7.5: Hyperbolic Functions

In some textbooks you might see the sine and cosine functions called circular functions, since any point on the unit circle \(x^2+y^2=1\) can be defined in terms of those functions (see Figure [fig:cirangle]). Those definitions motivate a similar idea for the unit hyperbola \(x^2-y^2=1\), whose points can be defined in terms of hyperbolic functions.

For a point \(P=(x,y)\) on the unit hyperbola \(x^2-y^2=1\), the hyperbolic angle \(a\) is twice the area of the shaded hyperbolic sector in Figure [fig:hypangle]).¹¹ The area \(\frac{a}{2}\) thus equals the area of the right triangle with hypotenuse \(OP\) and legs of length \(x\) and \(y\) (so that the triangle’s area is \(\frac{1}{2}xy\)) minus the area under the hyperbola over the interval \(\int_1^x\). So since the upper half of the hyperbola \(x^2-y^2=1\) is the function \(y=\sqrt{x^2-1}\),

 

\[\begin{aligned} \frac{a}{2} ~&=~ \text{(area of triangle)} ~-~ \text{(area under the hyperbola from $1$ to $x$)} \ &=~ \frac{1}{2}xy ~-~ \int_1^x  \sqrt{u^2-1}~\du\

 

\[4pt] \frac{a}{2}~&=~\frac{1}{2}x\sqrt{x^2-1} ~-~ \left(\frac{1}{2}x\sqrt{x^2-1} ~-~ \frac{1}{2}\ln\,\left(x + \sqrt{x^2-1}\,\right)\right) \quad\text{(by formula (\ref{eqn:sqrtu2a2sec}))}\

 

\[2pt] a ~&=~ \ln\,\left(x + \sqrt{x^2-1}\,\right)\\ e^a ~&=~ x + \sqrt{x^2-1}\\ (e^a - x)^2 ~&=~ \left(\sqrt{x^2-1}\,\right)^2\\ e^{2a} - 2xe^a + \cancel{x^2} ~&=~ \cancel{x^2} - 1\\ x ~&=~ \frac{e^{2a} + 1}{2e^a} ~=~ \frac{e^a + e^{-a}}{2} ~=~ \cosh\,a\end{aligned} \nonumber \]

where \(\cosh\,a\) is the hyperbolic cosine of \(a\). The \(y\)-coordinate of \(P\) can then be found:

 

\[\begin{aligned} y ~&=~ \sqrt{x^2 - 1} ~=~ \sqrt{\left(\frac{e^a + e^{-a}}{2}\right)^2 - 1} ~=~ \sqrt{\frac{e^{2a} + 2 + e^{-2a}}{4} - \frac{4}{4}}\

 

\[4pt] &=~ \sqrt{\frac{e^{2a} - 2 + e^{-2a}}{4}} ~=~ \sqrt{\left(\frac{e^a - e^{-a}}{2}\right)^2} \text{, so since $a \ge 0$}\

 

\[4pt] y ~&=~ \frac{e^a - e^{-a}}{2} ~=~ \sinh\,a\end{aligned} \nonumber \]

where \(\sinh\,a\) is the hyperbolic sine of \(a\). All six hyperbolic functions can now be defined in general, analogous to the trigonometric (circular) functions:

The graphs of the hyperbolic functions are shown below:

The graph of \(y=\cosh\,x\) in Figure [fig:hyperfcns](a) might look familiar: a catenary—a uniform cable hanging from two fixed points—has the shape of a hyperbolic cosine function. The hyperbolic functions satisfy the following identities:

The identity \(\cosh^2 x - \sinh^2 x = 1\) was proved when deriving the coordinates of points on the unit hyperbola \(x^2-y^2=1\) in terms of the hyperbolic angle (since such a point \((x,y) = (\cosh\,a,\sinh\,a)\) must satisfy \(x^2-y^2=1\)). The addition identities can be proved similarly using hyperbolic angles (i.e. areas).¹² However, it is simpler to use the definitions of \(\sinh\) and \(\cosh\) in terms of exponential functions. For example:

