# How do you express sin^4theta+cot^2theta -cos^4 theta in terms of non-exponential trigonometric functions?

Feb 1, 2016

$\left(\sin \theta + \cos \theta\right) \left(\sin \theta - \cos \theta\right) + \left(\csc \theta + 1\right) \left(\csc \theta - 1\right)$

#### Explanation:

Rewrite to group the sine and cosine terms.

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

Simplify the first two terms as a difference of squares.

$= \left({\sin}^{2} \theta + {\cos}^{2} \theta\right) \left({\sin}^{2} \theta - {\cos}^{2} \theta\right) + {\cot}^{2} \theta$

Note that ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$ through the Pythagorean Identity.

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

The first two terms can again be factored as a difference of squares.

$= \left(\sin \theta + \cos \theta\right) \left(\sin \theta - \cos \theta\right) + {\cot}^{2} \theta$

Use the identity: ${\cot}^{2} \theta + 1 = {\csc}^{2} \theta$ to say that ${\cot}^{2} \theta = {\csc}^{2} \theta - 1$.

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

Again, the last two terms can be factored as a difference of squares.

$= \left(\sin \theta + \cos \theta\right) \left(\sin \theta - \cos \theta\right) + \left(\csc \theta + 1\right) \left(\csc \theta - 1\right)$