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Math Help - arccos x + arccos y = ...

  1. #1
    Junior Member darence's Avatar
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    arccos x + arccos y = ...

    Proove that for real numbers x,y\in[-1,1] is \arccos x+\arccos y =\begin{cases}\arccos(xy-\sqrt{1-y^2}\sqrt{1-x^2} & ,x +y \geq0\\2\pi-\arccos(xy-\sqrt{1-y^2}\sqrt{1-x^2} & ,x+y< 0\end{cases}
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  2. #2
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    Re: arccos x + arccos y = ...

    Hey darence.

    Have you tried setting up two right hand triangles and using Pythagoras and trig identities?
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  3. #3
    Junior Member darence's Avatar
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    Re: arccos x + arccos y = ...

    I tried this :
    \alpha=\arccos x\\\beta=\arccos y\\\Rightarrow \cos\alpha=x, \cos\beta=y\\\cos(\alpha+\beta)=\cos\alpha\cos\bet  a-\sin\alpha\sin\beta\\\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-x^2}\\ \sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-y^2}\\ \Rightarrow\cos\gamma=xy-\sqrt{1-x^2}\sqrt{1-y^2}\\
    where
    \gamma=\arccos(xy-\sqrt{1-x^2}\sqrt{1-y^2})

    But, where to use given conditions ?
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