Reciprocal of Hyperbolic Cosine Plus One
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Theorem
- $\dfrac 1 {\cosh x + 1} = \dfrac 1 2 \sech^2 \dfrac x 2$
Proof
\(\ds \cosh x\) | \(=\) | \(\ds 2 \cosh^2 \frac x 2 - 1\) | Double Angle Formula for Hyperbolic Cosine: Corollary $1$ | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \cosh x + 1\) | \(=\) | \(\ds 2 \cosh^2 \frac x 2\) | adding $1$ to both sides | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \frac 1 {\cosh x + 1}\) | \(=\) | \(\ds \frac 1 2 \frac 1 {\cosh^2 \frac x 2}\) | taking the reciprocal of both sides | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 \sech^2 \frac x 2\) | Definition 2 of Hyperbolic Secant |
$\blacksquare$