Quadruple Angle Formulas/Sine/Corollary 1

Corollary to Quadruple Angle Formula for Sine

For all $\theta$ such that $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$

$\dfrac {\sin 4 \theta} {\sin \theta} = 8 \cos^3 \theta - 4 \cos \theta$

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 $\theta \ne 0, \pm \pi, \pm 2 \pi \ldots$

\(\ds \sin 4 \theta\)

\(=\)

\(\ds 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta\)

Quadruple Angle Formula for Sine

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\sin 4 \theta} {\sin \theta}\)

\(=\)

\(\ds 4 \cos \theta - 8 \paren {1 - \cos^2 \theta} \cos \theta\)

Sum of Squares of Sine and Cosine

\(\ds \)

\(=\)

\(\ds 8 \cos^3 \theta - 4 \cos \theta\)

multiplying out and gathering terms

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: De Moivre's Theorem: $93 \ \text{(a)}$

(although see Quadruple Angle Formulas/Sine/Mistake for analysis of an error in that work)