arctg

kid said:
yes, my question is which value is equal to arctg(2π)
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Do you mean arctan? If you do, you should know that it is only defined in (-π/2, π/2) for good reason. It is not unique outside the interval,
 
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kid said:
does someone know how to find arctg(2π)? thanks :)
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That's oddly written. Usually, we think of "\(\displaystyle 2\pi\)" as an angle but in "triangle trigonometry", it is tangent that is applied to angles to get a "number" result and then arctangent go back from the numbers to angles. That is, because \(\displaystyle tan(0)= tan(2\pi)= 0\), we have \(\displaystyle arctan(0)= 0\) or \(\displaystyle /arctan(0)= 2\pi\). It is the tangent that is restricted to between \(\displaystyle -\pi/2\) and \(\displaystyle \pi/2\) in order to get a unique value for arctangent.

However, when we extend "tangent" and "arctangent" to general "functions" then we can have such values. The simplest way to find \(\displaystyle arctan(2\pi)\) is to use a calculator: \(\displaystyle arctan(2\pi)= 1.412965\) approximately. That is, of course, slightly less than \(\displaystyle \pi/2= 1.57079\). That is the principal value. If you allow "multivalued" functions, add any integer multiple of \(\displaystyle \pi\) to that.
 
iocal said:
Do you mean arctan? If you do, you should know that it is only defined in (-π/2, π/2) for good reason. It is not unique outside the interval,
Click to expand...

You are confusing range with domain. In the posted problem 2π is a value associated with the "independent" variable - hence associated with the domain.

The OP's question can be translated as:

If tan(x) = 2π, what is the value of x?

As HoI showed above it can be calculated.
 
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