Hyperbolic Functions

I. Introduction

Earlier we saw how all the trigonometric ratios relate to the unit circle. A hyperbola has similar, interesting properties.

If we draw a unit hyperbola of the form $x^2 - y^2 = 1$ and select any random point on curve, draw the line joining the point $(0, 0)$ with the point we will see some interesting properties of the curve.

Try moving the point $A$ along the curve, observe its coordinate values, and also the value of the area and the two expressions shown in the figure.

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Let $\alpha$ be the area of the region shaded in green.

Then,

$\sinh \alpha = \dfrac{e^{\alpha} - e^{-\alpha}}{2}$

and

$\cosh \alpha = \dfrac{e^{\alpha} + e^{-\alpha}}{2}$

We will see the derivations in the following section.

II. Derivations Of The Hyperbolic Ratios

Let,

$\alpha = \text{total area of green region}$

$\beta = \text{total area of brown region}$

If we consider only the top half of the area in the widget, then the area in green represents $\left( \dfrac{\alpha}{2} \right)$, the corresponding area in brown $\left( \dfrac{\beta}{2} \right)$ is the area bounded by the curve $y = \sqrt{x^2 - 1}$ and the $x$-axis. (Note that the positive value of the $\sqrt{x^2 - 1}$ represents curve above the $x$-axis).

Therefore, $[\triangle APO] = \dfrac{\alpha}{2} + \dfrac{\beta}{2}$

$\dfrac{\alpha}{2} = [\triangle APO] - \dfrac{\beta}{2}$

Let the point $A$ be $(x, y)$.

$\therefore \dfrac{\alpha}{2} = \dfrac{xy}{2} - \displaystyle \int \limits_{1}^{x} (\sqrt{x^2 - 1})dx$ ($\because$ the hyperbola intersects the $y$-axis at $x = 1$)

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Since $P$ lies on the curve, it satisfies the equation: $y = \sqrt{x^2 - 1}$

$\therefore \dfrac{\alpha}{2} = \dfrac{x\sqrt{x^2 - 1}}{2} - \displaystyle \int \limits_{1}^{x} (\sqrt{x^2 - 1})dx$

We will evaluate $\displaystyle \int (\sqrt{x^2 - 1})dx$ using Euler substitution.

Let

$\sqrt{x^2 - 1} = t - x$

(Note that the above substitution implies $x = \dfrac{t^2 + 1}{2t}$)

$\Rightarrow x^2 - 1 = (t - x)^2$

$\Rightarrow \cancel{x^2} - 1 = t^2 + \cancel{x^2} - 2tx$

$\Rightarrow 2tx = t^2 + 1$

$\Rightarrow 2x = \dfrac{t^2 + 1}{t} = t + \dfrac{1}{t}$

$\therefore 2 dx = \left(1 - \dfrac{1}{t^2}\right) dt = \left(\dfrac{t^2 - 1}{t^2}\right)dt$

$\Rightarrow dx = \left(\dfrac{t^2 - 1}{2t^2}\right)dt$

Therefore, our integral becomes:

$\displaystyle \int (\sqrt{x^2 - 1})dx$

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$= \displaystyle \int (t - x) \left(\dfrac{t^2 - 1}{2t^2}\right)dt$

$= \displaystyle \int \left(t - \dfrac{t^2 + 1}{2t}\right) \left(\dfrac{t^2 - 1}{2t^2}\right)dt$

$= \displaystyle \int \left(\dfrac{2t^2 - t^2 - 1}{2t}\right) \left(\dfrac{t^2 - 1}{2t^2}\right)dt$

$= \displaystyle \int \left(\dfrac{t^2 - 1}{2t}\right) \left(\dfrac{t^2 - 1}{2t^2}\right)dt$

$= \displaystyle \int \left(\dfrac{(t^2 - 1)^2}{4t^3}\right)dt$

$= \displaystyle \dfrac{1}{4} \int \left(\dfrac{t^4 - 2t^2 + 1}{t^3}\right)dt$

$= \displaystyle \dfrac{1}{4} \int \left(t - 2t^{-1} + t^{-3}\right)dt$

$= \displaystyle \dfrac{1}{4} \left[\dfrac{t^2}{2} - 2 \ln (|t|) - \dfrac{1}{2t^2}\right] + C$

$= \displaystyle \dfrac{1}{4} \left[\dfrac{1}{2} \left(t^2 - \dfrac{1}{t^2}\right) - 2 \ln (|t|)\right] + C$

Now we will simplify the term $\left(t^2 - \dfrac{1}{t^2}\right)$ and plug it in the original expression.

