A student asks, can you prove the following identity, and is it well known?
- arctan((x2-y2)/(2xy))+2arctan(y/x)=pi/2
Graeme's answer:
It turns out this identity comes up quite naturally by expanding 2arctan(y/x) into a single arctan using the "sum of arctan" form of the "tan of sum" identity. I'm not sure if it's well known, though. Here's how I approached it:
General Tan of Sum formula:
- tan(A+B) = (tan(A)+tan(B)) / (1-tan(A)tan(B))
Now, let x=tan(A) and y=tan(B), so arctan(x)=A and arctan(y)=B, and then take the arctan of both sides, giving you
- arctan(x)+arctan(y) = arctan((x+y) / (1-xy))
Using this to find 2arctan(x), we get
- 2arctan(x) = arctan(2x/(1-x2))
Replacing x with y/x to find 2arctan(y/x), we get
- 2arctan(y/x) = arctan(2y/x/(1-y2/x2)) = arctan(2xy/(x2-y2))
So the left-hand-side of the identity to be proven is:
- arctan((x2-y2)/(2xy))+arctan(2xy/(x2-y2))
The sum of arctans of reciprocals is pi/2, proving the identity.
An interesting side-effect of this investigation is that
- arctan((x2-y2)/(2xy))+arctan(y/x)=arctan(x/y),
which you can see in two ways.
First, since we showed arctan((x2-y2)/(2xy))+2arctan(y/x)=pi/2, we can subtract arctan(y/x) from both sides, giving us
- arctan((x2-y2)/(2xy))+arctan(y/x)=pi/2-arctan(y/x)=arctan(x/y)
by the arctan-of-reciprocals rule
Second, we can start with the arctan identity, replacing x with (x2-y2)/(2xy) and replacing y with y/x, we get
- arctan((x2-y2)/(2xy))+arctan(y/x) = arctan(((x2-y2)/(2xy)+y/x) / (1-(x2-y2)/(2xy)(y/x))) = arctan(x/y)