Arctan(x) in terms of logs

  • Thread starter cragar
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  • #1
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my math teacher said that the arctan can be setup in terms of logs
does any one know how to do this.
 

Answers and Replies

  • #2
Hurkyl
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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.
 
  • #3
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can i use eulers formula to do it .

so would it be [(e^(ix)-e^(-ix)]/[(ie^(ix)+ie^(-ix))] = tan(x)
 
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  • #4
arildno
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Correct.

You'll need to solve a quadratic in the process.
 
  • #5
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what quadratic
 
  • #6
HallsofIvy
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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]
 
  • #7
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oh i see
 

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