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 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 $\dfrac {BA \cdot CA} 2$ |
$\blacksquare$
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): $\S 16.3 \ (3) \ \text{(iii)}$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Entry: double-angle formula (in trigonometry)
