Question

Tutorial on various triggy stuff @Jhannybean

Answer 1

\[\Large e^{i x}=\cos x + i \sin x\] From here we know cosine is an even function so cos(x)=cos(-x) and we know sine is an odd function so sin(-x)=-sin(x). We'll use these facts to figure something cool out.

First, we just plug in -x instead of x to euler's formula. \[\Large e^{-ix}=\cos(-x)+i \sin(-x)\] Now we invoke those fancy properties we just described to get \[\Large e^{-ix} = \cos x - i \sin x\] Now we can add the original to this or subtract this from the original to get the following two formulas \[\Large e^{ix}+e^{-ix} = 2 \cos x \\ \Large e^{ix}-e^{-ix}=2 i \sin x\] Just divide the 2 away to get the alternate representation, super nice and easy.

Now, this describes movement around a circle, so these satisfy the equation \[\Large x^2+y^2=1\] if \[\Large x= \cos \theta \\ \Large y = \sin \theta\] so it should be true that their complex versions do as well, test it out. Now let's go onto hyperbolic sine and cosine. Two different, but related functions that describe movement on a hyperbola... Not a circle.

Answer 2

So now let's talk about hyperbolic sine and cosine. These are just like the ones for the circular sine and cosine, except they describe the unit hyperbola \[\Large x^2-y^2=1\]