User:Timwi

Lemma

My name is Timwi.

Proof

Self-evident.

$\blacksquare$

Corollary

I started the following pages:

Trigonometry

Arccosine Logarithmic Formulation $\displaystyle \arccos x = -i \ln \left({ i \sqrt{1-x^2} + x }\right)$ $\displaystyle \arcsin x = -i \ln \left({ \sqrt{1-x^2} + ix }\right)$ $\displaystyle \arctan x = \frac 1 2 i \ln \left({ \frac{1-ix}{1+ix} }\right)$ $\displaystyle \operatorname{Area}\left({ABC}\right) = \frac 1 2 a b \sin\theta$ $\displaystyle \cos x = \dfrac {e^{i x} + e^{-i x} } 2$ $\displaystyle \cos \left({ix}\right) = \cosh x$ $\displaystyle \cos \left({a + b}\right) = \cos a \cos b - \sin a \sin b$ $\displaystyle \cosh \left({-x}\right) = \cosh x$ $\displaystyle \sinh \left({-x}\right) = -\sinh x$ $\displaystyle \tanh \left({-x}\right) = -\tanh x$ $\displaystyle \sin x = \frac 1 2 i \left({ e^{-i x} - e^{i x} }\right)$ $\displaystyle \sin \left({ix}\right) = i \sinh x$ $\displaystyle \sin \left({a + b}\right) = \sin a \cos b + \cos a \sin b$ $\displaystyle \tan x = i \frac { 1 - e^{2ix} }{ 1 + e^{2ix} }$ $\displaystyle \tan \left({ix}\right) = i \tanh x$