Double Angle Formulas/Sine/Proof 3

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Theorem

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


Proof

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 $\dfrac {BA \cdot CA} 2$

$\blacksquare$