Double Angle Formulas/Sine

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

where $\sin$ and $\cos$ denote sine and cosine respectively.


$\sin 2 \theta = \dfrac {2 \tan \theta} {1 + \tan^2 \theta}$

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


Proof 2

\(\displaystyle \sin 2 \theta\) \(=\) \(\displaystyle \map \sin {\theta + \theta}\)
\(\displaystyle \) \(=\) \(\displaystyle \sin \theta \cos \theta + \cos \theta \sin \theta\) Sine of Sum
\(\displaystyle \) \(=\) \(\displaystyle 2 \sin \theta \cos \theta\)


Proof 3

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:

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

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

And by cancelling out common terms:

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


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


Also see