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Math Help - Proof that two inverse trig functions are equal for all X

  1. #1
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    Proof that two inverse trig functions are equal for all X

    I seem to be having some issues.
    Prove that:

    2arcsin(x) = arccos(1-2x^2)     \forall     x>0

    If the above is true, then

    2arcsin(x) - arccos(1-2x^2) = f(x) = 0  \forall    x>0

    f(1) = 0

    but

    f(2) \not= 0
    by WolframAlpha/Mathematica
    http://www.wolframalpha.com/input/?i=2+arcsin%282%29++-+arccos%281-2%282%29^2%29


    Can anyone figure out what I am doing incorrectly?
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  2. #2
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    Quote Originally Posted by skyd171 View Post
    I seem to be having some issues.
    Prove that:

    2arcsin(x) = arccos(1-2x^2) \forall x>0

    If the above is true, then

    2arcsin(x) - arccos(1-2x^2) = f(x) = 0 \forall x>0

    f(1) = 0

    but

    f(2) \not= 0
    by WolframAlpha/Mathematica
    http://www.wolframalpha.com/input/?i=2+arcsin%282%29++-+arccos%281-2%282%29^2%29


    Can anyone figure out what I am doing incorrectly?
    Obviously the domain of f(x) is -1 \leq x \leq 1.
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  3. #3
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    Hello, skyd171!

    \text{Prove that: }\;2\arcsin(x) \:=\: \arccos(1-2x^2),\;\;\forall x>0

    \text{Let: }\:\alpha \,=\, \arcsin(x)

    The left side is: . 2\alpha

    Take the cosine: . \cos(2\alpha) \:=\:2\cos^2\alpha-1 .[1]


    \text{Since }\,\sin\alpha \:=\:x \;=\; \dfrac{x}{1} \:=\:\dfrac{opp}{hyp}

    . . \,\alpha is is a right triangle with: opp = x,\;hyp = 1

    \text{Pythagorus gives us: }\:adj \,=\,\sqrt{1-x^2}

    . . \text{Hence: }\:\cos\alpha \:=\:\sqrt{1-x^2}


    Substitute into [1]:

    . . \cos(2\alpha) \;=\;2(\sqrt{1-x^2})^2 - 1 \;=\;2(1-x^2)-1 \;=\;2-2x^2 - 1

    . . \cos(2\alpha) \;=\;1 - 2x^2

    \text{Then: }\:2\alpha \;=\;\arccos(1-2x^2)


    \text{Therefore: }\:2\arcsin(x) \;=\;\arccos(1-2x^2)

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  4. #4
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    Thanks all.
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