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

Oct 15, 2016

${\cot}^{3} \theta + {\tan}^{3} \theta$

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

$= \frac{{\cos}^{6} \theta + {\sin}^{6} \theta}{{\sin}^{3} \theta {\cos}^{3} \theta}$

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

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

$= \frac{8 - 24 \left({\cos}^{2} \theta {\sin}^{2} \theta\right)}{{2}^{3} {\sin}^{3} \theta {\cos}^{3} \theta}$

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

$= \frac{8 - 3 \cdot 2 {\sin}^{2} 2 \theta}{{\sin}^{3} 2 \theta}$

=(8-3*(1-cos4theta))/(1/4(3sin2theta-sin6theta)

=(5+3cos4theta)/(1/4(3sin2theta-sin6theta)

$= \frac{20 + 12 \cos 4 \theta}{3 \sin 2 \theta - \sin 6 \theta}$