Double Angle Formulas/Tangent/Proof 2
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Theorem
- $\tan 2 \theta = \dfrac {2 \tan \theta} {1 - \tan^2 \theta}$
Proof
| \(\ds \tan 2 \theta\) | \(=\) | \(\ds \map \tan {\theta + \theta}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\tan \theta + \tan \theta} {1 - \tan \theta \tan \theta}\) | Tangent of Sum | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}\) |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Tangents of sum and difference