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Math Help - arccos(1/3)+2*arccos(-2sqrt(2)/3)

  1. #1
    Junior Member darence's Avatar
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    arccos(1/3)+2*arccos(-2sqrt(2)/3)

    Without calculator, find value of:
    $\arccos\frac{1}{3}+2\arccos(\frac{-2\sqrt{2}}{3})$

    Please help.
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  2. #2
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    Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

    I presume you know that \left(\frac{1}{3}\right)^2= \frac{1}{9} but you might not have noticed that \left(\frac{2\sqrt{2}}{3}\right)^2= \frac{8}{9} and so \left(\frac{1}{3}\right)^2+ \left(\frac{2\sqrt{2}}{3}\right)^2= 1.
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    Junior Member darence's Avatar
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    Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

    I have try, but I think this is not correct. Look !



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    Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

    It is that "2" in " 2 arccos\left(\frac{-2\sqrt{2}}{3}\right) that is the problem. I am tempted to suggest that it is an error!
    Obviously arccos\left(\frac{1}{3}\right)+ arccos\left(\frac{-2\sqrt{2}}{3}\right) would be simple.
    Last edited by HallsofIvy; June 2nd 2013 at 01:09 PM.
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  5. #5
    Junior Member darence's Avatar
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    Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

    Yes, it is 2. It is not an error
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    Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

    Darn!
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  7. #7
    Junior Member Bradyns's Avatar
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    Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

    When in situations likes this, I love drawing pictures.

    For the second term, we have a triangle with a base of 1, a hypotenuse of 3, so the other side using pythagoras is \pm 2\sqrt{2} given, that we are talking distance, we take the positive.

    Hence, we can say.
    arccos(\frac{1}{3} = arcsin(\frac{2\sqrt(2)}{3})

    Now, we set the original equation to theta, as there is some angle that is works out to be,

    arccos(\frac{-2\sqrt(2)}{3}) = \theta - arcsin(\frac{2\sqrt(2)}{3})

    Take the inverse of the inverse.
    \frac{-2\sqrt(2)}{3} = cos(\theta - arcsin(\frac{2\sqrt(2)}{3}) )

    Recall cos(a-b) = cos(a)cos(b) - sin(a)sin(b)
    \frac{-2\sqrt(2)}{3} = \frac{cos(\theta)}{3} - sin(\theta)\frac{2\sqrt(2)}{3}

    Multiply through by 3

    -2\sqrt(2) = cos(\theta) - 2\sqrt(2)sin(\theta)

    -2\sqrt(2) = \frac{cos(\theta) }{1-sin(\theta)}

    8 = \frac{cos^2(\theta) }{(1-sin(\theta))^2}

    This is about all I can whittle it down to..

    Brutal question for non-calc.
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    Re: arccos(1/3)+2*arccos(-2sqrt(2)/3)

    It is really tough question for me and i am not able to solve this type of question.
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