Hyperbolic functions

sinh, cosh and tanh

csch, sech and coth

There are various equivalent ways to define the hyperbolic functions.

Exponential definitions

sinh x is half the difference of e^x and e^−x

cosh x is the average of e^x and e^−x

In terms of the exponential function:[1][4]

• Hyperbolic sine: the odd part of the exponential function, that is,
  $${\displaystyle \sinh x={\frac {e^{x}-e^{-x}}{2}}={\frac {e^{2x}-1}{2e^{x}}}={\frac {1-e^{-2x}}{2e^{-x}}}.}$$

  
• Hyperbolic cosine: the even part of the exponential function, that is,
  $${\displaystyle \cosh x={\frac {e^{x}+e^{-x}}{2}}={\frac {e^{2x}+1}{2e^{x}}}={\frac {1+e^{-2x}}{2e^{-x}}}.}$$

  
• Hyperbolic tangent:
  $${\displaystyle \tanh x={\frac {\sinh x}{\cosh x}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}={\frac {e^{2x}-1}{e^{2x}+1}}.}$$

  
• Hyperbolic cotangent: for x ≠ 0,
  $${\displaystyle \coth x={\frac {\cosh x}{\sinh x}}={\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}={\frac {e^{2x}+1}{e^{2x}-1}}.}$$

  
• Hyperbolic secant:
  $${\displaystyle \operatorname {sech} x={\frac {1}{\cosh x}}={\frac {2}{e^{x}+e^{-x}}}={\frac {2e^{x}}{e^{2x}+1}}.}$$

  
• Hyperbolic cosecant: for x ≠ 0,
  $${\displaystyle \operatorname {csch} x={\frac {1}{\sinh x}}={\frac {2}{e^{x}-e^{-x}}}={\frac {2e^{x}}{e^{2x}-1}}.}$$

  

Differential equation definitions

The hyperbolic functions may be defined as solutions of differential equations: The hyperbolic sine and cosine are the solution (s, c) of the system

$${\displaystyle {\begin{aligned}c'(x)&=s(x),\\s'(x)&=c(x),\\\end{aligned}}}$$

with the initial conditions ${\displaystyle s(0)=0,c(0)=1.}$ The initial conditions make the solution unique; without them any pair of functions ${\displaystyle (ae^{x}+be^{-x},ae^{x}-be^{-x})}$ would be a solution.

sinh(x) and cosh(x) are also the unique solution of the equation f ″(x) = f (x), such that f (0) = 1, f ′(0) = 0 for the hyperbolic cosine, and f (0) = 0, f ′(0) = 1 for the hyperbolic sine.

The hyperbolic functions satisfy many identities, all of them similar in form to the trigonometric identities. In fact, Osborn's rule[18] states that one can convert any trigonometric identity for ${\displaystyle \theta }$, ${\displaystyle 2\theta }$, ${\displaystyle 3\theta }$ or ${\displaystyle \theta }$ and ${\displaystyle \varphi }$ into a hyperbolic identity, by expanding it completely in terms of integral powers of sines and cosines, changing sine to sinh and cosine to cosh, and switching the sign of every term containing a product of two sinhs.

Odd and even functions:

$${\displaystyle {\begin{aligned}\sinh(-x)&=-\sinh x\\\cosh(-x)&=\cosh x\end{aligned}}}$$

Hence:

$${\displaystyle {\begin{aligned}\tanh(-x)&=-\tanh x\\\coth(-x)&=-\coth x\\\operatorname {sech} (-x)&=\operatorname {sech} x\\\operatorname {csch} (-x)&=-\operatorname {csch} x\end{aligned}}}$$

Thus, cosh x and sech x are even functions; the others are odd functions.

$${\displaystyle {\begin{aligned}\operatorname {arsech} x&=\operatorname {arcosh} \left({\frac {1}{x}}\right)\\\operatorname {arcsch} x&=\operatorname {arsinh} \left({\frac {1}{x}}\right)\\\operatorname {arcoth} x&=\operatorname {artanh} \left({\frac {1}{x}}\right)\end{aligned}}}$$

Hyperbolic sine and cosine satisfy:

$${\displaystyle {\begin{aligned}\cosh x+\sinh x&=e^{x}\\\cosh x-\sinh x&=e^{-x}\\\cosh ^{2}x-\sinh ^{2}x&=1\end{aligned}}}$$

the last of which is similar to the Pythagorean trigonometric identity.

One also has

$${\displaystyle {\begin{aligned}\operatorname {sech} ^{2}x&=1-\tanh ^{2}x\\\operatorname {csch} ^{2}x&=\coth ^{2}x-1\end{aligned}}}$$

for the other functions.

