Chapter 5, Problem 28RE

Solution

To find: The given expression, $\sin 3x-3\sin 2x$ in terms of $\sin x$ and $\cos x$ only.

The simplified form of the given expression is $3\sin x{\cos }^{2}x-{\sin }^{3}x+6\sin x\cos x$ .

Given information:

The given expression is $\sin 3x-3\sin 2x$ .

Formula used:

The formula of sin of sum is $\sin \left(u+v\right)=\sin u\cos v+\cos u\sin v$ .

The double angle formulas are, $\sin 2\theta =2\sin \theta \cos \theta$ and $\cos 2\theta =\left({\cos }^2\theta -{\sin }^2\theta \right)$ .

Calculation:

Apply formula of sin of sum, $\sin \left(u+v\right)=\sin u\cos v+\cos u\sin v$ in the given expression.

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

Apply double angle formulas, $\sin 2\theta =2\sin \theta \cos \theta$ and $\cos 2\theta =\left({\cos }^2\theta -{\sin }^2\theta \right)$ into obtained expression and simplify.

  $\begin{array}{c}\sin 2x\cos x+\cos 2x\sin x+3\sin 2x=\left(\begin{array}{l}2\sin x\cos x\cos x+\left({\cos }^{2}x-{\sin }^{2}x\right)\sin x\\ +3\left(2\sin x\cos x\right)\end{array}\right)\\ =2\sin x{\cos }^{2}x+\left({\cos }^{2}x\sin x-{\sin }^{3}x\right)+6\sin x\cos x\\ =3\sin x{\cos }^{2}x-{\sin }^{3}x+6\sin x\cos x\end{array}$

Thus, the simplified form of the given expression is $3\sin x{\cos }^{2}x-{\sin }^{3}x+6\sin x\cos x$ .