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
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 $
$\blacksquare$
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.38$
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 16.3 \ (3) \ \text{(iii)}$
