- #1

johnsmi

- 31

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I came across a post here about the proof for: arctan(x)+arctan(1/x)=sign(x)pi/2

and wanted to share a differnt point of view (not completely scientific but could be aranged)

OK. so the proof uses the Brewster angle in Optics.

Brewster summarization:

Brewster's angle is the angle where the wave is completely transferred from

one matter (n_1) to another (n_2): tan(Theta_B)=n_2/n_1

=>Theta_B=arctan(n_2/n_1)

Now Imagine two waves one from n_1 hitting the surface at brewster angle and another on the other side of the surface at it's own brewster angle (the fraction of n's is inverse)

it is pretty "clear" (unexplained here but true) that the first one will continue at the same angle as the second one hit the surface and therefore according to snell's law the angle between the first hitting wave and it's transferred wave will be pi/2. (same for the second wave)

===>arctan(x)+arctan(1/x)=sign(x)pi/2

Yeah, I know its not complete and it does not work for negative x's (no negative refractive index as far as I know) but it's a little 'out of the box'