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  1. #1
    Super Member fardeen_gen's Avatar
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    Find range of...?

    Find the range of (arcsin x)^2 + (arccos x)^2 ?
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    Quote Originally Posted by fardeen_gen View Post
    Find the range of (arcsin x)^2 + (arccos x)^2 ?
    The range of this function are all y\in \mathbb{R} such that the equation (\arcsin x)^2 + (\arccos x)^2 = y has a solution in the domain i.e. -1\leq x\leq 1. Now we use the identity that \arccos x = \tfrac{\pi}{2} - \arcsin x.

    This gives,
    (\arcsin x)^2 + \left( \tfrac{\pi}{2} - \arcsin x\right)^2 = y
    This becomes,
    2(\arcsin x)^2 - \pi \arcsin x + \left( \tfrac{\pi^2}{4} - y\right) = 0
    It order for this to have a solution we require,
    \pi^2 - 4(2)\left( \tfrac{\pi^2}{4} -y\right) \geq 0 \implies y\geq \frac{\pi^2}{8}

    But once we have y satisfy this inequality we have,
    \arcsin x = \frac{\pi \pm \sqrt{8y - \pi^2}}{4}
    Therefore,
    x = \sin \left( \frac{\pi \pm \sqrt{8y - \pi^2}}{4} \right)

    Certainly this is in the domain, but does it solve the equation?
    For that we need to substitute but we for we do that note,
    \arcsin(\sin x) = \left\{ \begin{array}{c} x \text{ for }-\pi \leq x\leq \pi \\ f(x+2\pi) = f(x) \end{array} \right.

    This means,
    \arcsin \left[ \sin \left( \frac{\pi + \sqrt{8y - \pi^2}}{4} \right) \right] = \left\{ \begin{array}{c} \frac{\pi+\sqrt{8y-\pi^2}}{4} \text{ for }y\leq \frac{5}{4}\pi^2 \\ f(y+2\pi) = f(y)\end{array} \right. . . . .[*]

    Now if y\leq \tfrac{5}{4}\pi^2 then,
    (\arcsin x)^2 + (\arccos x)^2 = (\arcsin x)^2 + (\tfrac{\pi}{2} - \arcsin x)^2 would give,
    \left( \frac{\pi+\sqrt{8y - \pi^2}}{4} \right)^2 + \left( \frac{\pi - \sqrt{8y - \pi^2}}{4} \right)^2 = y

    What happens if y > \tfrac{5}{4}\pi^2?
    The answer is sad things happen.
    The remaining step is to use[*] to show the solutions do not match upon substitution.

    Therefore the range is \left[ \frac{\pi^2}{8}, \frac{5\pi^2}{4} \right].
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  3. #3
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    Quote Originally Posted by fardeen_gen View Post
    Find the range of (arcsin x)^2 + (arccos x)^2 ?
    (\arcsin x )^2 + (\arccos x)^2 = (\arcsin x + \arccos x)^2 - 2 \arcsin x + \arccos x = \left( \frac{\pi}{2}\right)^2 - 2 ( \arcsin x ) ( \arccos x ) .

    So all you have to do is get the range of ( \arcsin x ) ( \arccos x ) .
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