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Math Help - identity

  1. #1
    Super Member Random Variable's Avatar
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    identity

    Show that  \displaystyle \arccos \Big(\frac{\cos x}{1+ 2 \cos x } \Big) = \arctan \sqrt{\frac{1 + \cos x}{1 + 3 \cos x}} .
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  2. #2
    A Plied Mathematician
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    This identity is not true, I don't think. Let x=\pi/4. Then the LHS is approximately 1.27, and the RHS is approximately 0.64.
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  3. #3
    Super Member Random Variable's Avatar
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    My mistake. It should be  \displaystyle \arccos \Big(\frac{\cos x}{1+ 2 \cos x } \Big) = 2 \arctan \sqrt{\frac{1 + \cos x}{1 + 3 \cos x}}
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  4. #4
    Super Member Random Variable's Avatar
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    The identity being used is  \displaystyle \arccos x = 2 \arctan \sqrt{\frac{1-x}{1+x}} . But from where does that identity come?
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Random Variable View Post
    The identity being used is  \displaystyle \arccos x = 2 \arctan \sqrt{\frac{1-x}{1+x}} . But from where does that identity come?
    Hang on. This is going to be a ride...

    Take the cosine of both sides:
     \displaystyle cos(acs x)  = x

    So
    \displaystyle cos \left ( 2 ~ atn  \left ( \sqrt{\frac{1-x}{1+x}} \right ) \right ) must be equal to x.

    To check this. let \theta = atn  \left ( \sqrt{\frac{1-x}{1+x}} \right ) \right )

    So we have
    \displaystyle cos \left ( 2 ~ atn  \left ( \sqrt{\frac{1-x}{1+x}} \right ) \right ) = cos( 2 \theta) = cos^2(\theta) - sin^2(\theta)

    Now sketch a right triangle. Label one angle \theta. The leg of the triangle across from \theta has a length of \sqrt{1 - x} and the other leg is \sqrt{1 + x}. This means the hypotenuse has a length of \sqrt{2}.

    So
    \displaystyle cos( \theta ) = \frac{\sqrt{1 + x}}{\sqrt{2}}

    and
    \displaystyle sin( \theta ) = \frac{\sqrt{1 - x}}{\sqrt{2}}

    Square both of these and subtract and you get
    \displaystyle cos^2( \theta ) - sin^2( \theta ) = \frac{1 + x}{2} - \frac{1 - x}{2} = x

    as advertised.

    -Dan
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  6. #6
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    Quote Originally Posted by Random Variable View Post
    The identity being used is  \displaystyle \arccos x = 2 \arctan \sqrt{\frac{1-x}{1+x}} . But from where does that identity come?
    I know this is in the trigonometry forum, but I'm sure you won't mind integral calculus!

    \displaystyle \cos^{-1}{x} = \int_{x}^{1}\frac{1}{\sqrt{1-t^2}}\;{dt} = \int_{x}^{1}\frac{1}{\sqrt{1-t}\sqrt{1+t}}\;{dt}

    Let u = \frac{\sqrt{1-t}}{\sqrt{1+t}} \Rightarrow dt = -\sqrt{1-t}\sqrt{1+t}(1+t)}\;{du} and (for simplicity) write \alpha = \frac{\sqrt{1-x}}{\sqrt{1+x}}, then:

    \displaystyle \begin{aligned} \cos^{-1}{x} & = -\int_{\alpha}^{0}\frac{\sqrt{1-t}\sqrt{1+t}(1+t)}{\sqrt{1-t}\sqrt{1+t}}\;{du} = \int_{0}^{\alpha} 1+t \;{du} = \int_{0}^{\alpha} 1+\frac{1-u^2}{1+u^2}\;{du} \\& =  \int_{0}^{\alpha} \frac{(1+u^2)+(1-u^2)}{1+u^2}\;{du} = \int_{0}^{\alpha} \frac{2}{1+u^2}\;{du} = 2\tan^{-1}{u}\bigg|_{0}^{\alpha} \\& = 2\tan^{-1}{\alpha}-2\tan^{-1}(0) = 2\tan^{-1}{\alpha} = 2\tan^{-1}\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right).\end{aligned}
    Last edited by TheCoffeeMachine; February 26th 2011 at 08:51 PM.
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  7. #7
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    Try this method.
    Let x = cosθ.

    Then sqrt[{1-x}/{1+x}] = sqrt[{1-cosθ}/{1+cosθ)] = sqrt[{2sin^2(θ/2)}/{2cos^2(θ/2)] = tan(θ/2)

    Rest is obvious.
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