What is cos^3theta-6sin^3theta in terms of non-exponential trigonometric functions?

Dec 31, 2015

${\cos}^{3} \left(\theta\right) - 6 {\sin}^{3} \left(\theta\right)$

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

Explanation:

Since this is of degree $3$, the place to start is the expansion of $\cos \left(3 \theta\right)$ and $\sin \left(3 \theta\right)$.

By De Moivre's formula:

$\cos \left(3 \theta\right) + i \sin \left(3 \theta\right)$

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

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

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

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

So equating Real and Imaginary parts we find:

$\cos \left(3 \theta\right) = 4 {\cos}^{3} \left(\theta\right) - 3 \cos \left(\theta\right)$

$\sin \left(3 \theta\right) = 3 \sin \left(\theta\right) - 4 {\sin}^{3} \left(\theta\right)$

Hence:

${\cos}^{3} \left(\theta\right) = \frac{1}{4} \left(\cos \left(3 \theta\right) + 3 \cos \left(\theta\right)\right)$

${\sin}^{3} \left(\theta\right) = \frac{1}{4} \left(3 \sin \left(\theta\right) - \sin \left(3 \theta\right)\right)$

So:

${\cos}^{3} \left(\theta\right) - 6 {\sin}^{3} \left(\theta\right)$

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