- #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.

does any one know how to do this.

- 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.

does any one know how to do this.

- #2

Hurkyl

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Other methods are possible (e.g. antidifferentiate

- #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)

so would it be [(e^(ix)-e^(-ix)]/[(ie^(ix)+ie^(-ix))] = tan(x)

Last edited:

- #4

arildno

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Correct.

You'll need to solve a quadratic in the process.

You'll need to solve a quadratic in the process.

- #5

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what quadratic

- #6

HallsofIvy

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[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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