Unifying the connections between the trigonometric and hyperbolic functions

Question

There are many, many connections between the trigonometric and hyperbolic functions, some of which are listed here. It is probably too optimistic to expect that a single insight could explain all of these connections, but is there a holistic way of seeing the parallels between $\sin$ and $\sinh$, $\cos$ and $\cosh$? Can all of these seemingly disparate connections be shown to be essentially the same, or at least very similar?

Geometric connections

• Sine and cosine parameterise the unit circle $x^2+y^2=1$, just as hyperbolic sine and cosine parameterise the 'unit hyperbola' $x^2-y^2=1$. Both circles and hyperbolas are conic sections.
• The sector of the circle connecting the points $(0,0)$, $(1,0)$, and $(\cos t,\sin t)$ has an area of $t/2$. The region of the hyperbola connecting the points $(0,0)$, $(1,0)$, and $(\cosh t,\sinh t)$ has an area of $t/2$. This can even be used to define the hyperbolic functions geometrically, and many authors do the same with the trigonometric functions.
• Sine and hyperbolic sine are odd, whereas cosine and hyperbolic cosine are even. But sine and cosine are periodic functions, unlike the hyperbolic counterparts.
• The analogue of the identity $\cos^2x+\sin^2x \equiv 1$ is $\cosh^2x-\sinh^2x \equiv 1$. The compound angle formulae are almost identical to their hyperbolic counterparts, save for a pesky minus sign: \begin{align} \sin(x+y) &= \sin(x)\cos(y)+\cos(x)\sin(y) \\ \sinh(x+y) &= \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \\[4pt] \cos(x+y) &= \cos(x)\cos(y) \color{red}{-} \sin(x)\sin(y) \\ \cosh(x+y) &= \cosh(x)\cosh(y) \color{blue}{+} \sinh(x)\sinh(y) \, . \end{align}
• In general, given a trigonometric identity, it is (usually) possible to find a corresponding hyperbolic identity using Osborn's rule: replace every occurrence of $\cos$ with $\cosh$; replace every occurrence of $\sin$ by $\sinh$, but negate the product of two $\sinh$ terms.

Analytic connections

• $\sin$ is the unique solution to the initial value problem \begin{align} f''(x) &= \color{red}{-}f(x) \\ f'(0) &= 1 \\ f(0) &= 0 \, , \end{align} and the corresponding initial value problem for $\sinh$ is the same, except $f''(x) = \color{blue}{+}f(x)$.
• Likewise, the initial value problem for $\cos$ is \begin{align} f''(x) &= \color{red}{-}f(x) \\ f'(0) &= 0 \\ f(0) &= 1 \, , \end{align} and again we see a mysterious sign change for $\cosh$: $f''(x) = \color{blue}{+}f(x)$.
• It follows that the higher-order derivatives of $\sin$ and $\sinh$ form periodic sequences.
• If we solve the initial value problems shown above, we obtain the exponential forms of all $4$ functions: \begin{align} \sin x &= \frac{e^{\color{green}{i}x}-e^{-\color{green}{i}x}}{2\color{green}{i}} \quad{} \cos x = \frac{e^{\color{\green}{i}x}+ e^{-\color{green}{i}x}}{2} \\[3pt] \sinh x &= \frac{e^{x}-e^{-x}}{2} \quad{} \cosh x = \frac{e^x + e^{-x}}{2} \, . \end{align}
• All $4$ functions are analytic, and their Taylor series bear a striking resemblance to each other: \begin{align} \sin x &= x \color{red}{-} \frac{x^3}{3!} + \frac{x^5}{5!} \color{red}{-} \frac{x^7}{7!} + \ldots \\[4pt] \sinh x &= x \color{blue}{+} \frac{x^3}{3!} + \frac{x^5}{5!} \color{blue}{+} \frac{x^7}{7!} + \ldots \\[4pt] \cos x &= 1 \color{red}{-} \frac{x^2}{2!} + \frac{x^4}{4!} \color{red}{-} \frac{x^6}{6!} + \ldots \\[4pt] \cosh x &= 1 \color{blue}{+} \frac{x^2}{2!} + \frac{x^4}{4!} \color{blue}{+} \frac{x^6}{6!} + \ldots \end{align}
• And Euler's formula $$ e^{ix} = \cos x + i \sin x $$ is replaced by the underwhelming $$ e^x = \cosh x + \sinh x \, . $$

Answer 1

I my opinion I think that when you know

$$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i} ~~~~~~~~~~~~~ \cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$

you can derive all the circular trigonometric identities.

If you add the "Wick" transformation $ x \to ix$ then you will step into the hyperbolic world, with all the consequent identities.

$$\sin(ix) = \frac{e^{-x} - e^{x}}{2i} = i\frac{e^x - e^{-x}}{2} = i\sinh(x) ~~~~~ \longrightarrow ~~~~~ \sinh(x) = \frac{e^x + e^{-x}}{2}$$

And similarly for $\cosh(x)$.

Answer 2

Yes; in fact they can all be traced to the same point, and it is a point that you brought up in your question. One important thing to note is that in Math there can be many equivalent definitions of concepts that can lead to each other, for instance one can define $\pi$ as the ratio of circumference to diameter; or area to radius squared. Both are equivalent.

Similarly we can define these functions geometrically,

And from that obtain differential equations that describe the functions, and from that obtain their exponential definitions; this can be used to unify all of the points you brought up, the Pythagorean Identity, sum of angles, Taylor Series, and exponential definition.

Alternatively we could have defined these functions in terms of their exponential definitions (As @Turing suggests), and from that obtain the Pythagorean Identity, sum of angles, Taylor Series, and geometric definition.

The latter is much easier, but the former is how we came to understanding this family of functions historically.

