The unit hyperbola has a sector with an area half of the hyperbolic angle
The independent variable of a hyperbolic function is called a hyperbolic angle . Just as the circular functions sine and cosine can be seen as projections from the unit circle to the axes, so the hyperbolic functions sinh and cosh are projections from a unit hyperbola to the axes.
The hyperbolic functions are defined in analogy with the trigonometric functions:
sinh
(
x
)
=
e
x
−
e
−
x
2
{\displaystyle \sinh(x)={\frac {e^{x}-e^{-x}}{2}}}
cosh
(
x
)
=
e
x
+
e
−
x
2
{\displaystyle \cosh(x)={\frac {e^{x}+e^{-x}}{2}}}
tanh
(
x
)
=
sinh
(
x
)
cosh
(
x
)
=
e
x
−
e
−
x
e
x
+
e
−
x
{\displaystyle \tanh(x)={\frac {\sinh(x)}{\cosh(x)}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}}
The reciprocal functions csch, sech, coth are defined from these functions:
c
s
c
h
(
x
)
=
1
sinh
(
x
)
{\displaystyle {\rm {csch}}(x)={\frac {1}{\sinh(x)}}}
s
e
c
h
(
x
)
=
1
cosh
(
x
)
{\displaystyle {\rm {sech}}(x)={\frac {1}{\cosh(x)}}}
coth
(
x
)
=
1
tanh
(
x
)
{\displaystyle \coth(x)={\frac {1}{\tanh(x)}}}
cosh
2
(
x
)
−
sinh
2
(
x
)
=
1
{\displaystyle \cosh ^{2}(x)-\sinh ^{2}(x)=1}
1
−
tanh
2
(
x
)
=
s
e
c
h
2
(
x
)
{\displaystyle 1-\tanh ^{2}(x)={\rm {sech}}^{2}(x)}
sinh
(
2
x
)
=
2
sinh
(
x
)
cosh
(
x
)
{\displaystyle \sinh(2x)=2\sinh(x)\cosh(x)}
cosh
(
2
x
)
=
cosh
2
(
x
)
+
sinh
2
(
x
)
{\displaystyle \cosh(2x)=\cosh ^{2}(x)+\sinh ^{2}(x)}
d
d
x
sinh
(
x
)
=
cosh
(
x
)
{\displaystyle {\frac {d}{dx}}\sinh(x)=\cosh(x)}
d
d
x
cosh
(
x
)
=
sinh
(
x
)
{\displaystyle {\frac {d}{dx}}\cosh(x)=\sinh(x)}
d
d
x
tanh
(
x
)
=
s
e
c
h
2
(
x
)
{\displaystyle {\frac {d}{dx}}\tanh(x)={\rm {sech}}^{2}(x)}
d
d
x
c
s
c
h
(
x
)
=
−
c
s
c
h
(
x
)
c
o
t
h
(
x
)
{\displaystyle {\frac {d}{dx}}{\rm {csch}}(x)=-{\rm {csch}}(x){\rm {coth}}(x)}
d
d
x
s
e
c
h
(
x
)
=
−
s
e
c
h
(
x
)
tanh
(
x
)
{\displaystyle {\frac {d}{dx}}{\rm {sech}}(x)=-{\rm {sech}}(x)\tanh(x)}
d
d
x
c
o
t
h
(
x
)
=
−
c
s
c
h
2
(
x
)
{\displaystyle {\frac {d}{dx}}{\rm {coth}}(x)=-{\rm {csch}}^{2}(x)}
There is no problem in defining principal braches for sinh and tanh because they are injective. We choose one of the principal branches for cosh.
sinh
:
R
→
R
{\displaystyle \sinh :\mathbb {R} \to \mathbb {R} }
cosh
:
[
0
,
∞
]
→
[
1
,
∞
]
{\displaystyle \cosh :[0,\infty ]\to [1,\infty ]}
tanh
:
R
→
(
−
1
,
1
)
{\displaystyle \tanh :\mathbb {R} \to (-1,1)}
With the principal values defined above, the definition of the inverse functions is immediate:
a
r
s
i
n
h
:
R
→
R
{\displaystyle {\rm {arsinh:\mathbb {R} \to \mathbb {R} }}}
a
r
c
o
s
h
:
[
1
,
∞
]
→
[
0
,
∞
]
{\displaystyle {\rm {arcosh:[1,\infty ]\to [0,\infty ]}}}
a
r
t
a
n
h
:
(
−
1
,
1
)
→
R
{\displaystyle {\rm {artanh:(-1,1)\to \mathbb {R} }}}
We can define
a
r
c
s
c
h
{\displaystyle {\rm {arcsch}}}
,
a
r
s
e
c
h
{\displaystyle {\rm {arsech}}}
and
a
r
c
o
t
h
{\displaystyle {\rm {arcoth}}}
similarly.
We can also write these inverses using the logarithm function,
a
r
s
i
n
h
(
x
)
=
ln
(
x
+
x
2
+
1
)
{\displaystyle {\rm {arsinh}}(x)=\ln {\big (}x+{\sqrt {x^{2}+1}}{\big )}}
a
r
c
o
s
h
(
x
)
=
ln
(
x
+
x
2
−
1
)
{\displaystyle {\rm {arcosh}}(x)=\ln {\big (}x+{\sqrt {x^{2}-1}}{\big )}}
a
r
t
a
n
h
(
x
)
=
ln
(
1
+
x
1
−
x
)
{\displaystyle {\rm {artanh}}(x)=\ln \left({\sqrt {\frac {1+x}{1-x}}}\right)}
These identities can simplify some integrals.
d
d
x
a
r
s
i
n
h
(
x
)
=
1
1
+
x
2
{\displaystyle {\frac {d}{dx}}{\rm {arsinh}}(x)={\frac {1}{\sqrt {1+x^{2}}}}}
d
d
x
a
r
c
o
s
h
(
x
)
=
1
x
2
−
1
,
x
>
1
{\displaystyle {\frac {d}{dx}}{\rm {arcosh}}(x)={\frac {1}{\sqrt {x^{2}-1}}}\ ,\ x>1}
d
d
x
a
r
t
a
n
h
(
x
)
=
1
1
−
x
2
,
|
x
|
<
1
{\displaystyle {\frac {d}{dx}}{\rm {artanh}}(x)={\frac {1}{1-x^{2}}}\ ,\ |x|<1}
d
d
x
a
r
c
s
c
h
(
x
)
=
−
1
|
x
|
1
+
x
2
,
x
≠
0
{\displaystyle {\frac {d}{dx}}{\rm {arcsch}}(x)=-{\frac {1}{|x|{\sqrt {1+x^{2}}}}}\ ,\ x\neq 0}
d
d
x
a
r
s
e
c
h
(
x
)
=
−
1
x
1
−
x
2
,
0
<
x
<
1
{\displaystyle {\frac {d}{dx}}{\rm {arsech}}(x)=-{\frac {1}{x{\sqrt {1-x^{2}}}}}\ ,\ 0<x<1}
d
d
x
a
r
c
o
t
h
(
x
)
=
1
1
−
x
2
,
|
x
|
>
1
{\displaystyle {\frac {d}{dx}}{\rm {arcoth}}(x)={\frac {1}{1-x^{2}}}\ ,\ |x|>1}
Hyperbolic functions are examples of transcendental functions -- they are not algebraic functions. They include trigonometric, inverse trigonometric, logarithmic and exponential functions.