 

\[\begin{aligned} \sinh\,u\;\cosh\,v + \cosh\,u\;\sinh\,v \;&=\; \frac{e^u - e^{-u}}{2} \cdot \frac{e^v + e^{-v}}{2} ~+~ \frac{e^u + e^{-u}}{2} \cdot \frac{e^v - e^{-v}}{2}\

 

\[4pt] &=\; \frac{e^{u+v}+e^{u-v}-e^{-u+v}-e^{-u-v}+e^{u+v}-e^{u-v}+e^{-u+v}-e^{-u-v}}{4}\\ &=\; \frac{2e^{u+v} - 2e^{-u-v}}{4}\

 

\[4pt] &=\; \frac{e^{u+v} - e^{-(u+v)}}{2}\

 

\[2pt] &=\; \sinh\,(u+v) \quad\checkmark\end{aligned} \nonumber \]

The identity for \(\sinh\,2x\) is then easy to prove by letting \(u=v=x\) in the above identity:

 

\[\sinh\,2x ~=~ \sinh\,(x+x) ~=~ \sinh\,x\;\cosh\,x + \cosh\,x\;\sinh\,x ~=~ 2\,\sinh\,x\;\cosh\,x \quad\checkmark \nonumber \]

Note that the identities \(\cosh\,(-x) = \cosh\,x\) and \(\sinh\,(-x) = -\sinh\,x\) mean that \(\cosh\) is an even function and \(\sinh\) is an odd function. Those two functions thus (sort of) serve as even and odd versions of the exponential function (which is neither even nor odd). Both \(\cosh\,x\) and \(\sinh\,x\) grow exponentially (the \(e^{-x}\) term for both functions becomes negligible as \(x \to \infty\)), while \(\sinh\,x\) decreases exponentially to \(-\infty\) as \(x \to -\infty\). The derivatives of the hyperbolic functions and their integral equivalents are:

For example, by definition of \(\cosh\,x\):

 

\[\ddx\,(\cosh\,x) ~=~ \ddx\,\left(\frac{e^x + e^{-x}}{2}\right) ~=~ \frac{e^x - e^{-x}}{2} ~=~ \sinh\,x \quad\checkmark \nonumber \]

Find the derivative of \(y=\sinh\,x^3\).

Solution: By the Chain Rule, \(\Dydx = 3x^2\,\cosh\,x^3\).

Evaluate \(\displaystyle\int\tanh\,x~\dx\).

Solution: Use the definition of \(\tanh\,x\) and the substitutions \(u=\cosh\,x\), \(\du=\sinh\,x\;\dx\):

 

\[\int\,\tanh\,x~\dx ~=~ \int\,\frac{\sinh\,x}{\cosh\,x}~\dx ~=~ \int\,\frac{\du}{u} ~=~ \ln\,\abs{u} ~+~ C ~=~ \ln\,(\cosh\,x) ~+~ C \nonumber \]

For any constant \(a>0\), both \(y=\cosh\,at\) and \(y=\sinh\,at\) satisfy the differential equation

 

\[y''(t) ~=~ a^2y(t) ~, \nonumber \]

which models rectilinear motion of a particle under a repulsive force proportional to the displacement. This is just one of the many reasons why hyperbolic functions appear in so many physical applications.

Example \(\PageIndex{1}\): paramag

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Solution

In the classical theory of paramagnetism, the total number \(n\) of molecules in a gas subject to a magnetic field of strength \(H\) is

 

\[n ~=~ 2\pi N\,\int_0^{\pi} e^{\frac{\mu H}{kT}\cos\,\theta}\sin\,\theta~\dtheta ~, \nonumber \]

where \(N\) is the number of molecules per unit solid angle having zero potential energy, \(\mu\) is the magnetic moment, \(k\) is Boltzmann’s constant, and \(T\) is the temperature of the gas. Show that

 

\[n ~=~ \frac{4\pi NkT}{\mu H}\,\sinh\,\left(\frac{\mu H}{kT}\right) ~. \nonumber \]

Solution: Let \(a=\frac{\mu H}{kT}\), and let \(u=\cos\,\theta\) so that \(\du=-\sin\,\theta\;\dtheta\):

