# How do you express cot^3theta-cos^2theta-tan^3theta  in terms of non-exponential trigonometric functions?

Jan 21, 2017

${\cos}^{3} \frac{\theta}{\sin} ^ 3 \theta - {\cos}^{2} \theta - {\sin}^{3} \frac{\theta}{\cos} ^ 3 \theta$

$\frac{{\cos}^{6} \theta - {\sin}^{6} \theta}{{\cos}^{3} \theta {\sin}^{3} \theta} - {\cos}^{2} \theta$

$\frac{{\cos}^{6} \theta - {\sin}^{6} \theta - {\cos}^{5} \theta {\sin}^{3} \theta}{\cos} ^ 3 \theta$

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

Now use the power reduction formulae ${\cos}^{2} x = \frac{1 + \cos 2 x}{2}$ and ${\sin}^{2} x = \frac{1 - \cos 2 x}{2}$

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

This can be simplified further, but I'll leave the algebra up to you.

Hopefully this helps!

Jan 21, 2017

${\cot}^{3} \theta - {\cos}^{2} \theta - {\tan}^{3} \theta = \frac{16 \cot 2 \theta}{1 - \cos 4 \theta} - 2 \cot 2 \theta - \frac{1 + \cos 2 \theta}{2}$

#### Explanation:

${\cot}^{3} \theta - {\cos}^{2} \theta - {\tan}^{3} \theta$

= ${\cot}^{3} \theta - {\tan}^{3} \theta - {\cos}^{2} \theta$

= $\left(\cot \theta - \tan \theta\right) \left({\cot}^{2} \theta + {\tan}^{2} \theta + \cot \theta \tan \theta\right) - \frac{1 + \cos 2 \theta}{2}$

= $\left(\cot \theta - \tan \theta\right) \left({\left(\cot \theta + \tan \theta\right)}^{2} - \cot \theta \tan \theta\right) - \frac{1 + \cos 2 \theta}{2}$

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

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

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

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

= $\frac{16 \cot 2 \theta}{1 - \cos 4 \theta} - 2 \cot 2 \theta - \frac{1 + \cos 2 \theta}{2}$