Chapter 5.4, Problem 19E

Solution

To prove: The identity $\sin 3x=\left(\sin x\right)\left(4{\cos }^{2}x-1\right)$ .

Given information:

The given equation is $\sin 3x=\left(\sin x\right)\left(4{\cos }^{2}x-1\right)$ .

Formula used:

The double angle identity for $\cos 2x$ is given by,

  $\cos 2x=2{\cos }^{2}x-1$

Also,

  $\cos 2x={\cos }^{2}x-{\sin }^{2}x$

The double angle identity for $\sin 2x$ is given by,

  $\sin 2x=2\sin x\cos x$

The double angle identity for $\tan 2u$ is given by,

  $\tan 2u=\frac{2\tan u}{1-{\tan }^{2}u}$

Calculation:

Consider the given equation.

  $\sin 3x=\left(\sin x\right)\left(4{\cos }^{2}x-1\right)$

Consider the left-hand side of the equation and simplify.

  $\begin{array}{c}\sin 3x=\sin \left(2x+x\right)\\ =\sin 2x\cos x+\cos 2x\sin x\\ =\left(2\sin x\cos x\right)\cos x+\left(2{\cos }^{2}x-1\right)\sin x\\ =2\sin x{\cos }^{2}x+2\sin x{\cos }^{2}x-\sin x\end{array}$

Simplify, further.

  $\begin{array}{c}\sin 3x=4\sin x{\cos }^{2}x-\sin x\\ =\sin x\left(4{\cos }^{2}x-1\right)\end{array}$

Therefore, it is proved that $\sin 3x=\left(\sin x\right)\left(4{\cos }^{2}x-1\right)$ .