derivation of formulas for hyperbolic
functions from definition of hyperbolic
angle∗
Wkbj79†
2013-03-22 0:34:34
integration formula
Let H1 be the branch (connected component) of the unit hyperbola x2 −y 2 =
1 with x > 0.
y
H1
.
x
.
Let α > 0. Then (cosh α, sinh α) is the point on H1 with hyperbolic angle
α.
In order to draw the hyperbolic angle, the line passing through (0, 0) and
(cosh α, sinh α) must be drawn. Recall that tanh α is defined by
tanh α :=
sinh α
.
cosh α
∗ hDerivationOfFormulasForHyperbolicFunctionsFromDefinitionOfHyperbolicAnglei
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1
Thus, the equation of the line passing through (0, 0) and (cosh α, sinh α) is
y = (tanh α)x.
Below is the graph of H1 and the line y = (tanh α)x.
y
H1
y = (tanh α)x
(cosh α, sinh α)
.
x
.
Observe also that sinh α > 0 and cosh α > 1.
In the hyperbolic angle entry, it is discussed that α is twice the area bounded
by the x axis, H1 , and the line y = (tanh α)x. We will use this fact to obtain
formulas for cosh α and sinh α.
In the calculations below, the following integration formula will be used:
Z p
p
xp 2
1 x2 − 1 dx =
x − 1 − ln x + x2 − 1 + C
2
2
Thus, we have
 cosh α

cosh
Z
Z αp
α = 2
(tanh α)x dx −
x2 − 1 dx
0
1
cosh α
p
p
cosh α
= (tanh α)x|0
− x x2 − 1 − ln x + x2 − 1 1
p
p
p
p
2
2
= tanh α cosh α − cosh α cosh α − 1 + ln cosh α + cosh2 α − 1 + 12 − 1 − ln 1 + 12 − 1
= cosh α sinh α − cosh α sinh α + ln(cosh α + sinh α)
= ln(cosh α + sinh α).
2
Thus, we have
eα = cosh α + sinh α.
A similar calculation yields
e−α = cosh α − sinh α.
The formulas for the hyperbolic functions are easily derived from the above
equations:
eα + e−α
2
eα − e−α
sinh α =
2
cosh α =
As mentioned in the hyperbolic angle entry, these formulas can be extended
to all α ∈ R and from there to all α ∈ C.
3