# Triple Angle Formulas/Sine/Proof 2

## Theorem

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

## Proof

We have:

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

Hence:

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

$\blacksquare$