 

\[\begin{aligned} n ~&=~ -2\pi N\,\int_1^{-1} e^{au}~\du ~=~ 2\pi N\,\int_{-1}^{1} e^{au}~\du ~=~ \frac{2\pi N}{a}\,e^{au}~\Biggr|_{-1}^1 ~=~ \frac{2\pi N}{a}\,\left(2 \cdot \frac{e^a - e^{-a}}{2}\right)\\ &=~ \frac{4\pi NkT}{\mu H}\,\sinh\,\left(\frac{\mu H}{kT}\right)\end{aligned} \nonumber \]

Since \(\ddx\,(\sinh\,x) = \cosh\,x = \frac{e^x+e^{-x}}{2} > 0\) for all \(x\), then \(y=\sinh\,x\) is an increasing function and thus its inverse function \(x=\sinh^{-1}y\) is defined. The remaining inverse hyperbolic functions can be defined similarly, with the following domains and ranges (switching the roles of \(x\) and \(y\), as usual) plus their graphs:

function	\(\sinh^{-1} x\)	\(\cosh^{-1} x\)	\(\tanh^{-1} x\)	\(\csch^{-1} x\)	\(\sech^{-1} x\)	\(\coth^{-1} x\)
domain	all \(x\)	\(x \ge 1\)	\(\abs{x} < 1\)	all \(x \ne 0\)	\(0 < x \le 1\)	\(\abs{x} > 1\)
range	all \(y\)	\(y \ge 0\)	all \(y\)	all \(y \ne 0\)	\(y \ge 0\)	all \(y \ne 0\)

The inverse hyperbolic functions can be expressed in terms of the natural logarithm:

Notice that the above formula for \(\cosh^{-1} x\) was actually proved at the beginning of this section. The formula for \(\sinh^{-1} x\) follows from the definition of an inverse function:

 

\[\begin{aligned} y ~=~ \sinh^{-1} x \quad&\Rightarrow\quad x ~=~ \sinh\,y ~=~ \frac{e^{y}-e^{-y}}{2}\

 

\[2pt] &\Rightarrow\quad e^y - 2x - e^{-y} ~=~ 0\\ &\Rightarrow\quad e^{2y} - 2xe^y - 1 ~=~ 0\\ &\Rightarrow\quad u^2 - 2xu - 1 ~=~ 0 \quad\text{for $u=e^y$}\\ &\Rightarrow\quad u ~=~ \frac{2x \pm \sqrt{4x^2 - 4(1)(-1)}}{2} ~=~ x \pm \sqrt{x^2 + 1}\

 

\[4pt] &\Rightarrow\quad e^y ~=~ u ~=~ x + \sqrt{x^2 + 1} \quad\text{since $e^y > 0$}\

 

\[2pt] &\Rightarrow\quad y ~=~ \ln\,(x + \sqrt{x^2 + 1}\,) \quad\checkmark\end{aligned} \nonumber \]

The remaining formulas can be proved similarly.

Example \(\PageIndex{1}\): hyperangleacosh

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Solution

Show that for the hyperbolic angle \(a\) of a point \(P=(x,y)\) on the unit hyperbola \(x^2-y^2=1\), the area \(\frac{a}{2}\) of the hyperbolic sector \(\hypsector\; OAP\) (the shaded region in the figure on the right) is

 

\[\frac{a}{2} ~=~ \frac{1}{2}\,\cosh^{-1} x ~. \nonumber \]

Solution: It was shown earlier that the area of \(\hypsector\; OAP\) is

 

\[\frac{a}{2} ~=~ \frac{1}{2}\,\ln\,(x + \sqrt{x^2 - 1}\,) \nonumber \]

so by the formula \(\cosh^{-1} x ~=~ \ln\,(x + \sqrt{x^2 - 1}\,)\) the result follows.

This makes sense, since \(x=\cosh\,a\) and so \(\cosh^{-1} x = \cosh^{-1} (\cosh\,a) = a\), by definition of an inverse.

