public-notes/Euler's Formula and Complex Trig Functions.md

#Math #Trig

Euler's Formula

Euler's formula states:

e^{i \theta} = i\sin \theta + \cos \theta

Proof

\frac d {d \theta} \frac {i \sin \theta + \cos \theta} {e^{i \theta}} \\
= e^{-i\theta}(i \sin \theta + \cos \theta) \\
= (e^{-i\theta})(i \sin \theta + \cos \theta)\prime + (e^{-i\theta}) \prime (i \sin \theta + \cos \theta) \\
= (e^{-i\theta})(i \cos \theta - \sin \theta) - i(e^{-i\theta})(i \sin \theta + \cos \theta) \\
= (e^{-i\theta})(i \cos \theta - \sin \theta) - (e^{-i\theta})(i \cos \theta - \sin \theta) \\
= 0

Therefore \frac {i \sin \theta + \cos \theta} {e^{i \theta}} is a constant. Plug in \theta = 0, to get \frac {i \sin \theta + \cos \theta} {e^{i \theta}} = 1. Multiply both sides by e^{i\theta} to get

e^{i \theta} = i\sin \theta + \cos \theta

Euler's Identity

Plug \theta = π into Euler's Formula

e^{i \pi} = i\sin \pi + \cos \pi \\
e^{i \pi} = -1

Trig Functions Redefined

Sine:

e^{i \theta} = i\sin \theta + \cos \theta \\
-e^{-i \theta} = -i\sin -\theta - \cos -\theta \\
-e^{-i \theta} = i\sin \theta - \cos \theta \\
e^{i\theta} - e^{-i\theta} = 2i \sin \theta \\
\sin \theta = \frac {e^{i\theta} - e^{-i\theta}} {2i}

Cosine:

e^{i \theta} = i\sin \theta + \cos \theta \\
e^{-i \theta} = i\sin -\theta + \cos -\theta \\
e^{-i \theta} = -i\sin \theta + \cos \theta \\
e^{i\theta} + e^{-i \theta} = 2\cos \theta \\
\cos \theta = \frac {e^{i\theta} + e^{-i \theta}} 2