## Theorem

$\sin 4 \theta = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$

where $\sin$ and $\cos$ denote sine and cosine respectively.

### Corollary 1

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.

### Corollary 2

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 1

 $\displaystyle \sin \left({4 \theta}\right)$ $=$ $\displaystyle \sin \left({3 \theta + \theta}\right)$ $\displaystyle$ $=$ $\displaystyle \sin 3 \theta \cos \theta + \cos 3 \theta \sin \theta$ Sine of Sum $\displaystyle$ $=$ $\displaystyle \left({3 \sin \theta - 4 \sin^3 \theta}\right) \cos \theta + \left({4 \cos^3 \theta - 3 \cos \theta}\right) \sin \theta$ Triple Angle Formula for Sine and Triple Angle Formula for Cosine $\displaystyle$ $=$ $\displaystyle 3 \sin \theta \cos \theta - 4 \sin^3 \theta \cos \theta + 4 \cos^3 \theta \sin \theta - 3 \cos \theta \sin \theta$ $\displaystyle$ $=$ $\displaystyle 3 \sin \theta \cos \theta - 4 \sin^3 \theta \cos \theta + 4 \cos \theta \left({1 - \sin^2 \theta}\right) \sin \theta - 3 \cos \theta \sin \theta$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta$ multiplying out and gathering terms

$\blacksquare$

## Proof 2

We have:

 $\displaystyle \cos 4 \theta + i \sin 4 \theta$ $=$ $\displaystyle \paren {\cos \theta + i \sin \theta}^4$ De Moivre's Formula $\displaystyle$ $=$ $\displaystyle \paren {\cos \theta}^4 + \binom 4 1 \paren {\cos \theta}^3 \paren {i \sin \theta} + \binom 4 2 \paren {\cos \theta}^2 \paren {i \sin \theta}^2$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \binom 4 3 \paren {\cos \theta} \paren {i \sin \theta}^3 + \paren {i \sin \theta}^4$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle \cos^4 \theta + 4 i \cos^3 \theta \sin \theta - 6 \cos^2 \theta \sin^2 \theta$ substituting for binomial coefficients $\displaystyle$  $\, \displaystyle - \,$ $\displaystyle 4 i \cos \theta \sin^3 \theta + \sin^4 \theta$ and using $i^2 = -1$ $\text {(1)}: \quad$ $\displaystyle$ $=$ $\displaystyle \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle 4 i \cos^3 \theta \sin \theta - 4 i \cos \theta \sin^3 \theta$ rearranging

Hence:

 $\displaystyle \sin 4 \theta$ $=$ $\displaystyle 4 \cos^3 \theta \sin \theta - 4 \cos \theta \sin^3 \theta$ equating imaginary parts in $(1)$ $\displaystyle$ $=$ $\displaystyle 4 \cos \theta \paren {1 - \sin^2 \theta} \sin \theta - 4 \cos \theta \sin^3 \theta$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle 4 \cos \theta \sin \theta - 8 \sin^3 \theta \cos \theta$ multiplying out and gathering terms

$\blacksquare$