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Math Help - arcus functions

  1. #1
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    arcus functions

    Hi again!
    Does anyone here know how to solve arctan(x)=arccos(2x)?
    For x of course.
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  2. #2
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    Quote Originally Posted by a4swe View Post
    Hi again!
    Does anyone here know how to solve arctan(x)=arccos(2x)?
    For x of course.
    Take the tangent of both sides,
    tan(arctan(x))=tan(arccos(2x))
    x=tan(arccos(2x))

    Let,
    arccos(2x)=u
    Then,
    cos u =2x=(2x)/1
    Thus,
    A right triangle angle is "u" with adjacent side 2x and hypotenuse 1.
    By Pythagorean theorem the opposite side is sqrt(1-4x^2)
    Then,
    tan u
    Is opposite over adjacent which is,
    sqrt(1-4x^2)/(2x)

    Thus we finally have,
    x=sqrt(1-4x^2)/(2x)
    Multiply by 2x,
    2x=sqrt(1-4x^2)
    Square,
    4x^2=1-4x^2
    Thus,
    8x^2=1
    Thus,
    x^2=1/8
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  3. #3
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    To me it seems like that is wrong (sorry) since I can't get that to be sqrt((1/2)(sqrt(2)-1)) witch my lecturer states as the correct answer.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Take the tangent of both sides,
    tan(arctan(x))=tan(arccos(2x))
    x=tan(arccos(2x))

    Let,
    arccos(2x)=u
    Then,
    cos u =2x=(2x)/1
    Thus,
    A right triangle angle is "u" with adjacent side 2x and hypotenuse 1.
    By Pythagorean theorem the opposite side is sqrt(1-4x^2)
    Then,
    tan u
    Is opposite over adjacent which is,
    sqrt(1-4x^2)/(2x)

    Thus we finally have,
    x=sqrt(1-4x^2)/(2x)
    Multiply by 2x,
    2x=sqrt(1-4x^2)
    You were correct up to this step. You only multiplied both sides by 2, not 2x.

    To continue the last line should have been:
    2x^2 = sqrt(1 - 4x^2)
    Square both sides:
    4x^4 = 1 - 4x^2
    4x^4 + 4x^2 - 1 = 0

    We may use the quadratic formula to solve this for x^2:
    x^2 = (-4 (+/-) sqrt[4^2 - 4*4*-1])/(2*4)

    x^2 = -1/2 (+/-) sqrt(2)/2

    Note we need to choose the "+" solution since the "-" solution gives a negative x^2.

    Thus x = (+/-)sqrt{-1/2 + sqrt(2)/2}

    (In case you saw the original, I found a mistake I made. This is the fixed copy.)

    I will note that the + solution works, but the - solution doesn't. Perhaps it has to do with the domains on arccos and arctan?

    -Dan
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  5. #5
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    Yes I was about to answer you on the incorrect one.
    Thanks both of you, now I can continue my work.
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