Quintuple Angle Formulas/Sine

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Theorem

$\sin 5 \theta = 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta$

where $\sin$ denotes sine.


Corollary

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 1

\(\displaystyle \sin 5 \theta\) \(=\) \(\displaystyle \map \sin {3 \theta + 2 \theta}\)
\(\displaystyle \) \(=\) \(\displaystyle \sin 3 \theta \cos 2 \theta + \cos 3 \theta \sin 2 \theta\) Sine of Sum
\(\displaystyle \) \(=\) \(\displaystyle \paren {3 \sin \theta - 4 \sin^3 \theta} \cos 2 \theta + \paren {4 \cos^3 \theta - 3 \cos \theta} \sin 2 \theta\) Triple Angle Formulas
\(\displaystyle \) \(=\) \(\displaystyle \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {\cos^2 \theta - \sin^2 \theta} + \paren {4 \cos^3 \theta - 3 \cos \theta} 2 \sin \theta \cos \theta\) Double Angle Formulas
\(\displaystyle \) \(=\) \(\displaystyle \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {\cos^2 \theta - \sin^2 \theta} + 8 \cos^4 \theta \sin \theta - 6 \cos^2 \theta \sin \theta\) multiplying out
\(\displaystyle \) \(=\) \(\displaystyle \paren {3 \sin \theta - 4 \sin^3 \theta} \paren {1 - 2 \sin^2 \theta} + 8 \paren {1 - \sin^2 \theta}^2 \sin \theta - 6 \paren {1 - \sin^2 \theta} \sin \theta\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle 5 \sin \theta - 20 \sin^3 \theta + 16 \sin ^5 \theta\) multiplying out and gathering terms

$\blacksquare$


Proof 2

We have:

\(\displaystyle \cos 5 \theta + i \sin 5 \theta\) \(=\) \(\displaystyle \paren {\cos \theta + i \sin \theta}^5\) De Moivre's Formula
\(\displaystyle \) \(=\) \(\displaystyle \paren {\cos \theta}^5 + \binom 5 1 \paren {\cos \theta}^4 \paren {i \sin \theta} + \binom 5 2 \paren {\cos \theta}^3 \paren {i \sin \theta}^2\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle \binom 5 3 \paren {\cos \theta}^2 \paren {i \sin \theta}^3 + \binom 5 4 \paren {\cos \theta} \paren {i \sin \theta}^4 + \paren {i \sin \theta}^5\) Binomial Theorem
\(\displaystyle \) \(=\) \(\displaystyle \cos^5 \theta + 5 i \cos^4 \theta \sin \theta - 10 \cos^3 \theta \sin^2 \theta\) substituting for binomial coefficients
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle 10 i \cos^2 \theta \sin^3 \theta + 5 \cos \theta \sin^4 \theta + i \sin^5 \theta\) and using $i^2 = -1$
\(\text {(1)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta\)
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle i \paren {5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta}\) rearranging


Hence:

\(\displaystyle \sin 5 \theta\) \(=\) \(\displaystyle 5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta\) equating imaginary parts in $(1)$
\(\displaystyle \) \(=\) \(\displaystyle 5 \paren {1 - \sin^2 \theta}^2 \sin \theta - 10 \paren {1 - \sin^2 \theta} \sin^3 \theta + \sin^5 \theta\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle 5 \sin \theta - 20 \sin^3 \theta + 16 \sin^5 \theta\) multiplying out and gathering terms

$\blacksquare$


Sources