# 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 a Isosceles Triangle $\triangle ABC$ with base $BC$, and head angle $\angle BAC = 2 \alpha$.

Draw a angle bisector to $\angle BAC$ and name it $AH$.

$\angle BAH = \angle CAH = \alpha$
$AH \perp BC$.
$\operatorname {Area} \left({\triangle BAH}\right) = \dfrac {BA \cdot AH \sin \alpha} 2$
$\operatorname {Area} \left({\triangle CAH}\right) = \dfrac {CA \cdot AH \sin \alpha} 2$

By definition of sine:

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

And so:

$\operatorname {Area} \left({\triangle BAH}\right) = \dfrac {BA \cdot CA \cos \alpha \sin \alpha} 2$
$\operatorname {Area} \left({\triangle CAH}\right) = \dfrac {CA \cdot BA \cos \alpha \sin \alpha} 2$

 $\displaystyle \operatorname {Area} \left({\triangle ABC}\right)$ $=$ $\displaystyle \operatorname {Area} \left({\triangle BAH}\right) + \operatorname {Area} \left({\triangle CAH}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac {BA \cdot CA \cos \alpha \sin \alpha} 2 + \dfrac {CA \cdot BA \cos \alpha \sin \alpha} 2$ $\displaystyle$ $=$ $\displaystyle BA CA \cos \alpha \sin \alpha$ $\displaystyle$ $=$ $\displaystyle \dfrac {BA \cdot CA \sin 2 \alpha} 2$ Area of Triangle in Terms of Two Sides and Angle ($\triangle ABC$)

And by cancelling out common terms:

$\sin 2 \alpha = 2 \cos \alpha \sin \alpha$

$\blacksquare$

## 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

$\blacksquare$