Deriving Two Inverse Functions

In “Instrumental Flying“, we defined function y=\sinh^{-1}(x) as

\{(x, y) | x = \frac{e^y-e^{-y}}{2}\}.\quad\quad\quad(1)

From x = \frac{e^y-e^{-y}}{2}, we obtain

(e^y)^2-2x\cdot e^y-1=0.

It means either e^y = x-\sqrt{x^2+1} or e^y = x+\sqrt{x^2+1}.

But e^y = x-\sqrt{x^2+1} suggests e^y < 0 (see Exercise-1), contradicts the fact that \forall t \in R, e^t > 0 (see “Two Peas in a Pod, Part 2“).

Therefore,

e^y = x+\sqrt{x^2+1} \implies y = \log(x + \sqrt{x^2+1}).

i.e.,

\sinh^{-1}(x) = \log(x + \sqrt{x^2+1}), \;\;x \in (-\infty, +\infty).

We also defined a non-negative valued function y = \cosh^{-1}(x):

\{(x, y) | x = \frac{e^y + e^{-y}}{2}, y \ge 0\}.\quad\quad\quad(2)

Simplifying x=\frac{e^y+e^{-y}}{2} yields

(e^y)^2-2x\cdot e^y+1=0.

It follows that either e^y = x-\sqrt{x^2-1} or e^y=x+\sqrt{x^2-1}.

For both expressions’ right-hand side to be valid requires that x \ge 1. However, when x = 2,

e^y = x-\sqrt{x^2-1} = 2 - \sqrt{3} < 1

suggests that y < 0 (see Exercise-2,3), contradicts (2).

Hence,

e^y = x+\sqrt{x^2-1} \implies y = \log(x+\sqrt{x^2-1}).

i.e.,

\cosh^{-1}(x) = \log(x+\sqrt{x^2-1}), \;\;x \in [1, +\infty).

See also “A Relentless Pursuit“.


Exercise-1 Show that \forall x \in R, x-\sqrt{x^2+1} < 0.

Exercise-2 Without calculator or CAS, show that 2-\sqrt{3} < 1.

Exercise-3 Prove \forall t \ge 0. e^t \ge 1 (hint: see “Two Peas in a Pod, Part 2“)

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