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