# What is cos^2thetasin^2theta in terms of non-exponential trigonometric functions?

Dec 25, 2015

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

#### Explanation:

First express purely in terms of $\cos \theta$:

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

Since this is of degree $4$, a good place to start might be $\cos 4 \theta$ and $\sin 4 \theta$.

Using De Moivre's formula:

$\cos 4 \theta + i \sin 4 \theta$

$= {\left(\cos \theta + i \sin \theta\right)}^{4}$

$= {\cos}^{4} \theta + 4 i {\cos}^{3} \theta \sin \theta - 6 {\cos}^{2} \theta {\sin}^{2} \theta - 4 i \cos \theta {\sin}^{3} \theta + {\sin}^{4} \theta$

$= \left({\cos}^{4} \theta - 6 {\cos}^{2} \theta {\sin}^{2} \theta + {\sin}^{4} \theta\right) + \left(4 {\cos}^{3} \theta \sin \theta - 4 \cos \theta {\sin}^{3} \theta\right) i$

So looking at the Real part, we find:

$\cos 4 \theta$

$= {\cos}^{4} \theta - 6 {\cos}^{2} \theta {\sin}^{2} \theta + {\sin}^{4} \theta$

$= {\cos}^{4} \theta - 6 {\cos}^{2} \theta \left(1 - {\cos}^{2} \theta\right) + \left(1 - {\cos}^{2} \theta\right) \left(1 - {\cos}^{2} \theta\right)$

$= {\cos}^{4} \theta - 6 {\cos}^{2} \theta + 6 {\cos}^{4} \theta + 1 - 2 {\cos}^{2} \theta + {\cos}^{4} \theta$

$= 8 {\cos}^{4} \theta - 8 {\cos}^{2} \theta + 1$

$= - 8 \left({\cos}^{2} \theta - {\cos}^{4} \theta\right) + 1$

Subtract $1$ from both ends to get:

$\cos 4 \theta - 1 = - 8 \left({\cos}^{2} \theta - {\cos}^{4} \theta\right)$

Divide both sides by $- 8$ to get:

$\frac{1}{8} \left(1 - \cos 4 \theta\right) = {\cos}^{2} \theta - {\cos}^{4} \theta$

Voila!