(a) cosh(x) = (ex+e-x)/2,
(b) sinh(x) = (ex-e-x)/2,
(c) tanh(x) = sinh(x)/cosh(x) (= (ex-e-x)/(ex+e-x)).
From the definitions, we can easily deduce
(1) cosh2(x) - sinh2(x) = 1,
(2) sinh(2x) = 2cosh(x)sinh(x),
Using (1) and (2),(3), we get
(4) sinh(2x) = 2tanh(x)/(1-tanh2(x)),
Again from the definition, after some calculation,
(7) tanh(x+y) = (tanh(x)+tanh(y))/(1+tanh(x)tanh(y)).
Note that, by (c),
tanh(x) = (1-e-2x)/(1+e-2x) = 1 - 2/(e2x+1), so that
tanh(0) = 0.
Also, for x > 0, e2x > 1, and increases with x, so that
tanh(x) increases on [0,∞), and
tanh(x) tends to 1 as x tends to ∞.
Thus tanh(x) is increasing, and maps [0,∞) to [0,1).
It follows that the inverse function arctanh(x) is an