public-notes/sin x = 2.md

#Math #Trig

\sin x = 2

\frac {e^{ix} - e^{-ix}} {2i} = 2

e^{ix} - e^{-ix} = 4i \\

e^{ix} - (e^{ix})^{-1} = 4i

Let u = e^{ix}:

u - u^{-1} = 4i

u^2 - 1 = 4iu

u^2 - 4iu - 1 = 0

u^2 - 4iu - 4 = -3

(u - 2i)^2 = -3

u - 2i = \pm \sqrt {-3} 

u = 2i \pm \sqrt {-3}

u = i(2 \pm \sqrt 3)

Substitute back into u, for n \in \mathbb{Z}:

e^{ix} = i(2 \pm \sqrt 3) \\ 

ix = \ln (i(2 \pm \sqrt 3)) \\ 

ix = \ln i + 2\pi n+ \ln(2 \pm \sqrt 3) \\

ix = \frac {i\pi} 2 + 2\pi n + \ln(2 \pm \sqrt 3)

x = \frac \pi 2 - i\ln(2 \pm \sqrt 3) + 2\pi n