You can use either the general formula for the derivative of an inverse function or the above formulas to find the derivatives of the inverse hyperbolic functions: For example, here is one way to find the derivative of \(\tanh^{-1} x\):

 

\[\begin{aligned} \ddx\,(\tanh^{-1} x) ~&=~ \ddx\,\left(\frac{1}{2}\,\ln\,\frac{1+x}{1-x}\right) ~=~ \frac{1}{2}\,\ddx\,(\ln\,(1+x) ~-~ \ln\,(1-x))\

 

\[6pt] &=~ \frac{1}{2}\,\left(\frac{1}{1+x} - \frac{-1}{1-x}\right) ~=~ ~=~ \frac{1-x + (1+x)}{2 (1+x)(1-x)} ~=~ \frac{1}{1-x^2} \quad\checkmark\end{aligned} \nonumber \]

[sec7dot5]

Prove the identities for \(\sinh\,(-x)\), \(\cosh\,(-x)\), \(\tanh\,(-x)\), \(\coth\,(-x)\), \(\sech\,(-x)\), and \(\csch\,(-x)\) on p.232.

Prove the identities for \(\cosh\,(u \pm v)\), \(\tanh\,(u \pm v)\), and \(\tanh\,2x\) on p.232. For Exercises 3-15 prove the given identity. Your proofs can use other identities. [[1.]]

2

\((\cosh\,x + \sinh\,x)^r = \cosh\,rx + \sinh\,rx\) for all \(r\)

\(\sinh\,3x ~=~ 3\,\sinh\,x + 4\,\sinh^{3} x\)

\(\sinh\,A\;\cosh\,B ~=~ \frac{1}{2}\left(\sinh\,(A+B) \;+\; \sinh\,(A-B)\right)\)	\(\tanh^2 x + \sech^2 x ~=~ 1\)
[-4pt] \(\sinh\,A\;\sinh\,B ~=~ \frac{1}{2}\left(\cosh\,(A+B) \;-\; \cosh\,(A-B)\right)\)	\(\coth^2 x - \csch^2 x ~=~ 1\)
[-4pt] \(\cosh\,A\;\cosh\,B ~=~ \frac{1}{2}\left(\cosh\,(A+B) \;+\; \cosh\,(A-B)\right)\)	[exer:coth1overx] \(\coth^{-1}\left(\frac{1}{x}\right) ~=~ \tanh^{-1} x\)

\(\cosh\,2x ~=~ \cosh^2 x + \sinh^2 x ~=~ 2\,\cosh^2 x - 1 ~=~ 1 + 2\,\sinh^2 x\vphantom{\dfrac{\cosh\,x-1}{2}}\)

\(\sinh^2 \frac{x}{2} ~=~ \dfrac{\cosh\,x \;-\; 1}{2}\)

\(\cosh^2 \frac{x}{2} ~=~ \dfrac{\cosh\,x \;+\; 1}{2}\)

\(\tanh^2 \frac{x}{2} ~=~ \dfrac{\cosh\,x \;-\; 1}{\cosh\,x \;+\; 1}\)

\(\tanh\,\frac{x}{2} ~=~ \dfrac{\cosh\,x \;-\; 1}{\sinh\,x} ~=~ \dfrac{\sinh\,x}{\cosh\,x \;+\; 1}\)

Prove the identities for \(\tanh^{-1} x\), \(\coth^{-1} x\), \(\sech^{-1} x\), and \(\csch^{-1} x\) on p.235.

Prove the derivative formulas for \(\sinh\,x\), \(\tanh\,x\), \(\coth\,x\), \(\sech\,x\), and \(\csch\,x\) on p.233.

Prove the derivative formulas for \(\sinh^{-1} x\), \(\cosh^{-1} x\), \(\coth^{-1} x\), \(\sech^{-1} x\), and \(\csch^{-1} x\) on p.236.

2

Show that \(\cosh\,x = O(e^x)\) and \(\sinh\,x = O(e^x)\vphantom{\Ddx}\).

Show that \(\Ddx\,(\tan^{-1} (\sinh\,x)) ~=~ \sech\,x~\).

Verify that the curve \(y=\tanh\,x\) has asymptotes \(y=\pm 1\).