Answer 3

The trigonometric functions are related to the imaginary exponential $e^{iv}=\cos v+i\sin v$.

The hyperbolic functions are related to the real exponential $e^{u}=\cosh u+\sinh u$.

They both are special cases of the complex exponential $e^z=e^{u+iv}$.

The associated conics are obtained by

$$1=e^{iv}e^{-iv}=(\cos v+i\sin v)(\cos v-i\sin v)=\cos^2u-i^2\sin^2u,$$

$$1=e^{v}e^{-v}=(\cosh v+\sinh v)(\cosh v-\sinh v)=\cosh^2u-\sinh^2u.$$

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A unifying ODE is

$$z''''=z$$ giving the characteristic polynomial $$\omega^4=1,\\\omega=\pm1,\pm i$$

and the solution

$$z=a\cos t+b\sin t+c\cosh t+d\sinh t.$$

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Just think

$$\text{real}\leftrightarrow\text{imaginary}.$$

Answer 4

The connection between the trigonometric and hyperbolic functions becomes more intimate when one introduces split-complex numbers: the numbers of the form $a+bj$, where $j^2=1$ and $j\ne \pm 1$.

Using Taylor series you can easily find that $$ e^{jx} = \cosh x + j\sinh x , \quad \sinh x = \frac{e^{jx}-e^{-jx}}{2j}, \quad \cosh x = \frac{e^{jx} + e^{-jx}}{2}. $$

One can understand $\cos x$ and $\sin x$ as the real and imaginary parts of the complex exponential $e^{ix}$. Likewise, $\cosh x$ and $\sinh x$ are the real and 'imaginary' parts of the split-complex exponential $e^{jx}$.

The analogy can be summarized as follows: $$ \begin{align} & &&\textbf{Complex numbers} && \textbf{Split-complex numbers} \\ &\text{Imaginary unit:} && i^2 = -1, && j^2 = 1, \\ &\text{Numbers:} && z = a+bi, && z = a+bj, \\ &\text{Real part:} && \mathfrak{R}(a+bi)=a, && \mathfrak{R}(a+bj)=a, \\ &\text{Conjugation:} && z=a+bi\mapsto \bar{z}=a-bi, && z=a+bj\mapsto \bar{z}=a-bj, \\ &\text{Sesquilinear form:} && \langle z|w\rangle= z\bar{w}, && \langle z|w\rangle= z\bar{w} , \\ &\text{Real bilinear form:} && \mathfrak{R}(z\bar{w})\quad \text{(positive definite)}, && \mathfrak{R}(z\bar{w}) \quad\text{(of indefinite signature)} , \\ &\text{Other real bilinear form:} && \mathfrak{R}(zw)\quad \text{(of indefinite signature)}, && \mathfrak{R}(zw) \quad\text{(positive definite)} , \\ &\text{Squared modulus:} && |z|^2=\langle z|z\rangle = a^2+b^2, && |z|^2=\langle z|z\rangle = a^2-b^2, \\ &\text{Euler's formula:} && e^{ix} = \cos x + i \sin x, && e^{jx} = \cosh x + j \sinh x, \\ &\text{Pythagorean theorem:} && |e^{ix}| = \cos^2 x + \sin^2 x =1, && |e^{jx}| = \cosh^2 x - \sinh^2 x = 1, \\ &\text{Geometric meaning:} && \text{Unit circle,} && \text{Unit hyperbola (right branch)}. \\ \end{align} $$

The connection is in the correspondence: $$ \begin{pmatrix} \text{complex} \; i \\ \sin, \; \cos \end{pmatrix} \leftrightarrow \begin{pmatrix} \text{split-complex} \; j \\ \sinh, \; \cosh \end{pmatrix}. $$

One can use imagination to say that $\sin$, $\cos$ and complex numbers come from the elliptic world, while $\sinh$, $\cosh$ and split-complex numbers are their counterparts from the hyperbolic world.

Answer 5

Closed relation of the trigonometrical and hyperbolic functions can be demonstrated on the example of the Chebyshev Polyhomials of the First Kind $$T_n(x) = \begin{cases} \cos(n\arccos x),\;\text{if}\;|x|\le 1\\ \cosh(n\text{ arccosh } x),\;\text{otherwize}. \end{cases} $$ This example shows that hyperbolic functions can be suitable addition to the trigonometrical ones in the real analysis.

Answer 6

The Jacobi elliptic functions, in particular the sine and cosine of the amplitude are also connected to the hyperbolic trigonometric functions using the Gudermannian function which relates the circular functions and hyperbolic functions without explicitly using complex numbers. More precisely, if the parameter $\,m=0\,$ then $$ \text{sn}(x,0) = \sin(x)\quad \text{ and } \quad \text{cn}(x,0) = \cos(x) $$ while if $\,m=1\,$ then $$ \text{sn}(x,1) = \tanh(x)\quad \text{ and } \quad \text{cn}(x,1) = \text{sech}(x). $$ The Gudermannian function links the two cases with the identity $$ \sin(\text{gd}(x)) = \tanh(x) \quad \text{ and } \quad \tan(\text{gd}(x)) = \sinh(x) $$ and the related identity $$ \cos(\text{gd}(x)) = \text{sech}(x) \quad \text{ and } \quad \sec(\text{gd}(x)) = \cosh(x). $$ In some sense, the Jacobi functions generalize the circular and hyperbolic trigonometric functions. Note the fundamental identity $$ \text{sn}(x,m)^2 + \text{cn}(x,m)^2 = 1. $$ Whem $\,m=0\,$ this becomes $$ \sin(x)^2+\cos(x)^2=1 $$ and when $\,m=1\,$ this becomes $$ \tanh(x)^2+\text{sech}(x)^2=1 $$ which is equivalent to $$\cosh(x)^2-\sinh(x)^2=1. $$