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I learned about hyperbolic functions recently so I'll try my best to word out my question.

We define $\sinh(x) = \frac{e^x-e^{-x}}{2}$. But how do we know that $\sinh(x)$ has a connection or relation to the unit hyperbola? ($x^2 - y^2 = 1$).

To understand my question more, assume we define $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ then how do we know that $\sin(x)$ has a connection to the unit circle?

For me, it looks like there is a missing step. How do we know that $\frac{e^x-e^{-x}}{2}$ has a connection to the unit hyperbola? Am I missing a definition?

Gary
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Hayst
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1 Answers1

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In $\mathbb{R}^2$, you can parametrize the line $x+y=0$ with $(t,-t)$. This is like an "antidiagonal" embedding of $\mathbb{R}$ in the direct product $\mathbb{R}\times\mathbb{R}$ (compared to the diagonal embedding given by $(t,t)$ whose range is the perpendicular diagonal line with the equation $y=x$).

You can do this in any group $G$: there is a "diagonal" embedding $G\to G\times G$ given by $g\mapsto(g,g)$, and there is an "antidiagonal" embedding $g\mapsto (g,g^{-1})$ (which is only a homomorphism if $G$ is abelian).

When we use the multiplicative group $\mathbb{R}^\times$ of nonzero reals, the image of the antidiagonal embedding is the graph of the reciprocal function, i.e. $xy=1$. But we can also parametrize the positive component of $\mathbb{R}^\times$ using the exponential function, i.e. $\exp:(\mathbb{R},+)\to(\mathbb{R}^\times,\cdot)$ is a group homomorphism. This means we can parametrize our graph using $(e^t,e^{-t})$.

If we rotate the graph of $xy=1$ by $45^\circ$ and rescale we get the graph of $u^2-v^2=1$, where $u,v=x\pm y$. When applied to the exponential parametrization, we get the hyperbolic parametrization $(\cosh t,\sinh t)$.

In the complex domain, we can multiply $y$ by $\pm i$ and the graph of $x^2-y^2=1$ becomes the graph of $x^2+y^2=1$ (or vice-versa), which is why the trig functions $(\cos\theta,\sin\theta)$ that parametrize the unit circle are related to the hyperbolic trig functions with the imaginary unit $i$.

anon
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