Latest revision as of 23:21, 6 December 2015
Hyperbolic Trigonometric Functions
Hyperbolic trigs are analogs of usual Trigonometric Functions
 $\cosh x = \cfrac{e^x + e^{x}}{2}$
 $\sinh x = \cfrac{e^x  e^{x}}{2}$
 $\tanh x = \cfrac{\sinh x}{\cosh x} = \cfrac{e^x  e^{x}}{e^x + e^{x}}$
(source: [1])
 $\cosh^2 x  \sinh^2 x = 1$

Taylor Expansion:
 $\cosh x = \cfrac{e^x + e^{x}}{2} = \cfrac{1}{2}\, e^x + \cfrac{1}{2}\, e^{x} = \ ...$
 $... \ = \cfrac{1}{2}\, \left(1 + x + \cfrac{1}{2!}\, x^2 + \cfrac{1}{3!}\, x^3 + \ ... \ \right) + \cfrac{1}{2}\, \left(1  x + \cfrac{1}{2!}\, x^2  \cfrac{1}{3!}\, x^3 + \ ... \ \right) = \ ...$
 odd degree terms cancel out
 $... \ = 1 + \cfrac{1}{2!}\, x^2 + + \cfrac{1}{4!}\, x^4 + \ ... \ = \sum\limits_{k=0}^\infty \cfrac{x^2}{(2k)!}$
 co we have even powers, like for $\cos x$, just without $(1)^k$ term
 same for $\sinh x = \cfrac{1}{2}\, e^x  \cfrac{1}{2}\, e^{x} = \sum\limits_{k=0}^\infty \cfrac{x^{2k + 1}}{(2k+1)!}$
 if we Taylor expand $e^x$, even powers cancel out, and we're left only with odd powers
 just like usual $\sin x$, but without alternating sing
Derivatives
 $\cfrac{d}{dx}\, \sinh x = \cosh x$
 $\cfrac{d}{dx}\, \cosh x = \sinh x$
Can show that with Taylor expansions:
 $\cfrac{d}{dx}\, \sinh x = \cfrac{d}{dx} \sum\limits_{k=0}^\infty \cfrac{x^{2k + 1}}{(2k+1)!} = \sum\limits_{k=0}^\infty (2k+1)\, \cfrac{x^{2k}}{(2k+1)!} = \sum\limits_{k=0}^\infty \cfrac{x^{2k}}{(2k)!} = \cosh x$
 same with $\cosh x$
Sources