How can arctan(x) be expressed using logs?

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SUMMARY

The arctan function can be expressed in terms of logarithms through the relationship tan(y) = x, which can be rewritten using Euler's formula. Specifically, the expression involves the equation tan(x) = (e^(ix) - e^(-ix)) / (i(e^(ix) + e^(-ix))). To derive this, one must manipulate the equation by solving a quadratic equation formed during the process. Understanding how to express tan(y) in terms of exponentials is essential for this derivation.

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  • Understanding of trigonometric functions, specifically the tangent function.
  • Familiarity with Euler's formula and complex exponentials.
  • Ability to solve quadratic equations.
  • Knowledge of logarithmic identities and properties.
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  • Study the derivation of arctan using logarithmic identities.
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  • Practice solving quadratic equations in the context of trigonometric identities.
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cragar
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my math teacher said that the arctan can be setup in terms of logs
does anyone know how to do this.
 
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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.
 
can i use eulers formula to do it .

so would it be [(e^(ix)-e^(-ix)]/[(ie^(ix)+ie^(-ix))] = tan(x)
 
Last edited:
Correct.

You'll need to solve a quadratic in the process.
 
what quadratic
 
You have
tan x= \frac{e^x- e^{-x}}{e^x+ e^{-x}}= y
First multiply on both sides of the equation by e^x+ e^{-x}.
Then multiply both sides o the equation by e^x
 
oh i see
 

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