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

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

$\blacksquare$

Proof 2

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

$\blacksquare$

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 $AH$:

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

By definition of sine:

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

and so:

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

$\blacksquare$

Proof 4

 $\ds \sin 2 \theta$ $=$ $\ds \frac 1 {2 i} \paren {e^{2 i \theta} - e^{-2 i \theta} }$ Sine Exponential Formulation $\ds$ $=$ $\ds \frac 1 {2 i} \paren {e^{i \theta} + e^{-i \theta} } \paren {e^{i \theta} - e^{-i \theta} }$ Difference of Two Squares $\ds$ $=$ $\ds 2 \paren {\frac {e^{i \theta} - e^{-i \theta} } {2 i} \cdot \frac {e^{i \theta} + e^{-i \theta} } 2}$ $\ds$ $=$ $\ds 2 \sin \theta \cos \theta$ Sine Exponential Formulation, Cosine Exponential Formulation

$\blacksquare$