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

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