What is cos^4theta+sin^3theta in terms of non-exponential trigonometric functions?

1 Answer
Jul 8, 2016

$= \frac{1}{8} \left(3 + 4 \cos 2 \theta + 2 \cos 4 \theta + 6 \sin \theta - 2 \sin 3 \theta\right)$

Explanation:

We will use the identities

$\textcolor{red}{> 2 {\cos}^{2} \theta = \left(1 + \cos 2 \theta\right)}$

$\textcolor{b l u e}{> 4 {\sin}^{3} \theta = 3 \sin \theta - \sin 3 \theta}$

The given expression

$= {\cos}^{4} \theta + {\sin}^{3} \theta$

$= \frac{1}{4} \left(4 {\cos}^{4} \theta + 4 {\sin}^{3} \theta\right)$

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

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

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

$= \frac{1}{4} \left(1 + 2 \cos 2 \theta + \frac{1}{2} \left(1 + \cos 4 \theta\right) + 3 \sin \theta - \sin 3 \theta\right)$

$= \frac{1}{4} \left(\frac{3}{2} + 2 \cos 2 \theta + \cos 4 \theta + 3 \sin \theta - \sin 3 \theta\right)$

$= \frac{1}{8} \left(3 + 4 \cos 2 \theta + 2 \cos 4 \theta + 6 \sin \theta - 2 \sin 3 \theta\right)$