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1. Re: Anybody good at trig identities?
kglad Dec 12, 2012 10:13 PM (in response to Rothrock)that's not a valid expression. there are mismatched parantheses.

2. Re: Anybody good at trig identities?
Rothrock Dec 13, 2012 6:41 AM (in response to kglad)whoops. I had actually replaced
var len:Number=Math.sqrt(d*d+h*h);
var angle:Number=Math.asin(1/(2*len))
var newX:Number = h + d*Math.tan(Math.atan(h/d) + 2*n*angle)
So I only got one extra there. I need two more at the end.
var newX:Number = h + d*Math.tan(Math.atan(h/d) + 2*n*Math.asin(1/(2*Math.sqrt(d*d + h*h))))

3. Re: Anybody good at trig identities?
kglad Dec 13, 2012 10:00 PM (in response to Rothrock)1 person found this helpfuli don't think much can be done with that though this can be used:
var newX:Number=h+(h+Math.tan(2*n*angle))/(dh*Math.tan(2*n*angle))
if angle were Math.asin(d/Math.sqrt(d*d+h*h)) or Math.asin(h/Math.sqrt(d*d+h*h)) much more could be done.

4. Re: Anybody good at trig identities?
Rothrock Dec 14, 2012 7:54 PM (in response to kglad)Thanks I figured it was about as simple as it could be.