Sums of arguments

$${\displaystyle {\begin{aligned}\sinh(x+y)&=\sinh x\cosh y+\cosh x\sinh y\\\cosh(x+y)&=\cosh x\cosh y+\sinh x\sinh y\\[6px]\tanh(x+y)&={\frac {\tanh x+\tanh y}{1+\tanh x\tanh y}}\\\end{aligned}}}$$

particularly

$${\displaystyle {\begin{aligned}\cosh(2x)&=\sinh ^{2}{x}+\cosh ^{2}{x}=2\sinh ^{2}x+1=2\cosh ^{2}x-1\\\sinh(2x)&=2\sinh x\cosh x\\\tanh(2x)&={\frac {2\tanh x}{1+\tanh ^{2}x}}\\\end{aligned}}}$$

Also:

$${\displaystyle {\begin{aligned}\sinh x+\sinh y&=2\sinh \left({\frac {x+y}{2}}\right)\cosh \left({\frac {x-y}{2}}\right)\\\cosh x+\cosh y&=2\cosh \left({\frac {x+y}{2}}\right)\cosh \left({\frac {x-y}{2}}\right)\\\end{aligned}}}$$

Subtraction formulas

$${\displaystyle {\begin{aligned}\sinh(x-y)&=\sinh x\cosh y-\cosh x\sinh y\\\cosh(x-y)&=\cosh x\cosh y-\sinh x\sinh y\\\tanh(x-y)&={\frac {\tanh x-\tanh y}{1-\tanh x\tanh y}}\\\end{aligned}}}$$

Also:[19]

$${\displaystyle {\begin{aligned}\sinh x-\sinh y&=2\cosh \left({\frac {x+y}{2}}\right)\sinh \left({\frac {x-y}{2}}\right)\\\cosh x-\cosh y&=2\sinh \left({\frac {x+y}{2}}\right)\sinh \left({\frac {x-y}{2}}\right)\\\end{aligned}}}$$

Half argument formulas

$${\displaystyle {\begin{aligned}\sinh \left({\frac {x}{2}}\right)&={\frac {\sinh x}{\sqrt {2(\cosh x+1)}}}&&=\operatorname {sgn} x\,{\sqrt {\frac {\cosh x-1}{2}}}\\[6px]\cosh \left({\frac {x}{2}}\right)&={\sqrt {\frac {\cosh x+1}{2}}}\\[6px]\tanh \left({\frac {x}{2}}\right)&={\frac {\sinh x}{\cosh x+1}}&&=\operatorname {sgn} x\,{\sqrt {\frac {\cosh x-1}{\cosh x+1}}}={\frac {e^{x}-1}{e^{x}+1}}\end{aligned}}}$$

where sgn is the sign function.

If x ≠ 0, then[20]

$${\displaystyle \tanh \left({\frac {x}{2}}\right)={\frac {\cosh x-1}{\sinh x}}=\coth x-\operatorname {csch} x}$$

Square formulas

$${\displaystyle {\begin{aligned}\sinh ^{2}x&={\tfrac {1}{2}}(\cosh 2x-1)\\\cosh ^{2}x&={\tfrac {1}{2}}(\cosh 2x+1)\end{aligned}}}$$

Each of the functions sinh and cosh is equal to its second derivative, that is:

$${\displaystyle {\frac {d^{2}}{dx^{2}}}\sinh x=\sinh x}$$

$${\displaystyle {\frac {d^{2}}{dx^{2}}}\cosh x=\cosh x\,.}$$

All functions with this property are linear combinations of sinh and cosh, in particular the exponential functions ${\displaystyle e^{x}}$ and ${\displaystyle e^{-x}}$.[22]

$${\displaystyle {\begin{aligned}\int \sinh(ax)\,dx&=a^{-1}\cosh(ax)+C\\\int \cosh(ax)\,dx&=a^{-1}\sinh(ax)+C\\\int \tanh(ax)\,dx&=a^{-1}\ln(\cosh(ax))+C\\\int \coth(ax)\,dx&=a^{-1}\ln \left|\sinh(ax)\right|+C\\\int \operatorname {sech} (ax)\,dx&=a^{-1}\arctan(\sinh(ax))+C\\\int \operatorname {csch} (ax)\,dx&=a^{-1}\ln \left|\tanh \left({\frac {ax}{2}}\right)\right|+C=a^{-1}\ln \left|\coth \left(ax\right)-\operatorname {csch} \left(ax\right)\right|+C=-a^{-1}\operatorname {arcoth} \left(\cosh \left(ax\right)\right)+C\end{aligned}}}$$

