<s>[INST] Unifying the connections between the trigonometric and hyperbolic functions 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 function, it is possible to write down the 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
$$
e^x = \cosh x + \sinh x \, .
$$
 [/INST] 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:
$$
\begin{align}
& &&\textbf{Complex numbers} && \textbf{Split-complex numbers} \\
&\text{Imaginary unit:} &&  i^2 = -1, &&    j^2 = 1, \\
&\text{Numbers:}  && z = a+bi,  &&    z = a+bj,  \\
&\text{Modulus:}  && |z| = \sqrt{a^2+b^2}, &&  |z| = \sqrt{a^2-b^2} \; (\text{for } a^2 \ge b^2), \\
&\text{Euler's formula:}  &&  e^{ix} = \cos x + i \sin x, &&   e^{jx} = \cosh x + j \sinh x, \\
&\text{Pythagorean theorem:} &&  |e^{ix}| = \cos^2 x + \sin^2 x =1,  &&   |e^{jx}| = \cosh^2 x - \sinh^2 x = 1,  \\
&\text{Geometric meaning:} &&  \text{Unit circle,}  &&   \text{Unit hyperbola (right branch)}.  \\
\end{align}
$$
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.
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