Double Angle Formulas/Sine

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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

Double angle sin.png

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 $AH$:

$\angle BAH = \angle CAH = \alpha$

From Bisector of Apex of Isosceles Triangle is Perpendicular to Base:

$AH \perp BC$

From Area of Triangle in Terms of Two Sides and Angle:

\(\displaystyle \map \Area {\triangle BAH}\) \(=\) \(\displaystyle \dfrac {BA \cdot AH \sin \alpha} 2\)
\(\displaystyle \map \Area {\triangle CAH}\) \(=\) \(\displaystyle \dfrac {CA \cdot AH \sin \alpha} 2\)


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

$\blacksquare$


Also see


Sources