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Hyperbolic Trigonometric Functions
Just as trigonometry can be performed on the unit circle, it can also be performed on the unit hyperbola:
x 2 − y 2 = 1.
Define the hyperbolic cosine function by
cosh( x ) =
e x
and the hyperbolic sine function by
sinh( x ) =
e x − e − x
A simple computation then shows that
cosh 2 ( x ) − sinh 2 ( x ) = 1 ,
which is the hyperbolic trigonometric identity analogous to the Pythagorean identity
cos 2 ( x ) + sin 2 ( x ) = 1
from ordinary (circular) trigonometry.
Next, as in the case of ordinary trigonometry, we define the remaining four hyperbolic trigonometric functions
from these first two:
Define the hyperbolic tangent function by
tanh( x ) =
sinh( x )
cosh( x )
and the hyperbolic cotangent function by
coth( x ) =
cosh( x )
sinh( x )
Define the hyperbolic secant function by
sech( x ) =
cosh( x )
and the hyperbolic cosecant function by
csch( x ) =
sinh( x )
Hyperbolic Pythagorean Identities
In addition to the identity derived earlier,
cosh
2 ( x ) − sinh
2 ( x ) = 1 ,
there are two other hyperbolic identities obtained by dividing this one by sinh 2 ( x ) and cosh 2 ( x ), respectively:
cosh 2 ( x ) − sinh 2 ( x ) = 1
cosh 2 ( x )
sinh 2 ( x )
sinh 2 ( x )
sinh 2 ( x )
sinh 2 ( x ) ( cosh( x )
sinh( x )
sinh 2 ( x )
coth 2 ( x ) − 1 = csch 2 ( x ).
cosh 2 ( x ) − sinh 2 ( x ) = 1
cosh 2 ( x )
cosh 2 ( x )
sinh 2 ( x )
cosh 2 ( x )
cosh 2 ( x )
sinh( x )
cosh( x )
cosh 2 ( x )
1 − tanh
2 ( x ) = sech
2 ( x ).
We summarize these here:
cosh 2 ( x ) − sinh 2 ( x ) = 1
1 − tanh 2 ( x ) = sech 2 ( x )
coth 2 ( x ) − 1 = csch 2 ( x )
Derivatives of Hyperbolic Functions
By using the definition of the hyperbolic sine function, we have
d
dx
(sinh( x )) =
d
d x
ex^ − e − x
2
ex^ + e − x
2
= cosh( x ).
Similarly,
d
dx
(cosh( x )) =
d
dx
e x
e x − e − x
= sinh( x ).
The derivatives of the remaining hyperbolic trigonometric functions are obtained from their definitions using the
Pythagorean hyperbolic trigonometric identities, the definitions of the hyperbolic functions, and the quotient rule:
d
d x
(tanh( x )) =
d
dx
sinh( x )
cosh( x )
d d x (sinh( x )) (cosh( x ))^ −^ (sinh( x ))^
d d x (cosh( x )) cosh 2 ( x )
cosh( x ) cosh( x ) − sinh( x ) sinh( x )
cosh 2 ( x )
cosh 2 ( x ) − sinh 2 ( x )
cosh 2 ( x )
cosh 2 ( x )
= sech 2 ( x ).
Similarly,
d
d x
(coth( x )) =
d
dx
cosh( x )
sinh( x )
d d x (cosh( x )) (sinh( x ))^ −^ (cosh( x ))^
d dx (sinh( x )) sinh 2 ( x )
sinh( x ) sinh( x ) − cosh( x ) cosh( x )
sinh 2 ( x )
sinh 2 ( x ) − cosh 2 ( x )
sinh 2 ( x )
sinh 2 ( x )
= − csch 2 ( x ).
Clearing fractions and moving everything to one side, we get
e y − 2 x − e − y = 0 ,
and multiplying by e y , we get e 2 y − 2 xe y − 1 = 0.
This is a quadratic equation in ey^ , so we may solve it by the quadratic formula:
e y =
2 x ±
(− 2 x )^2 − 4 ( 1 )(− 1 )
2
2 x ±
4 x^2 + 4
2
2 x ± 2
x^2 + 1
2
x ±
x^2 + 1
= x ±
x^2 + 1
Since ey^ must be positive, we can discard the negative square root here to get
e y = x +
x^2 + 1 ,
so,
sinh − 1 ( x ) = ln
x +
x^2 + 1
Logarithmic form for cosh − 1 ( x ) ( x ≥ 1)
Let y = cosh
− 1 ( x ).
Then
x = cosh( y ),
so
x =
e y
Clearing fractions and moving everything to one side, we get
e y − 2 x + e − y = 0 ,
and multiplying by e y , we get e 2 y − 2 xe y
This is a quadratic equation in ey^ , so we may solve it by the quadratic formula:
e y =
2 x ±
(− 2 x )^2 − 4 ( 1 )( 1 )
2
2 x ±
4 x^2 − 4
2
2 x ± 2
x^2 − 1
2
x ±
x^2 − 1
= x ±
x^2 − 1
Since ey^ must be positive, we can discard the negative square root here to get
e y = x +
x^2 − 1 ,
so,
cosh − 1 ( x ) = ln
x +
x^2 − 1
Logarithmic form for tanh − 1 ( x )
Let y = tanh
− 1 ( x ).
Then
x = tanh( y ),
so
x =
e y − e − y
ey^ + e − y^
Clearing fractions and moving everything to one side, we get
x
e y
= e y − e − y
x
e 2 y
= e 2 y − 1
xe 2 y
1 + x = e 2 y − xe 2 y
1 + x = e 2 y ( 1 − x )
e
2 y
1 + x
1 − x
2 y = ln
1 + x
1 − x
y =
ln
1 + x
1 − x
So,
tanh − 1 ( x ) =
ln
1 + x
1 − x
Logarithmic form for coth − 1 ( x )
Let y = coth − 1 ( x ).
Then
x = coth( y ),
so
x =
e y
ey^ − e − y^
Clearing fractions and moving everything to one side, we get
x
e y − e − y
= e y
x
e 2 y − 1
= e 2 y
xe 2 y − x = e 2 y
x + 1 = xe 2 y − e 2 y
x + 1 = e 2 y ( x − 1 )
e 2 y =
x + 1
x − 1
2 y = ln
x + 1
x − 1
y =
ln
x + 1
x − 1
So,
coth − 1 ( x ) =
ln
1 + x
x − 1
This is a quadratic equation in e y , so we can solve this using the quadratic formula:
e y =
(− 2 )^2 − 4 ( x )(− x )
2 x
4 + 4 x^2
2 x
1 + x^2
2 x
1 + x^2
2 x
1 + x^2
x
Since e y must be positive, we must have
e y =
1 + x^2
x
for x < 0
1 + x^2
x
for x > 0
We can write these two equations as one as follows:
e y =
x
1 + x^2
| x |
Taking the natural logarithm, we get
csch − 1 ( x ) = ln
x
1 + x^2
| x |
Derivatives of Inverse Hyperbolic Functions
Verify the following derivatives:
d
d x
sinh − 1 ( x )
1 + x^2 d
d x
cosh − 1 ( x )
x^2 − 1 d
d x
tanh − 1 ( x )
1 − x^2 d
d x
coth − 1 ( x )
1 − x^2 d
d x
sech
− 1 ( x )
x
1 − x^2 d
d x
csch − 1 ( x )
| x |
1 + x^2
