Double Angle Formulas/Tangent/Proof 3
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Theorem
- $\tan 2 \theta = \dfrac {2 \tan \theta} {1 - \tan^2 \theta}$
Proof
| \(\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}\) | \(=\) | \(\ds \dfrac {2 i \dfrac {1 - e^{2 i \theta} } {1 + e^{2 i \theta} } } {1 - \paren {i \dfrac {1 - e^{2 i \theta} } {1 + e^{2 i \theta} } }^2}\) | Euler's Tangent Identity | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {2 i \paren {1 - e^{2 i \theta} } \paren {1 + e^{2 i \theta} } } {\paren {1 + e^{2 i \theta} }^2 + \paren {1 - e^{2 i \theta} }^2}\) | multiplying both numerator and denominator by $\paren {1 + e^{2 i \theta} }^2$; $i$ is the imaginary unit | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {2 i \paren {1 - e^{4 i \theta} } } {1 + 2 e^{2 i \theta} + e^{4 i \theta} + 1 - 2 e^{2 i \theta} + e^{4 i \theta} }\) | Difference of Two Squares, Square of Sum, Square of Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {2 i \paren {1 - e^{4 i \theta} } } {2 + 2 e^{4 i \theta} }\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds i \dfrac {1 - e^{4 i \theta} } {1 + e^{4 i \theta} }\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \tan 2 \theta\) | Euler's Tangent Identity |
$\blacksquare$