[exer:hypdx] Show that \(\sinh\,\dx = \dx\) and \(\cosh\,\dx=1\) for any infinitesimal \(\dx\). (Hint: Use \(\tfrac{1}{1+\dx} = 1 - \dx\) from Exercise [exer:1over1plusdx] in Section 1.3 along with \(e^{\dx}=1+\dx\) from Exercise [exer:expdx] in Section 2.3.)

Use Exercise [exer:hypdx] and the addition formula for \(\sinh\,x\) to show that \(\ddx\,(\sinh\,x) = \cosh\,x\).

Sketch the graph of \(f(x)=e^{-2x} \sinh\,x\). Find all local maxima and minima, inflection points, and vertical or horizontal asymptotes.

Denoting the speed of light by \(c\), the Lorentz transformations for two inertial frames \(S\) and \(S'\) are

 

\[x' ~=~ \frac{x \;-\; vt}{\sqrt{1 \;-\; (v/c)^2}} \quad\text{and}\quad t' ~=~ \frac{t \;-\; vx/c^2}{\sqrt{1 \;-\; (v/c)^2}} \nonumber \]

where \(S'\) moves with speed \(v>0\) parallel to the \(x\)-axis of \(S\). Let \(v/c = \tanh\,\xi\). Show that

 

\[x' ~=~ -ct\,\sinh\,\xi ~+~ x\,\cosh\,\xi \quad\text{and}\quad t' ~=~ t\,\cosh\,\xi ~-~ (x/c)\,\sinh\,\xi ~. \nonumber \]

Evaluate the integral \(I=\int_2^3 \frac{\dx}{1-x^2}\) in two different ways:

1. Use \(\ddx\,(\coth^{-1} x) = \frac{1}{1-x^2}\) for \(\abs{x} > 1\) to show that \(I = \coth^{-1} 3 \;-\; \coth^{-1} 2\).
2. Use the substitution \(u = \frac{1}{x}\) in the integral, then use \(\ddx\,(\tanh^{-1} x) = \frac{1}{1-x^2}\) for \(\abs{x} < 1\) to show that \(I = \tanh^{-1} \frac{1}{3} \;-\; \tanh^{-1} \frac{1}{2}\). Is this answer equivalent to the answer from part (a)? Explain.

[[1.]]

The general solution of the differential equation \(y''=a^2y\) is \(y(t) = y_1(t) = c_1e^{at} + c_2e^{-at}\), where \(a\) is a positive constant, and \(c_1\) and \(c_2\) are arbitrary constants.

1. Verify that \(y(t) = y_2(t) = k_1\cosh\,at + k_2\sinh\,at\) is also a solution of \(y''=a^2y\).
2. Show that for any \(c_1\) and \(c_2\), \(y_1(t) = c_1e^{at} + c_2e^{-at}\) can be written as \(y_1(t) = k_1\cosh\,at + k_2\sinh\,at\) for some constants \(k_1\) and \(k_2\) in terms of \(c_1\) and \(c_2\).

Verify that for positive constants \(\beta\), \(p\), \(l\) and \(c\) the function

 

\[\eta(x) ~=~ \frac{\beta x}{p} ~-~ \frac{\beta c\,\sinh\,(px/c)}{p^2 \,\cosh\,(pl/c)} \nonumber \]

is a solution of the differential equation (related to water displacement in a canal of length \(2l\))

 

\[\frac{d^2 \eta}{\dx^2} ~-~ \frac{p^2}{c^2}\,\eta ~=~ -\frac{\beta xp}{c^2} ~. \nonumber \]

For any constant \(a\) and for \(s > \abs{a}\) the Laplace transform \(\mathcal{L}(s)\) of the function \(f(t) = \sinh\,at\) is

 

\[\mathcal{L}(s) ~=~ \int_0^{\infty} e^{-st} \sinh\,at~\dt ~. \nonumber \]

Show that \(\mathcal{L}(s) = \frac{a}{s^2 - a^2}\). [[1.]]

