<s>[INST] Showing that hyperbolic trigonometric functions parameterize the unit hyperbola I know that the same way circular trigonometry is defined over the circle $ x^2 + y^2 = 1 $, hyperbolic trigonometry is defined over the hyperbola $ x^2 - y^2 = 1 $.
What I don't know is how deduced the formulas
$$ \sinh x = \frac {e^x - e^{-x}} {2} \quad \text{and} \quad \cosh x = \frac {e^x + e^{-x}} {2} $$
are deduced.

My question is: How are the formulas for $ \sinh x $ and $ \cosh x $ deduced from the equation $ x^2 - y^2 = 1 $ of the unit hyperbola?

 [/INST] I'll prove $\sinh(x)$ = $\frac {e^x - e^{-x}} {2}$ and leave the proof of $\cosh(x)$ as an exercise for you. So, $\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$
And we know that, $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
And we also know that $$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ...$$
Therefore, $$e^x - e^{-x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... - (1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + ...)$$
Now, after we open the brackets, the negative terms will become positive and positive terms will become negative, so let's just do that
$$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... - 1 + x - \frac{x^2}{2!} + \frac{x^3}{3!} - ...$$
And we are left with : $$2x + 2\frac{x^3}{3!} + 2\frac{x^5}{5!} + ...$$ and we'll take the $2$ common so we get,
$$2(x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...)$$
And, we saw earlier that $$\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$$.
So we get that $$e^x - e^{-x} = 2 \times \sinh(x)$$ and hence we get that $$\sinh(x) = \frac {e^x - e^{-x}}{2}$$
$Q.E.D$
Hope it helps
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