Why are the hyperbolic functions defined the way they are?

Question

I know that $\sinh(x) := \frac{e^x-e^{-x}}{2}$ and $\cosh(x) := \frac{e^x + e^{-x}}{2}$ by definition. But what exactly is the significance of this and how is it related to trigonometry? I get that $\sinh(x)$ is just the difference between the graphs $\frac{e^x}{2}$ and $\frac{e^{-x}}{2}$ and that $\cosh(x)$ is just the sum of $\frac{e^x}{2}$ and $\frac{e^x}{2}$ when viewed geometrically.

I would like to know why mathematicians have used half of both $e^x$ and $e^{-x}$ instead of defining say $\sinh(x)$ to be equal to just $e^x-e^{-x}$ without needing to divide by $2$.

Thanks in advance.

Answer 1

\begin{align*} \sin x &= \frac{\mathrm{e}^{\mathrm{i} x} - \mathrm{e}^{-\mathrm{i} x}}{2\mathrm{i}} & \cos x &= \frac{\mathrm{e}^{\mathrm{i} x} + \mathrm{e}^{-\mathrm{i} x}}{2} \\ \sinh x &= \frac{\mathrm{e}^{x} - \mathrm{e}^{- x}}{2} & \cosh x &= \frac{\mathrm{e}^{x} + \mathrm{e}^{-x}}{2} \\ \end{align*}

The halves ultimately come from Euler's formula, which allows us to write $$ \cos x = \Re (\mathrm{e}^{\mathrm{i}x}) = \Re( \mathrm{e}^{\mathrm{i}x} + \overline{\mathrm{e}^{\mathrm{i}x}})/2 \text{,} $$ where overline means complex conjugation and $\overline{\mathrm{e}^{\mathrm{i} x}} = \mathrm{e}^{-\mathrm{i}x}$, and similarly $\sin x = \Im (\mathrm{e}^{\mathrm{i}x})$. Then through \begin{align*} \sinh x &= - \mathrm{i} \sin(\mathrm{i}x) \\ \cosh x &= \cos(\mathrm{i}x) \\ \end{align*} we get the hyperbolic functions above.

Answer 2

In regular trigonometry, $\sin t$ and $\cos t$ are define with the unit circle $x^2+y^2=1$, where $t$ is the lenght of the arc from the positive $x$-axis, $x=\cos t$ and $y=\sin t$.

Similarly hyperbolic trigonometry is define with the unit hyperbola $x^2-y^2=1$. We define $x=\cosh t$ and $y=\sinh t$, where $t$ is twice the area between the ray, the hyperbola and the $x$-axis.

See this Wikipedia article for more details.

Answer 3

Significance of Hyperbolic functions: Hyperbolic functions are analogs of the ordinary trigonometric functions defined for the hyperbola rather than on the circle: just as the points $(\cos t, \sin t)$ form a circle with a unit radius $~(x^2+y^2=1)~$, the points $(\cosh t, \sinh t)$ form the right half of the equilateral hyperbola $~(x^2-y^2=1)~$. Hyperbolic functions also have important physical applications. These functions occur in the solutions of many linear differential equations [for example, the hyperbolic cosine function may be used to describe the shape of the curve formed by a high-voltage line suspended between two towers (catenary)], of some cubic equations, in calculations of angles and distances in hyperbolic geometry, and of Laplace's equation in Cartesian coordinates. Laplace's equations are important in many areas of physics, including electromagnetic theory, heat transfer, fluid dynamics, and special relativity. Hyperbolic functions may also be used to define a measure of distance in certain kinds of non-Euclidean geometry.

How is hyperbolic function related to trigonometry ?

The followings are the graphical representation between geometric and hyperbolic functions (especially for $\sin x,~~\cos x,~~\tan x,~~\sinh x,~~\cosh x,~~\tanh x$ ).

• Relation between Hyperbolic function and Trigonometric function:

Hyperbolic sine: $~\sinh x=-i\sin(ix)~$

Hyperbolic cosine: $~\cosh x=\cos(ix)~$

Hyperbolic tangent: $~\tanh x=-i\tan(ix)~$

Hyperbolic cotangent: $~\coth x=i\cot(ix)~$

Hyperbolic secant: $~\operatorname {sech} x=\sec(ix)~$

Hyperbolic cosecant: $~\operatorname {csch} x=i\csc(ix)~$

where $~i~$ is the imaginary unit with the property that $~i^2 = −1~$.

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For more details find the following:

https://brilliant.org/wiki/hyperbolic-trigonometric-functions/

https://en.wikipedia.org/wiki/Hyperbolic_functions

Answer 4

Surprisingly, none of the above answers touched upon your question

I would like to know why mathematicians have used half of both $e^x$ and $e^{−x}$ instead of defining say $\sinh(x)$ to be equal to just $e^x−e^{−x}$ without needing to divide by $2$.

Here's an intuitive explanations for this. Defining $\sinh$ and $\cosh$ in this way gives us the identity $$ e^x = \cosh x + \sinh x $$

which let's us trivially separate an expression of $e^x$ into even and odd components. This manipulation becomes very useful in some areas of mathematics. For example, at the undergraduate level, this identity makes solving many basic Sturm-Liouville boundary value problems algebraically simpler.

The key takeaway here is that we want to isolate the even and odd behavior of $e^x$ into two functions whose net value is still $e^x$. Defining these functions are you described would force us to carry around extra constants all over the place as we work with these functions. This is somewhat analogous to one of the motivations for measuring angles in radians as opposed to degrees, since $$ \sin'(x) = \cos(x) $$ is much nicer to work with than $$ \sin'(x^\circ) = \frac{\pi}{180}\cos(x^\circ). $$ By using the natural definition upfront, we save ourselves from unnecessary algebraic burdens.