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Substituting $t = \sqrt{x^2 - 1} + x$ we get:

$\left(t^2 - \dfrac{1}{t^2}\right)$

$= (\sqrt{x^2 - 1} + x)^2 - \left( \dfrac{1}{\sqrt{x^2 - 1} + x} \right)^2$

$= (\sqrt{x^2 - 1} + x)^2 - \left[ \dfrac{\sqrt{x^2 - 1} - x}{(\sqrt{x^2 - 1} + x)(\sqrt{x^2 - 1} - x)} \right]^2$

$= (\sqrt{x^2 - 1} + x)^2 - \left[ \dfrac{\sqrt{x^2 - 1} - x}{(\cancel{x^2} - 1 - \cancel{x^2})} \right]^2$

$= (\sqrt{x^2 - 1} + x)^2 - (\sqrt{x^2 - 1} - x)^2$

$= 4x \sqrt{x^2 - 1}$

Therefore, we get:

$\displaystyle \int (\sqrt{x^2 - 1})dx = \dfrac{1}{4} \left[\dfrac{1}{2} \left(t^2 - \dfrac{1}{t^2}\right) - 2 \ln (|t|)\right] + C$

$= \dfrac{1}{4} \left[\dfrac{1}{2} \left(4x \sqrt{x^2 - 1}\right) - 2 \ln (|\sqrt{x^2 - 1} + x|)\right] + C$

$= \dfrac{x \sqrt{x^2 - 1}}{2} - \dfrac{1}{2} \ln (|\sqrt{x^2 - 1} + x|) + C$

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Applying the limits $(1, x)$ to our integral we get,

$\displaystyle \int \limits_{1}^{x} (\sqrt{x^2 - 1})dx$

$= \left.\dfrac{x \sqrt{x^2 - 1}}{2} - \dfrac{1}{2} \ln (|\sqrt{x^2 - 1} + x|)\right\rvert_{1}^{x}$

$= \left[\dfrac{x \sqrt{x^2 - 1}}{2} - \dfrac{1}{2} \ln (|\sqrt{x^2 - 1} + x|)\right] - \left[\dfrac{1 \sqrt{0}}{2} - \dfrac{1}{2} \ln (|\sqrt{0} + 1|)\right]$

$= \dfrac{x \sqrt{x^2 - 1}}{2} - \dfrac{1}{2} \ln (|\sqrt{x^2 - 1} + x|)$

From our previous equation:

$\dfrac{\alpha}{2} = \dfrac{x\sqrt{x^2 - 1}}{2} - \displaystyle \int \limits_{1}^{x} (\sqrt{x^2 - 1})dx$

$\Rightarrow \dfrac{\alpha}{2} = \dfrac{x\sqrt{x^2 - 1}}{2} - \left[\dfrac{x \sqrt{x^2 - 1}}{2} - \dfrac{1}{2} \ln (|\sqrt{x^2 - 1} + x|)\right]$

$\Rightarrow \dfrac{\alpha}{2} = \cancel{\dfrac{x\sqrt{x^2 - 1}}{2}} - \cancel{\dfrac{x \sqrt{x^2 + 1}}{2}} + \dfrac{1}{2} \ln (|\sqrt{x^2 - 1} + x|)$

$\Rightarrow \alpha = \ln (|\sqrt{x^2 - 1} + x|)$

$\therefore \sqrt{x^2 - 1} + x = e^\alpha$

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$\Rightarrow \sqrt{x^2 - 1} = e^\alpha - x$

