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Mathematics LibreTexts

4.11: Hyperbolic Functions

The hyperbolic functions appear with some frequency in applications, and are quite similar in many respects to the trigonometric functions. This is a bit surprising given our initial definitions.

Definition 4.11.1: Hyperbolic Cosines and Sines

The hyperbolic cosine is the function

\[\cosh x ={e^x +e^{-x }\over2},\]

and the hyperbolic sine is the function

\[\sinh x ={e^x -e^{-x}\over 2}.\]

Notice that \(\cosh\) is even (that is, \(\cosh(-x)=\cosh(x)\)) while \(\sinh\) is odd (\(\sinh(-x)=-\sinh(x)\)), and \( \cosh x + \sinh x = e^x\). Also, for all \(x\), \(\cosh x >0\), while \(\sinh x=0\) if and only if \( e^x -e^{-x }=0\), which is true precisely when \(x=0\).

Lemma 4.11.2

The range of \(\cosh x\) is \([1,\infty)\).

Proof

Let \(y= \cosh x\). We solve for \(x\):

\[\eqalign{y&={e^x +e^{-x }\over 2}\cr 2y &= e^x + e^{-x }\cr 2ye^x &= e^{2x} + 1\cr 0 &= e^{2x}-2ye^x +1\cr e^{x} &= {2y \pm \sqrt{4y^2 -4}\over 2}\cr e^{x} &= y\pm \sqrt{y^2 -1}\cr} \]

From the last equation, we see \( y^2 \geq 1\), and since \(y\geq 0\), it follows that \(y\geq 1\).

Now suppose \(y\geq 1\), so \( y\pm \sqrt{y^2 -1}>0\). Then \( x = \ln(y\pm \sqrt{y^2 -1})\) is a real number, and \(y =\cosh x\), so \(y\) is in the range of \(\cosh(x)\).

\(\square\)

Definition 4.11.3: Hyperbolic Tangent and Cotangent

The other hyperbolic functions are

\[\eqalign{\tanh x &= {\sinh x\over\cosh x}\cr \coth x &= {\cosh x\over\sinh x}\cr \text{sech} x &= {1\over\cosh x}\cr \text{csch} x &= {1\over\sinh x}\cr} \]

The domain of \(\coth\) and \(\text{csch}\) is \(x\neq 0\) while the domain of the other hyperbolic functions is all real numbers. Graphs are shown in figure 4.11.1

Figure 4.11.1. The hyperbolic functions.

Certainly the hyperbolic functions do not closely resemble the trigonometric functions graphically. But they do have analogous properties, beginning with the following identity.

Theorem 4.11.4

For all \(x\) in \(\mathbb{R}\), \( \cosh ^2 x -\sinh ^2 x = 1\).

Proof

The proof is a straightforward computation:

\[\cosh ^2 x -\sinh ^2 x = {(e^x +e^{-x} )^2\over 4} -{(e^x -e^{-x} )^2\over 4}= {e^{2x} + 2 + e^{-2x } - e^{2x } + 2 - e^{-2x}\over 4}= {4\over 4} = 1. \]

\(\square\)

This immediately gives two additional identities:

\[1-\tanh^2 x =\text{sech}^2 x\qquad\hbox{and}\qquad \coth^2 x - 1 =\text{csch}^2 x.\]

The identity of the theorem also helps to provide a geometric motivation. Recall that the graph of \( x^2 -y^2 =1\) is a hyperbola with asymptotes \(x=\pm y\) whose \(x\)-intercepts are \(\pm 1\). If \((x,y)\) is a point on the right half of the hyperbola, and if we let \(x=\cosh t\), then \( y=\pm\sqrt{x^2-1}=\pm\sqrt{\cosh^2x-1}=\pm\sinh t\). So for some suitable \(t\), \(\cosh t\) and \(\sinh t\) are the coordinates of a typical point on the hyperbola. In fact, it turns out that \(t\) is twice the area shown in the first graph of figure 4.11.2. Even this is analogous to trigonometry; \(\cos t\) and \(\sin t\) are the coordinates of a typical point on the unit circle, and \(t\) is twice the area shown in the second graph of figure 4.11.2.

Figure 4.11.2. Geometric definitions of sin, cos, sinh, cosh: \(t\) is twice the shaded area in each figure.

Given the definitions of the hyperbolic functions, finding their derivatives is straightforward. Here again we see similarities to the trigonometric functions.

Theorem 4.11.5

 \( {d\over dx}\cosh x=\sinh x\) and \thmrdef{thm:hyperbolic derivatives} \( {d\over dx}\sinh x = \cosh x\).

Proof

\[ {d\over dx}\cosh x= {d\over dx}{e^x +e^{-x}\over 2} = {e^x- e^{-x}\over 2} =\sinh x,\]

and

\[ {d\over dx}\sinh x = {d\over dx}{e^x -e^{-x}\over 2} = {e^x +e^{-x }\over 2} =\cosh x.\]

\(\square\)

Since \(\cosh x > 0\), \(\sinh x\) is increasing and hence injective, so \(\sinh x\) has an inverse, \(\text{arcsinh} x\). Also, \(\sinh x > 0\) when \(x>0\), so \(\cosh x\) is injective on \([0,\infty)\) and has a (partial) inverse, \(\text{arccosh} x\). The other hyperbolic functions have inverses as well, though \(\text{arcsech} x\) is only a partial inverse. We may compute the derivatives of these functions as we have other inverse functions.

Theorem 4.11.6

\( {d\over dx}\text{arcsinh} x = {1\over\sqrt{1+x^2}}\).

Proof

Let \(y=\text{arcsinh} x\), so \(\sinh y=x\). Then

\[ {d\over dx}\sinh y = \cosh(y)\cdot y' = 1,\]

and so

\[ y' ={1\over\cosh y} ={1\over\sqrt{1 +\sinh^2 y}} = {1\over\sqrt{1+x^2}}.\]

\(\square\)

The other derivatives are left to the exercises.

Contributors

  • Integrated by Justin Marshall.