Half Angle Formula for Hyperbolic Tangent/Corollary 2

Theorem

For $x \ne 0$:

$\tanh \dfrac x 2 = \dfrac {\cosh x - 1} {\sinh x}$

where $\tanh$ denotes hyperbolic tangent, $\sinh$ denotes hyperbolic sine and $\cosh$ denotes hyperbolic cosine.

$\ds \tanh \frac x 2$

$=$

$\ds \pm \sqrt {\frac {\cosh x - 1} {\cosh x + 1} }$

$\ds $

$=$

$\ds \pm \sqrt {\frac {\paren {\cosh x - 1}^2} {\paren {\cosh x + 1} \paren {\cosh x - 1} } }$

multiplying numerator and denominator by $\sqrt {\cosh x - 1}$

$\ds $

$=$

$\ds \pm \sqrt {\frac {\paren {\cosh x - 1}^2} {\cosh^2 x - 1} }$

$\ds $

$=$

$\ds \pm \sqrt {\frac {\paren {\cosh x - 1}^2} {\sinh^2 x} }$

$\ds $

$=$

$\ds \pm \frac {\cosh x - 1} {\sinh x}$

Since $\cosh x \ge 1$, it follows that $\cosh x - 1 \ge 0$, with equality happening at $x = 0$.

We also have that:

when $x > 0$, $\tanh \dfrac x 2 > 0$ and $\sinh x > 0$

when $x < 0$, $\tanh \dfrac x 2 < 0$ and $\sinh x < 0$.

Thus:

$\tanh \dfrac x 2 = \dfrac {\cosh x - 1} {\sinh x}$

$\blacksquare$