Double Angle Formulas/Hyperbolic Tangent/Proof 3
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Theorem
- $\tanh 2 x = \dfrac {2 \tanh x} {1 + \tanh^2 x}$
Proof
Starting from the right, we have
\(\ds \dfrac {2 \tanh x} {1 + \tanh^2 x}\) | \(=\) | \(\ds \dfrac {2 \paren {\dfrac {e^x - e^{-x} } {e^x + e^{-x} } } } {1 + \paren {\dfrac{e^x - e^{-x} } {e^x + e^{-x} } }^2}\) | Definition 1 of Hyperbolic Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 \paren {e^x + e^{-x} } \paren {e^x - e^{-x} } } {\paren {e^x + e^{-x} }^2 + \paren {e^x - e^{-x} }^2}\) | multiplying both numerator and denominator by $\paren {e^x + e^{-x} }^2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 \paren {e^{2 x} - e^{-2 x} } } {e^{2 x} + 2 + e^{-2 x} + e^{2 x} - 2 + e^{-2 x} }\) | Difference of Two Squares, Square of Sum, Square of Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 \paren {e^{2 x} - e^{-2 x} } } {2 e^{2 x} + 2 e^{-2 x} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {e^{2 x} - e^{-2 x} } {e^{2 x} + e^{-2 x} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \tanh {2 x}\) | Definition 1 of Hyperbolic Tangent |
$\blacksquare$