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Math Help - Maximum value

  1. #1
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    Maximum value

    find the maximum value of
    (cos(cos x))^2 +(sin(sin x))^2
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  2. #2
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    Quote Originally Posted by ayushdadhwal View Post
    find the maximum value of
    (cos(cos x))^2 +(sin(sin x))^2
    u = \cos{x} , v = \sin{x}

    y = \cos^2(u) + \sin^2(v)

    \dfrac{dy}{dx} = -2\cos(u)\sin(u) \cdot \dfrac{du}{dx}  + 2\sin(v)\cos(v) \cdot \dfrac{dv}{dx}

    take it from here?
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  3. #3
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    Quote Originally Posted by skeeter View Post
    u = \cos{x} , v = \sin{x}

    y = \cos^2(u) + \sin^2(v)

    \dfrac{dy}{dx} = -2\cos(u)\sin(u) \cdot \dfrac{du}{dx}  + 2\sin(v)\cos(v) \cdot \dfrac{dv}{dx}

    take it from here?
    not getting
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  4. #4
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    Quote Originally Posted by ayushdadhwal View Post
    not getting
    Are you even trying? You were told that u= cos(x). What is du/dx? You were told that v= sin(x). What is dv/dx?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by ayushdadhwal View Post
    find the maximum value of (cos(cos x))^2 +(sin(sin x))^2

    The function

    ^2+(\;\sin (\sin x)\^2" alt="f(x)= (\;\cos (\cos x)\^2+(\;\sin (\sin x)\^2" />

    is even and periodic with period \pi so, we only need to find the maximum on [0,\pi/2] .

    The values of f at the endpoints of [0,\pi/2] are:

    f(0)=\cos^2 1<\sin^2 1=f(\pi/2)

    Prove that there are no singular points in (0,\pi/2) so, the absolute maximun of f is \sin^2 1 and the absolute minimum \cos^2 1 .
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