Chapter 5.4, Problem 13E

Solution

To write: The given equation in terms of $\sin \theta$ and $\cos \theta$ .

The given equation can be written as $2\sin \theta \cos \theta +{\cos }^3\theta -3{\sin }^2\theta \cos \theta$ .

Given information:

The given equation is $\sin 2\theta +\cos 3\theta$ .

Formula used:

The double angle identity for $\cos 2x$ is $\cos 2x=2{\cos }^{2}x-1$ and $\cos 2x={\cos }^{2}x-{\sin }^{2}x$ .

The double angle identity for $\sin 2x$ is $\sin 2x=2\sin x\cos x$ .

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

Calculation:

Consider the given equation.

  $\sin 2\theta +\cos 3\theta$

Apply the double angle identities in the given equation.

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

Further, simplify.

  $\sin 2\theta +\cos 3\theta =2\sin \theta \cos \theta +{\cos }^3\theta -3{\sin }^2\theta \cos \theta$

Therefore, the given equation in terms of $\sin \theta$ and $\cos \theta$ is $2\sin \theta \cos \theta +{\cos }^3\theta -3{\sin }^2\theta \cos \theta$ .