In quantum mechanics the scaling factor \(e^{\pi/s_0}\) of the three-particle Efimov trimer is the solution \(s=s_0 > 0\) of the equation

 

\[s\,\cosh\,\left(\tfrac{\pi s}{2}\right) ~=~ \frac{8\,\sinh\,\left(\tfrac{\pi s}{6}\right)}{\sqrt{3}} ~, \nonumber \]

Use a numerical method to approximate \(s_0\), then calculate \(e^{\pi/s_0}\).

Continuing Example

Example \(\PageIndex{1}\): paramag

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Solution

, the total magnetic moment \(M\) is defined as

 

\[M ~=~ 2\pi N\mu\,\int_0^{\pi} e^{\frac{\mu H}{kT}\cos\,\theta}\sin\,\theta\;\cos\,\theta~\dtheta ~. \nonumber \]

1. Use the value of \(n\) from Example

   Example \(\PageIndex{1}\): paramag

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   Solution

   to show that \(\frac{M}{n\mu} ~=~ L(a)\), where \(L(a) = \coth\,a - \frac{1}{a}\) is the Langevin function and \(a=\frac{\mu H}{kT}\).
2. Show that for \(a=\frac{\mu H}{kT}\), \(\mu_{\text{max}} > 0\), and \(x = \frac{\mu_{\text{max}} H}{kT}\),

    

   \[\int_0^{\mu_{\text{max}}} L(a)~d\!\mu ~=~ \frac{1}{x}\,\ln\,\left(\frac{\sinh\,x}{x}\right) ~. \nonumber \]

The age \(t_0\) of the universe (in years) is given by

 

\[t_0 ~=~ \tau_0 \bigintss_{\,0}^1 \frac{\dx}{\sqrt{\frac{2q_0}{x} + 1 - 2q_0}} ~, \nonumber \]

where \(\tau_0 = 2 \times 10^{10}\) years is the Hubble time and \(q_0 \ge 0\) is the deceleration parameter. For the cosmological model with \(0 < q_0 < \frac{1}{2}\), use the substitution \(x = \frac{2q_0}{1-2q_0}\,\sinh\,\theta\) to show that

 

\[t_0 ~=~ \tau_0\,\left(\frac{1}{1-2q_0} ~-~ \frac{2q_0}{(1-2q_0)^{3/2}}\,\cosh^{-1}\left(\frac{1}{\sqrt{2q_0}}\right)\right) ~. \nonumber \]

What fraction of \(\tau_0\) is \(t_0\) (i.e. what is \(\frac{t_0}{\tau_0}\,\)) when \(q_0=\frac{1}{4}\)? (Hint: Use Exercise 11 or 12.)

Circular rotations preserve the area of circular sectors (see Figure [fig:hyprotate](a)). For any constant \(c>0\) the hyperbolic rotation \(\phi: (x,y) \mapsto (cx,y/c)\) moves points along the hyperbola \(xy=k\) (for \(k>0\)), as shown in Figure [fig:hyprotate](b) for \(c>1\). This hyperbolic rotation preserves the area of hyperbolic sectors.

As an example of why this is true, let \(a>0\) and consider the unit hyperbola in Figure [fig:hyprotate](c). Then:

1. Let \(c>0\). When \(c \ne 1\) the mapping \(\phi: (x,y) \mapsto (cx,y/c)\) does not move points along the unit hyperbola. Find the formula for \(\phi\) on the unit hyperbola as follows: use the rotation equations from Section 7.4 to rotate the unit hyperbola \(45\Degrees\) to a hyperbola of the form \(xy=k\), apply \(\phi\) to a generic point on that hyperbola, then rotate that hyperbola by \(-45\Degrees\) back to the unit hyperbola.
2. Use part (a) to find the value of \(c\) such that \(\phi\) maps \(A=(1,0)\) to \(P=(\cosh a,\sinh a)\).
3. Let \(P'\) be the point that \(\phi\) maps \(P\) to, and let \(A'=P\). Use part (b) to find the coordinates of \(P'\), then show that the hyperbolic sectors \(\;\hypsector\;OAP\) and \(\;\hypsector\;OA'P'\) have the same area \(a/2\).