Quadruple Angle Formulas/Sine/Corollary 2

Corollary to Quadruple Angle Formula for Sine
For all $\theta$ such that $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$
 * $\dfrac {\sin 4 \theta} {\sin \theta} = 2 \cos 3 \theta + 2 \cos \theta$

where $\sin$ denotes sine and $\cos$ denotes cosine.

Proof
First note that when $\theta = 0, \pm \pi, \pm 2 \pi \ldots$:
 * $\sin \theta = 0$

so $\dfrac {\sin 4 \theta} {\sin \theta}$ is undefined.

Therefore for the rest of the proof it is assumed that $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$

{{eqn | ll= \implies | l = \frac {\sin 4 \theta}} {\sin \theta} | r = 8 \paren {\dfrac {3 \cos \theta + \cos 3 \theta} 4} - 4 \cos \theta | c = Power Reduction Formulas: Cosine Cubed }}