$\Rightarrow \cancel{x^2} - 1 = e^{2 \alpha} + \cancel{x^2} - 2xe^\alpha$

$\Rightarrow x = \dfrac{e^{2 \alpha} + 1}{2e^\alpha}$

$\Rightarrow x = \dfrac{e^{\alpha} + e^{-\alpha}}{2}$

Therefore

$y = \sqrt{x^2 - 1}$

$\Rightarrow y = \sqrt{\left(\dfrac{e^{\alpha} + e^{-\alpha}}{2}\right)^2 - 1}$

$\Rightarrow y = \sqrt{\left(\dfrac{(e^{\alpha} + e^{-\alpha})^2 - 4}{4}\right)}$

$\Rightarrow y = \sqrt{\left(\dfrac{(e^{\alpha} + e^{-\alpha})^2 - 4e^{\alpha}e^{-\alpha}}{4}\right)}$

$\Rightarrow y = \sqrt{\dfrac{(e^{\alpha} - e^{-\alpha})^2}{4}}$

$\Rightarrow y = \dfrac{e^{\alpha} - e^{-\alpha}}{2}$

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Like the unit circle and trigonometric ratios, the $x$ and $y$ values for a unit hyperbola are defined as $\cosh \alpha$ and $\sinh \alpha$ respectively.

III. Definition Of Standard Hyperbolic Functions

As we saw in the previous section, the two functions $\sinh \alpha$ and $\cosh \alpha$ are defined as:

$\cosh \alpha = \dfrac{e^{\alpha} + e^{-\alpha}}{2}$

$\sinh \alpha = \dfrac{e^{\alpha} - e^{-\alpha}}{2}$

$\tanh \alpha = \dfrac{\sinh \alpha}{\cosh \alpha} = \dfrac{e^{\alpha} - e^{-\alpha}}{e^{\alpha} + e^{-\alpha}}$

$\csch \alpha = \dfrac{1}{\sinh \alpha} = \dfrac{2}{e^{\alpha} - e^{-\alpha}}$

$\sech \alpha = \dfrac{1}{\cosh \alpha} = \dfrac{2}{e^{\alpha} + e^{-\alpha}}$

$\coth \alpha = \dfrac{1}{\tanh \alpha} = \dfrac{e^{\alpha} + e^{-\alpha}}{e^{\alpha} - e^{-\alpha}}$

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IV. Identities Formed By Hyperbolic Functions

In this section we will look at a few identites involving hyperbolic functions. These identities are similar, but not the same as circular trigonometric functions.

$\cosh^2 x - \sinh^2 x$

$= \left(\dfrac{e^{x} + e^{-x}}{2}\right)^2 - \left(\dfrac{e^{x} - e^{-x}}{2}\right)^2$

$= \dfrac{e^{2x} + e^{-2x} + 2e^{x}e^{-x}}{4} - \dfrac{e^{2x} + e^{-2x} - 2e^{x}e^{-x}}{4}$

$= \dfrac{4}{4} = 1$

$\sech^2 x + \tanh^2 x$

$= \left(\dfrac{2}{e^{x} + e^{-x}}\right)^2 + \left(\dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}}\right)^2$

$= \dfrac{4}{(e^{x} + e^{-x})^2} + \dfrac{(e^{x} - e^{-x})^2}{(e^{x} + e^{-x})^2}$

$= \dfrac{4e^{x}e^{-x} + (e^{x} - e^{-x})^2}{(e^{x} + e^{-x})^2}$

$= \dfrac{(e^{x} + e^{-x})^2}{(e^{x} + e^{-x})^2}$

$= 1$

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$\coth^2 x - \csch^2 x$

$= \left( \dfrac{e^{x} + e^{-x}}{e^{x} - e^{-x}} \right)^2 - \left( \dfrac{2}{e^{x} - e^{-x}} \right)^2$

$= \dfrac{(e^{x} + e^{-x})^2}{(e^{x} - e^{-x})^2} - \dfrac{4}{(e^{x} - e^{-x})^2}$

$= \dfrac{(e^{x} + e^{-x})^2 - 4}{(e^{x} - e^{-x})^2}$

$= \dfrac{(e^{x} + e^{-x})^2 - 4e^{x}e^{-x}}{(e^{x} - e^{-x})^2}$

$= \dfrac{(e^{x} - e^{-x})^2}{(e^{x} - e^{-x})^2}$

$= 1$