The following integrals can be proved using hyperbolic substitution:

$${\displaystyle {\begin{aligned}\int {{\frac {1}{\sqrt {a^{2}+u^{2}}}}\,du}&=\operatorname {arsinh} \left({\frac {u}{a}}\right)+C\\\int {{\frac {1}{\sqrt {u^{2}-a^{2}}}}\,du}&=\operatorname {sgn} {u}\operatorname {arcosh} \left|{\frac {u}{a}}\right|+C\\\int {\frac {1}{a^{2}-u^{2}}}\,du&=a^{-1}\operatorname {artanh} \left({\frac {u}{a}}\right)+C&&u^{2}<a^{2}\\\int {\frac {1}{a^{2}-u^{2}}}\,du&=a^{-1}\operatorname {arcoth} \left({\frac {u}{a}}\right)+C&&u^{2}>a^{2}\\\int {{\frac {1}{u{\sqrt {a^{2}-u^{2}}}}}\,du}&=-a^{-1}\operatorname {arsech} \left|{\frac {u}{a}}\right|+C\\\int {{\frac {1}{u{\sqrt {a^{2}+u^{2}}}}}\,du}&=-a^{-1}\operatorname {arcsch} \left|{\frac {u}{a}}\right|+C\end{aligned}}}$$

where C is the constant of integration.

It is possible to express explicitly the Taylor series at zero (or the Laurent series, if the function is not defined at zero) of the above functions.

$${\displaystyle \sinh x=x+{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+{\frac {x^{7}}{7!}}+\cdots =\sum _{n=0}^{\infty }{\frac {x^{2n+1}}{(2n+1)!}}}$$

This series is convergent for every complex value of x. Since the function sinh x is odd, only odd exponents for x occur in its Taylor series.

$${\displaystyle \cosh x=1+{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}+{\frac {x^{6}}{6!}}+\cdots =\sum _{n=0}^{\infty }{\frac {x^{2n}}{(2n)!}}}$$

This series is convergent for every complex value of x. Since the function cosh x is even, only even exponents for x occur in its Taylor series.

The sum of the sinh and cosh series is the infinite series expression of the exponential function.

The following series are followed by a description of a subset of their domain of convergence, where the series is convergent and its sum equals the function.

$${\displaystyle {\begin{aligned}\tanh x&=x-{\frac {x^{3}}{3}}+{\frac {2x^{5}}{15}}-{\frac {17x^{7}}{315}}+\cdots =\sum _{n=1}^{\infty }{\frac {2^{2n}(2^{2n}-1)B_{2n}x^{2n-1}}{(2n)!}},\qquad \left|x\right|<{\frac {\pi }{2}}\\\coth x&=x^{-1}+{\frac {x}{3}}-{\frac {x^{3}}{45}}+{\frac {2x^{5}}{945}}+\cdots =\sum _{n=0}^{\infty }{\frac {2^{2n}B_{2n}x^{2n-1}}{(2n)!}},\qquad 0<\left|x\right|<\pi \\\operatorname {sech} x&=1-{\frac {x^{2}}{2}}+{\frac {5x^{4}}{24}}-{\frac {61x^{6}}{720}}+\cdots =\sum _{n=0}^{\infty }{\frac {E_{2n}x^{2n}}{(2n)!}},\qquad \left|x\right|<{\frac {\pi }{2}}\\\operatorname {csch} x&=x^{-1}-{\frac {x}{6}}+{\frac {7x^{3}}{360}}-{\frac {31x^{5}}{15120}}+\cdots =\sum _{n=0}^{\infty }{\frac {2(1-2^{2n-1})B_{2n}x^{2n-1}}{(2n)!}},\qquad 0<\left|x\right|<\pi \end{aligned}}}$$

where:

• ${\displaystyle B_{n}}$ is the nth Bernoulli number
• ${\displaystyle E_{n}}$ is the nth Euler number

The following expansions are valid in the whole complex plane:

${\displaystyle \sinh x=x\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{n^{2}\pi ^{2}}}\right)={\cfrac {x}{1-{\cfrac {x^{2}}{2\cdot 3+x^{2}-{\cfrac {2\cdot 3x^{2}}{4\cdot 5+x^{2}-{\cfrac {4\cdot 5x^{2}}{6\cdot 7+x^{2}-\ddots }}}}}}}}}$

${\displaystyle \cosh x=\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{(n-1/2)^{2}\pi ^{2}}}\right)={\cfrac {1}{1-{\cfrac {x^{2}}{1\cdot 2+x^{2}-{\cfrac {1\cdot 2x^{2}}{3\cdot 4+x^{2}-{\cfrac {3\cdot 4x^{2}}{5\cdot 6+x^{2}-\ddots }}}}}}}}}$

${\displaystyle \tanh x={\cfrac {1}{{\cfrac {1}{x}}+{\cfrac {1}{{\cfrac {3}{x}}+{\cfrac {1}{{\cfrac {5}{x}}+{\cfrac {1}{{\cfrac {7}{x}}+\ddots }}}}}}}}}$

Circle and hyperbola tangent at (1,1) display geometry of circular functions in terms of circular sector area u and hyperbolic functions depending on hyperbolic sector area u.

