cragar Messages 2,546 Reaction score 3 Thread starter Jul 2, 2009 #1 my math teacher said that the arctan can be setup in terms of logs does anyone know how to do this.
Hurkyl Staff Emeritus Science Advisor Gold Member Messages 14,922 Reaction score 28 Jul 2, 2009 #2 Solve tan(y)=x. (Prerequisite: you must know how to write tan(y) in terms of exponentials) Other methods are possible (e.g. antidifferentiate f(x)=1/(1+x²)), but there are more technical details involved.
Solve tan(y)=x. (Prerequisite: you must know how to write tan(y) in terms of exponentials) Other methods are possible (e.g. antidifferentiate f(x)=1/(1+x²)), but there are more technical details involved.
cragar Messages 2,546 Reaction score 3 Jul 2, 2009 #3 can i use eulers formula to do it . so would it be [(e^(ix)-e^(-ix)]/[(ie^(ix)+ie^(-ix))] = tan(x) Last edited: Jul 2, 2009
arildno Science Advisor Homework Helper Gold Member Dearly Missed Messages 10,165 Reaction score 138 Jul 3, 2009 #4 Correct. You'll need to solve a quadratic in the process.
HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 983 Jul 3, 2009 #6 You have [itex]tan x= \frac{e^x- e^{-x}}{e^x+ e^{-x}}= y[/itex] First multiply on both sides of the equation by [itex]e^x+ e^{-x}[/itex]. Then multiply both sides o the equation by [itex]e^x[/itex]
You have [itex]tan x= \frac{e^x- e^{-x}}{e^x+ e^{-x}}= y[/itex] First multiply on both sides of the equation by [itex]e^x+ e^{-x}[/itex]. Then multiply both sides o the equation by [itex]e^x[/itex]