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Hyperbolic Trig.md

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+#Math #Trig
+
+# Definition
+
+## Definition in terms of $e$
+
+We define $\cosh$ and $\sinh$ to be the even and odd parts of $e^x$ respectively:
+
+$$
+\cosh x = \frac {e^x + e^{-x}} 2 \\
+\sinh x = \frac {e^x - e^{-x}} 2
+$$
+
+Note this gives us:
+
+$$
+\sinh x + \cosh x = e^x
+$$
+
+similar to Euler's Formula for circular trig functions.
+
+## Definition in terms of a hyperbola
+
+[https://www.desmos.com/calculator/ixmjpfmukk](https://www.desmos.com/calculator/ixmjpfmukk)
+
+Know that the geometric definition of $\cosh$ is that $B = \cosh 2b$, where $b$ is the blue area. To find $b$, we can use:
+
+$$
+b = \frac {B\sqrt{B^2 - 1}} 2 -\int _1^B \sqrt {x^2 - 1} dx \\
+= \frac {B\sqrt{B^2 - 1}} 2 - \frac {B\sqrt {B^2 - 1} - \ln(B + \sqrt {B^2 - 1})} 2\\
+= \frac {\ln(B + \sqrt {B^2 - 1})} 2
+$$
+
+Now let $a = 2b = -\ln(B + \sqrt {B^2 - 1})$. Now we can solve for $B$ in terms of $a$ to define $\cosh$:
+
+$$
+a = \ln(B + \sqrt {B^2 - 1}) \\
+B = \frac {e^a + e^{-a}} 2 \\
+\cosh x = \frac {e^x + e^{-x}} 2
+$$
+
+Now using the fact $\cosh$ and $\sinh$ lie on a hyperbola (can be proved algebraically) we get:
+
+$$
+\sinh x = \frac {e^x - e^{-x}} 2
+$$