# Quintuple Angle Formulas/Sine

## 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

$\blacksquare$