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Math Help - Just solving inverse trigometric function

  1. #1
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    Just solving inverse trigometric function

    arctan(tan(11pi/5))

    The textbook I have doesn't cover inverse trigonometric functions, at least for just solving them. I thought you would use (2pi - 11pi/5) but I get - pi/5 which is 36 degrees, as far as I know there isn't an angle on the unit circle that is 36 degrees.
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  2. #2
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     \arctan(\tan\left(\frac{11\pi}{5}\right)= \frac{11\pi}{5}

    as \arctan and \tan are inverse operations.
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  3. #3
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    the answer here says its pi/5?
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  4. #4
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    ok, what is the domain of \arctan ?
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  5. #5
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    [-infinity, infinity]
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    bump
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  7. #7
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  9. #9
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    The whole point of this problem is that for x between -\pi/2 and \pi/2, the domian of "principle value" of tangent, arctan(tan(x))= x. What angle between -\pi/2 and \pi/2 gives the same tangent as 11\pi/5?


    (11/5= 2+ 1/5 so 11\pi/5= 2\pi+ \pi/5.)
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