Half Angle Formula for Hyperbolic Tangent/Corollary 1
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Theorem
Let $x \in \R$.
Then:
- $\tanh \dfrac x 2 = \dfrac {\sinh x} {\cosh x + 1}$
where $\tanh$ denotes hyperbolic tangent, $\sinh$ denotes hyperbolic sine and $\cosh$ denotes hyperbolic cosine.
Proof
\(\ds \tanh \frac x 2\) | \(=\) | \(\ds \pm \sqrt {\frac {\cosh x - 1} {\cosh x + 1} }\) | Half Angle Formula for Hyperbolic Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \sqrt {\frac {\paren {\cosh x - 1} \paren {\cosh x + 1} } {\paren {\cosh x + 1}^2} }\) | multiplying top and bottom by $\sqrt {\cosh x + 1}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \sqrt {\frac {\cosh x - 1} {\paren {\cosh x + 1}^2} }\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \sqrt {\frac {\sinh^2 x} {\paren {\cosh x + 1}^2} }\) | Difference of Squares of Hyperbolic Cosine and Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \frac {\sinh x} {\cosh x + 1}\) |
Since $\cosh x > 0$, it follows that $\cosh x + 1 > 0$.
We also have that:
- when $x \ge 0$, $\tanh \dfrac x 2 \ge 0$ and $\sinh x \ge 0$
- when $x \le 0$, $\tanh \dfrac x 2 \le 0$ and $\sinh x \le 0$.
Thus:
- $\tanh \dfrac x 2 = \dfrac {\sinh x} {\cosh x + 1}$
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 8$: Hyperbolic Functions: $8.29$: Double Angle Formulas