# Double Angle Formulas/Sine

## Theorem

$\sin 2 \theta = 2 \sin \theta \cos \theta$

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

### Corollary

$\sin 2 \theta = \dfrac {2 \tan \theta} {1 + \tan^2 \theta}$

## Proof 1

 $\displaystyle \cos 2 \theta + i \sin 2 \theta$ $=$ $\displaystyle \paren {\cos \theta + i \sin \theta}^2$ De Moivre's Formula $\displaystyle$ $=$ $\displaystyle \cos^2 \theta + i^2 \sin^2 \theta + 2 i \cos \theta \sin \theta$ $\displaystyle$ $=$ $\displaystyle \cos^2 \theta - \sin^2 \theta + 2 i \cos \theta \sin \theta$ $\displaystyle \leadsto \ \$ $\displaystyle \sin 2 \theta$ $=$ $\displaystyle 2 \cos \theta \sin \theta$ equating imaginary parts

$\blacksquare$

## Proof 2

 $\displaystyle \sin 2 \theta$ $=$ $\displaystyle \map \sin {\theta + \theta}$ $\displaystyle$ $=$ $\displaystyle \sin \theta \cos \theta + \cos \theta \sin \theta$ Sine of Sum $\displaystyle$ $=$ $\displaystyle 2 \sin \theta \cos \theta$

$\blacksquare$

## Proof 3

Consider an isosceles triangle $\triangle ABC$ with base $BC$ and apex $\angle BAC = 2 \alpha$.

Construct the angle bisector to $\angle BAC$ and name it $AH$:

$\angle BAH = \angle CAH = \alpha$
$AH \perp BC$
 $\displaystyle \map \Area {\triangle BAH}$ $=$ $\displaystyle \dfrac {BA \cdot AH \sin \alpha} 2$ $\displaystyle \map \Area {\triangle CAH}$ $=$ $\displaystyle \dfrac {CA \cdot AH \sin \alpha} 2$

By definition of sine:

 $\displaystyle AH$ $=$ $\displaystyle CA \cos \alpha$ $\displaystyle AH$ $=$ $\displaystyle BA \cos \alpha$

and so:

 $\displaystyle \map \Area {\triangle BAH}$ $=$ $\displaystyle \dfrac {BA \cdot CA \cos \alpha \sin \alpha} 2$ $\displaystyle \map \Area {\triangle CAH}$ $=$ $\displaystyle \dfrac {CA \cdot BA \cos \alpha \sin \alpha} 2$ $\displaystyle \map \Area {\triangle ABC}$ $=$ $\displaystyle \map \Area {\triangle BAH} + \map \Area {\triangle CAH}$ $\displaystyle$ $=$ $\displaystyle \frac {BA \cdot CA \cos \alpha \sin \alpha} 2 + \frac {CA \cdot BA \cos \alpha \sin \alpha} 2$ $\displaystyle$ $=$ $\displaystyle BA \cdot CA \cos \alpha \sin \alpha$ $\displaystyle$ $=$ $\displaystyle \frac {BA \cdot CA \sin 2 \alpha} 2$ Area of Triangle in Terms of Two Sides and Angle ($\triangle ABC$) $\displaystyle \leadsto \ \$ $\displaystyle \sin 2 \alpha$ $=$ $\displaystyle 2 \cos \alpha \sin \alpha$ dividing both sides by $\dfrac {BA \cdot CA} 2$

$\blacksquare$

## Proof 4

 $\displaystyle \sin 2\theta$ $=$ $\displaystyle \frac{1}{2i} \left(e^{2i\theta}-e^{-2i\theta}\right)$ Sine Exponential Formulation $\displaystyle$ $=$ $\displaystyle \frac{1}{2i} \left(e^{i\theta}+e^{-i\theta}\right) \left(e^{i\theta}-e^{-i\theta}\right)$ Difference of Two Squares $\displaystyle$ $=$ $\displaystyle 2 \left(\frac{e^{i\theta}-e^{-i\theta} }{2i} \cdot \frac{e^{i\theta} + e^{-i\theta} }{2}\right)$ $\displaystyle$ $=$ $\displaystyle 2 \sin \theta \cos \theta$ Sine Exponential Formulation, Cosine Exponential Formulation

$\blacksquare$