RE: The Priest is IN
01-24-2016, 09:46 AM
(This post was last modified: 01-24-2016, 09:51 AM by a52.)
e^ix=cos x + i*sin x
(e^-ix)/2 + (e^ix/2) = (cos x)/2 - (i*sin x)/2 + (cos x)/2 + (i*sin x)/2
= cos x
d/dx sin x = cos x
therefore sin x = the integral of (e^-ix)/2 + (e^ix/2)
the other one involves fourier analysis and that's quite a bit harder to explain but it's basically f^-1(f(sin x)), just using a very difficult function for f.