Double Angle Formulas/Sine/Proof 3

Theorem

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

Proof

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:

\(\ds \map \Area {\triangle BAH}\)

\(=\)

\(\ds \dfrac {BA \cdot AH \sin \alpha} 2\)

\(\ds \map \Area {\triangle CAH}\)

\(=\)

\(\ds \dfrac {CA \cdot AH \sin \alpha} 2\)

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 $\dfrac {BA \cdot CA} 2$

$\blacksquare$