Chapter 5, Problem 26RE

Solution

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

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

Given information:

The given expression is $\sin 2x+\cos 3x$ .

Formula used:

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

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

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

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.

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

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