Quintuple Angle Formulas/Sine/Corollary

Theorem

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

$\dfrac {\sin 5 \theta} {\sin \theta} = 16 \cos^4 \theta - 12 \cos^2 \theta + 1$

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 $\dfrac {\sin 5 \theta} {\sin \theta}$ is undefined.

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

\(\ds \sin 5 \theta\)

\(=\)

\(\ds 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta\)

Quintuple Angle Formula for Sine

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\sin 5 \theta} {\sin \theta}\)

\(=\)

\(\ds 5 - 20 \paren {1 - \cos^2 \theta} + 16 \paren {1 - \cos^2 \theta}^2\)

Sum of Squares of Sine and Cosine

\(\ds \)

\(=\)

\(\ds 16 \cos^4 \theta - 12 \cos^2 \theta + 1\)

multiplying out and gathering terms

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $21 \ \text{(b)}$