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Trigonometric question 2

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Write arctan8 + arctan11 as an expression containing max one term with arctan.


    2. Relevant equations
    tan(u+v) = [itex]\frac{tanu+tanv}{1-tanutanv}[/itex]
    arctan(tanx) = x


    3. The attempt at a solution
    u = arctan8, v = arctan11
    tan(u+v) = [itex]\frac{tanu+tanv}{1-tanutanv}[/itex] = [itex]\frac{8+11}{1-8*11}[/itex] = -(19/87) = tan(arctan8 + arctan11)

    arctan(tan(arctan8 + arctan11)) = arctan(-(19/87)) = arctan8 + arctan11

    arctan8 + arctan11 = arctan(-/19/87)) = -arctan(19/87)

    But (arctan8 + arctan11) > 0 and -arctan(19/87) < 0

    Where have I gone wrong?
     
    Last edited: Dec 2, 2011
  2. jcsd
  3. Dec 2, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    Remember, the tangent function is periodic with pi: tanθ=tan(θ±π)
    The range of the inverse function arctan is (-π/2,π/2). When you type in a number and hit tan-1, the calculator gives the principal value, an angle in the interval (-π/2,π/2).

    You got it right, tan(arctan8 + arctan11)= -(19/87).

    arctan8 + arctan11 is an angle in the second quadrant, and its tangent is the same as that of (arctan8 + arctan11-pi), an angle between -pi/2 and pi/2: That is what you get as result. Add pi to have the the real sum: arctan8 + arctan11=arctan(-19/87)+pi

    ehild
     
  4. Dec 2, 2011 #3
    @ehild

    Alright. I get it now. Tanv will produce the same result for all angles v + n*pi, where n=1, 2, 3... Thank you!
     
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