Explore the basic tenets of hyperbolic functions (definition, conversion into logarithmic form, series of hyperbolic cosine and sine functions, derivatives and integration)

From the definitions, we derive several important identities:

The fundamental identity: $\cosh^2 x - \sinh^2 x = 1$

Double-angle formulas: $\sinh 2x = 2\sinh x \cosh x$ $\cosh 2x = \cosh^2 x + \sinh^2 x$

Addition formulas: $\sinh(A \pm B) = \sinh A \cosh B \pm \cosh A \sinh B$ $\cosh(A \pm B) = \cosh A \cosh B \pm \sinh A \sinh B$

Worked Example

Given that sinh x = 3/4, find (a) cosh x, (b) x, (c) sinh 2x.

Solution

(a) Using $\cosh^2 x - \sinh^2 x = 1$: $\cosh^2 x = 1 + \sinh^2 x = 1 + \frac{9}{16} = \frac{25}{16}$ $\cosh x = \frac{5}{4}$ (cosh x is always positive)

(b) Since $\cosh x + \sinh x = e^x$: $e^x = \frac{5}{4} + \frac{3}{4} = 2$ $x = \ln 2 \approx 0.6931$

(c) Using $\sinh 2x = 2\sinh x \cosh x$: $\sinh 2x = 2 \times \frac{3}{4} \times \frac{5}{4} = \frac{15}{8}$

The remaining four hyperbolic functions are defined as ratios of sinh and cosh:

$\tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

$\coth x = \frac{\cosh x}{\sinh x} = \frac{e^x + e^{-x}}{e^x - e^{-x}}$

$\operatorname{sech} x = \frac{1}{\cosh x} = \frac{2}{e^x + e^{-x}}$

$\operatorname{cosech} x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}$

Derived Identities

From the fundamental identity, we obtain:

$1 - \tanh^2 x = \operatorname{sech}^2 x$

$\coth^2 x - 1 = \operatorname{cosech}^2 x$

Inverse hyperbolic functions can be expressed in logarithmic form.

For $y = \sinh^{-1} x$: $x = \sinh y = \frac{e^y - e^{-y}}{2}$

Solving this quadratic equation in $e^y$: $e^y = x \pm \sqrt{x^2 + 1}$

Since $\sqrt{x^2 + 1} > x$, the negative solution is invalid, giving: $\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)$

Similarly: $\cosh^{-1} x = \ln\left(x + \sqrt{x^2 - 1}\right), \quad x \geq 1$

$\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right), \quad |x| < 1$

$\coth^{-1} x = \frac{1}{2}\ln\left(\frac{x+1}{x-1}\right), \quad |x| > 1$

Using Maclaurin series, the hyperbolic sine and cosine functions can be expressed as infinite series.

Hyperbolic cosine: $\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$

Hyperbolic sine: $\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$

Worked Example

Using the power series expansion of cosh x up to the term in x^4, approximate cosh 1 to four decimal places.

Solution

$\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots$

At x = 1: $\cosh 1 = 1 + \frac{1^2}{2} + \frac{1^4}{24} = 1 + 0.5 + 0.0417 = 1.5417$

The derivatives follow directly from the definitions:

$\frac{d}{dx}(\sinh x) = \cosh x$

$\frac{d}{dx}(\cosh x) = \sinh x$

$\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x$

$\frac{d}{dx}(\operatorname{sech} x) = -\operatorname{sech} x \tanh x$

$\frac{d}{dx}(\operatorname{cosech} x) = -\operatorname{cosech} x \coth x$

$\frac{d}{dx}(\coth x) = -\operatorname{cosech}^2 x$

Derivatives of Inverse Hyperbolic Functions

$\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$

$\frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2 - 1}}, \quad x > 1$

$\frac{d}{dx}(\tanh^{-1} x) = \frac{1}{1 - x^2}, \quad |x| < 1$

The basic integrals are:

$\int \sinh x \, dx = \cosh x + c$

$\int \cosh x \, dx = \sinh x + c$

$\int \tanh x \, dx = \ln|\cosh x| + c$

$\int \operatorname{sech}^2 x \, dx = \tanh x + c$

$\int \operatorname{cosech}^2 x \, dx = -\coth x + c$

Important Integrals Using Hyperbolic Substitution

The following integrals are efficiently evaluated using hyperbolic substitutions:

$\int \frac{1}{\sqrt{a^2 + x^2}} \, dx = \sinh^{-1}\left(\frac{x}{a}\right) + c = \ln\left(\frac{x + \sqrt{x^2 + a^2}}{a}\right) + c$

$\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \cosh^{-1}\left(\frac{x}{a}\right) + c = \ln\left(\frac{x + \sqrt{x^2 - a^2}}{a}\right) + c$

Worked Example

Find $\int \sinh^2 x \, dx$.

Solution

Using the identity $\cosh 2x = \cosh^2 x + \sinh^2 x = 1 + 2\sinh^2 x$:

$\sinh^2 x = \frac{1}{2}(\cosh 2x - 1)$

$\int \sinh^2 x \, dx = \int \frac{1}{2}(\cosh 2x - 1) \, dx = \frac{1}{4}\sinh 2x - \frac{1}{2}x + c$

In Tanzania, hyperbolic functions are used in engineering and construction. For example, when designing suspension bridges like the Kilombero Bridge or cable car systems, engineers use the catenary curve (y = a cosh(x/a)) described by the hyperbolic cosine function to determine the exact shape that a hanging cable assumes under its own weight, ensuring structural stability and calculating the correct length of cable needed.