# What is 4cos^5thetasin^5theta in terms of non-exponential trigonometric functions?

Dec 28, 2015

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

#### Explanation:

We know that $\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$. We apply this formula here!

$4 {\cos}^{5} \left(\theta\right) {\sin}^{5} \left(\theta\right) = 4 {\left(\sin \left(\theta\right) \cos \left(\theta\right)\right)}^{5} = 4 {\left(\sin \frac{2 \theta}{2}\right)}^{5} = {\sin}^{5} \frac{2 \theta}{8}$.

We also know that ${\sin}^{2} \left(\theta\right) = \frac{1 - \cos \left(2 \theta\right)}{2}$ and ${\cos}^{2} \left(\theta\right) = \frac{1 + \cos \left(2 \theta\right)}{2}$.

So ${\sin}^{5} \frac{2 \theta}{8} = \sin \frac{2 \theta}{8} \cdot {\left(\frac{1 - \cos \left(4 \theta\right)}{2}\right)}^{2} = \sin \frac{2 \theta}{8} \cdot \frac{1 - 2 \cos \left(4 \theta\right) + {\cos}^{2} \left(4 \theta\right)}{4} = \sin \frac{2 \theta}{8} \cdot \left(\frac{1 - 2 \cos \left(4 \theta\right)}{4} + \frac{1 + \cos \left(8 \theta\right)}{8}\right) = \frac{1}{8} \sin \left(2 \theta\right) \left(3 - 4 \cos \left(4 \theta\right) + \cos \left(8 \theta\right)\right)$