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 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 $AH$.

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

From Angler Bisector and Altitude coincide iff triangle is isosceles:

$AH \perp BC$.

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

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

By definition of sine:

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


And so:

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


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

And by cancelling out common terms:

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

$\blacksquare$