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In order to compute $\cos(\sin^{-1}(x))$ or $\sin(\cos^{-1}(x))$, all one has to do is draw an appropriate triangle to find that $$\cos(\sin^{-1}(x))=\sqrt{1-x^2}=\sin(\cos^{-1}(x)).$$ Using algebra, one can find that $\cosh(\sinh^{-1}(x))=\sqrt{x^2+1}$ and $\sinh(\cosh^{-1}(x))=\sqrt{x^2-1}$. Is there a simple geometric picture we can draw in order to conclude these facts about hyperbolic trig functions that doesn't rely on triangles with imaginary angles or side lengths (since we could just write $\sinh$ and $\cosh$ in terms of $\cos,\sin,i$)?

Maxim Gilula
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    this is a nice question – Jesse Madnick Mar 20 '25 at 15:55
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    One could do so using the geometric definitions of hyperbolic functions (see this post, for instance, for more on that). – Ben Grossmann Mar 20 '25 at 15:56
  • Thanks! I did take a quick look about how $\cosh$ and $\sinh$ relate to the area of "that hyperbolic triangle" but it's not clear to me how to get inverse functions from there or how to take composition of hyperbolic and inverse hyperbolic from that picture. I guess I didn't think enough about that yet, thanks. – Maxim Gilula Mar 20 '25 at 15:57
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    $\tanh x$ and $\operatorname{sech} x$ form a regular unit hypotenuse right triangle. Use the same ratio logic you use for regular trig to define the relations you want. – Ninad Munshi Mar 20 '25 at 15:57
  • @NinadMunshi that's a great idea. – Maxim Gilula Mar 20 '25 at 16:01
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    The problem is that the "geometry" of $\cosh x$ and $\sinh x$ is not symmetric. The pairs $(u,v)=(\cosh x,\sinh x)$ are the points points of a hyperbola $u^2-v^2=1.$ So if you know $u$ then $v=\sqrt{u^2-1}$ and if you know $v$ then $u=\sqrt{1+v^2}.$'Imsuppose $u^2=1+v^2$ gives you a right triangle with one leg $1$ instead of hypotenuse $1.$ – Thomas Andrews Mar 20 '25 at 16:16
  • So if $\theta=\arctan(\sinh x)$ then $\cosh x=\sec\theta$ – Thomas Andrews Mar 20 '25 at 16:20
  • @ThomasAndrews It's ok if it's not symmetric, to me that just implies $\cosh(\sinh^{-1}(x))\neq \sinh(\cosh^{-1}(x))$. – Maxim Gilula Mar 20 '25 at 16:21

2 Answers2

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With regular trig (aka “circular”) functions, the parametric equation pair $(x(\theta) = \cos(\theta), y(\theta)= \sin(\theta))$ forms a circle centered at the origin, which can be expressed as $x^2 + y^2 = 1$.

Graph of the unit circle

Geometrically, $\sin(\cos^{-1}(t))$ asks “If you know that $x = t$ on the unit circle, what is $y$?” And $\cos(\sin^{-1}(t))$ asks “If you know that $y = t$ on the unit circle, what is $x$?” Both questions have the same answer, $\sqrt{1-t^2}$, due to the circle being symmetric.

Similarly, from $x(\theta) = \cosh(\theta) = \frac{e^{\theta} + e^{-\theta}}{2}$ and $y(\theta) = \sinh(\theta) = \frac{e^{\theta} - e^{-\theta}}{2}$, it can be shown that $x^2 - y^2 = 1$. The graph of this equation forms a hyperbola (which, in case you didn't know, is the origin of the term “hyperbolic functions”). Actually, since the definition of $\cosh$ forces $x > 0$, you only get half of a hyperbola.

Graph of the unit hyperbola

For your question, it doesn't matter what the parameter $\theta$ means geometrically (it's not the angle from the $x$ axis, which is $\arctan\left(1 - \frac{2}{e^{2\theta} + 1}\right)$), just how $x$ and $y$ are related to each other.

  • $\sinh(\cosh^{-1}(t))$: Given $x = t$, what is $y$? It's $\sqrt{t^2 - 1}$.
  • $\cosh(\sinh^{-1}(t))$: Given $y = t$, what is $x$? It's $\sqrt{t^2 + 1}$.
Dan
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    Awesome! Thanks. Yes, I was focused too much on what the parameter represents, and this is a very clear and clean way to answer my question. – Maxim Gilula Mar 21 '25 at 05:26
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$\tanh t$ and $\operatorname{sech} t$ form a right triangle of hypotenuse $1$ just like $\sin\theta$ and $\cos\theta$ do. Defining the following for a right triangle

$$\tanh t = \frac{\text{opp}}{\text{hyp}}$$

$$\operatorname{sech} t = \frac{\text{adj}}{\text{hyp}}$$

we get that

$$\sinh t = \frac{\text{opp}}{\text{adj}}$$

$$\cosh t = \frac{\text{hyp}}{\text{adj}} $$

Let's try an example. For $\sinh^{-1}(x)$ we label the opposite side as $x$ and the adjacent side as $1$ which leaves us with the hypotenuse as $\sqrt{1+x^2}$ and thus

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

Ninad Munshi
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