a_5=\frac{1}{8},n=5$

$\Sigma a_m=a_1+a_2+a_3+a_4+a_5$

$=\frac{1}{4}+\frac{3}{8}+\frac{1}{8}$

$=\frac{1}{4}+\frac{1}{2}$

$=\frac{3}{4}$