The hyperbolic functions represent an expansion of trigonometry beyond the circular functions. Both types depend on an argument, either circular angle or hyperbolic angle.

Since the area of a circular sector with radius r and angle u (in radians) is r²u/2, it will be equal to u when r = √2. In the diagram, such a circle is tangent to the hyperbola xy = 1 at (1,1). The yellow sector depicts an area and angle magnitude. Similarly, the yellow and red regions together depict a hyperbolic sector with area corresponding to hyperbolic angle magnitude.

The legs of the two right triangles with hypotenuse on the ray defining the angles are of length √2 times the circular and hyperbolic functions.

The hyperbolic angle is an invariant measure with respect to the squeeze mapping, just as the circular angle is invariant under rotation.[23]

The Gudermannian function gives a direct relationship between the circular functions and the hyperbolic functions that does not involve complex numbers.

The graph of the function a cosh(x/a) is the catenary, the curve formed by a uniform flexible chain, hanging freely between two fixed points under uniform gravity.

The decomposition of the exponential function in its even and odd parts gives the identities

$${\displaystyle e^{x}=\cosh x+\sinh x,}$$

and

$${\displaystyle e^{-x}=\cosh x-\sinh x.}$$

Combined with Euler's formula

$${\displaystyle e^{ix}=\cos x+i\sin x,}$$

this gives

$${\displaystyle e^{x+iy}=(\cosh x+\sinh x)(\cos y+i\sin y)}$$

for the general complex exponential function.

Additionally,

$${\displaystyle e^{x}={\sqrt {\frac {1+\tanh x}{1-\tanh x}}}={\frac {1+\tanh {\frac {x}{2}}}{1-\tanh {\frac {x}{2}}}}}$$

Hyperbolic functions in the complex plane

${\displaystyle \sinh(z)}$	${\displaystyle \cosh(z)}$	${\displaystyle \tanh(z)}$	${\displaystyle \coth(z)}$	${\displaystyle \operatorname {sech} (z)}$	${\displaystyle \operatorname {csch} (z)}$

Since the exponential function can be defined for any complex argument, we can also extend the definitions of the hyperbolic functions to complex arguments. The functions sinh z and cosh z are then holomorphic.

Relationships to ordinary trigonometric functions are given by Euler's formula for complex numbers:

$${\displaystyle {\begin{aligned}e^{ix}&=\cos x+i\sin x\\e^{-ix}&=\cos x-i\sin x\end{aligned}}}$$

so:

$${\displaystyle {\begin{aligned}\cosh(ix)&={\frac {1}{2}}\left(e^{ix}+e^{-ix}\right)=\cos x\\\sinh(ix)&={\frac {1}{2}}\left(e^{ix}-e^{-ix}\right)=i\sin x\\\cosh(x+iy)&=\cosh(x)\cos(y)+i\sinh(x)\sin(y)\\\sinh(x+iy)&=\sinh(x)\cos(y)+i\cosh(x)\sin(y)\\\tanh(ix)&=i\tan x\\\cosh x&=\cos(ix)\\\sinh x&=-i\sin(ix)\\\tanh x&=-i\tan(ix)\end{aligned}}}$$

Thus, hyperbolic functions are periodic with respect to the imaginary component, with period ${\displaystyle 2\pi i}$ (${\displaystyle \pi i}$ for hyperbolic tangent and cotangent).

• e (mathematical constant)
• Equal incircles theorem, based on sinh
• Hyperbolic growth
• Inverse hyperbolic functions
• List of integrals of hyperbolic functions
• Poinsot's spirals
• Sigmoid function
• Soboleva modified hyperbolic tangent
• Trigonometric functions

Wikimedia Commons has media related to Hyperbolic functions.

• "Hyperbolic functions", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Hyperbolic functions on PlanetMath
• GonioLab: Visualization of the unit circle, trigonometric and hyperbolic functions (Java Web Start)
• Web-based calculator of hyperbolic functions