Double Angle Formulas/Tangent

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Theorem

$\tan 2 \theta = \dfrac {2 \tan \theta} {1 - \tan^2 \theta}$

where $\tan$ denotes tangent.


Corollary

Let $u = \tan \dfrac \theta 2$.

Then:

$\tan \theta = \dfrac {2 u} {1 - u^2}$


Proof 1

\(\ds \tan 2 \theta\) \(=\) \(\ds \frac {\sin 2 \theta} {\cos 2 \theta}\) Tangent is Sine divided by Cosine
\(\ds \) \(=\) \(\ds \frac {2 \cos \theta \sin \theta} {\cos^2 \theta - \sin^2 \theta}\) Double Angle Formula for Sine and Double Angle Formula for Cosine
\(\ds \) \(=\) \(\ds \frac {\frac {2 \cos \theta \sin \theta} {\cos^2 \theta} } {\frac {\cos^2 \theta - \sin^2 \theta} {\cos^2 \theta} }\) dividing numerator and denominator by $\cos^2 \theta$
\(\ds \) \(=\) \(\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}\) simplifying: Tangent is Sine divided by Cosine

$\blacksquare$


Proof 2

\(\ds \tan 2 \theta\) \(=\) \(\ds \map \tan {\theta + \theta}\)
\(\ds \) \(=\) \(\ds \frac {\tan \theta + \tan \theta} {1 - \tan \theta \tan \theta}\) Tangent of Sum
\(\ds \) \(=\) \(\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}\)

$\blacksquare$


Proof 3

\(\ds \frac {2 \tan \theta} {1 - \tan^2 \theta}\) \(=\) \(\ds \dfrac {2 i \dfrac {1 - e^{2 i \theta} } {1 + e^{2 i \theta} } } {1 - \paren {i \dfrac {1 - e^{2 i \theta} } {1 + e^{2 i \theta} } }^2}\) 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

$\blacksquare$


Sources

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