Pornrat Ruengrot	Posted on Friday, 07 February, 2003 - 01:55 pm: Hi, I'm trying to find the relationship between cos-1 (x) and cosh-1 (x) but I got a contradiction. Can anyone point out mistakes? Here it goes... From; sin(ix) = isinh(x) , sinh(ix) = isin(x) cos(ix) = cosh(x) , cosh(ix) = cos(x) Let cosh-1 (x) = A cosh(A) = x cos(iA) = x iA = cos-1 (x) cos-1 (x) = icosh-1 (x) ......(*) Let cos-1 (x) = A cos(A) = x cosh(iA) = x iA = cosh-1 (x) cosh-1 (x) = icos-1 (x) .....(**) But when considering (*); cos-1 (x) = icosh-1 (x) (-i)cos-1 (x) = (-i)icosh-1 (x) (-i)cos-1 (x) = cosh-1 (x) cosh-1 (x) = (-i)cos-1 (x) ......(***) Comparing with (**); cosh-1 (x) = icos-1 (x) icos-1 (x) = (-i)cos-1 (x) How can it be?!?!?
Alex Fletcher	Posted on Friday, 07 February, 2003 - 03:04 pm: hint: does cos(a)=cos(b) imply a=b?
Dan Goodman	Posted on Friday, 07 February, 2003 - 03:06 pm: Do you know that $\cosh (-x)=\cosh (x)$? If so, this should explain what is going on here. You are treating ${\cosh }^{-1}$ as if it had only one value, but it actually has more than that. Whenever $y$ is a value of ${\cosh }^{-1}$, $-y$ is also a value. Incidentally, can you show that whenever $y$ is a value of ${\cosh }^{-1}$, $y+2n\pi i$, where $n$ is an integer, is also a value?
Pornrat Ruengrot	Posted on Friday, 07 February, 2003 - 05:32 pm: Oh...I forgot completely that cosh(-x) = cosh(x) If A = cosh(x) then cosh-1 (A) = ±x Thank you, Alex and Dan, I got it now.