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Math Help - hard 2

  1. #1
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    Forum Admin topsquark's Avatar
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    There has to be something more behind this. What chapter of what book is it taken from, or what part of what course are you working with right now?

    -Dan
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  3. #3
    MHF Contributor red_dog's Avatar
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    a^2-b^2+ac=\sin ^25-\sin ^249+\sin 5\sin 87=
    =(\sin 5-\sin 49)(\sin 5+\sin 49)+\sin 5\sin 87=
    =-2\sin 22\cos 27\cdot 2\sin 27\cos 22+\sin 5\sin 87=
    =-\sin 44\sin 54+\sin 5\sin 87=-\frac{1}{2}(\cos 10-\cos 98-\cos 82+\cos 92)=
    =-\frac{1}{2}(2\cos 51\cos 41-2\cos 90\cos 5)=-\cos 41\cos 51

    Now the equality becomes
    \sin 73=\frac{\cos 41\cos 51}{4\sin 5\cos 51\cos 41+\sin 5-\sin 49+\sin 87}

    We have 4\sin 5\cos 51\cos 41=2(\sin 56-\sin 46)\cos 41=2\sin 56\cos 41-2\sin 46\cos 41=
    =\sin 15+\sin 97-\sin 5-\sin 87
    The equality becomes
    \sin 73=\frac{\cos 41\cos 51}{\sin 15+\sin 97-\sin 49}\Leftrightarrow
    \Leftrightarrow \cos 41\cos 51=\sin 15\sin 73+\sin 97\sin 73-\sin 49\sin 73\Leftrightarrow
    \Leftrightarrow \cos 92+\cos 10=\cos 58-\cos 88+\cos 24-\cos 170-\cos 24+\cos 122\Leftrightarrow
    \Leftrightarrow \cos 92+\cos 10=-\cos 88-\cos 170
    which is true because \cos 92=\cos (180-88)=-\cos 88
    and \cos 10=\cos (180-170)=-\cos 170
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by red_dog View Post
    a^2-b^2+ac=\sin ^25-\sin ^249+\sin 5\sin 87=
    =(\sin 5-\sin 49)(\sin 5+\sin 49)+\sin 5\sin 87=
    =-2\sin 22\cos 27\cdot 2\sin 27\cos 22+\sin 5\sin 87=
    =-\sin 44\sin 54+\sin 5\sin 87=-\frac{1}{2}(\cos 10-\cos 98-\cos 82+\cos 92)=
    =-\frac{1}{2}(2\cos 51\cos 41-2\cos 90\cos 5)=-\cos 41\cos 51

    Now the equality becomes
    \sin 73=\frac{\cos 41\cos 51}{4\sin 5\cos 51\cos 41+\sin 5-\sin 49+\sin 87}

    We have 4\sin 5\cos 51\cos 41=2(\sin 56-\sin 46)\cos 41=2\sin 56\cos 41-2\sin 46\cos 41=
    =\sin 15+\sin 97-\sin 5-\sin 87
    The equality becomes
    \sin 73=\frac{\cos 41\cos 51}{\sin 15+\sin 97-\sin 49}\Leftrightarrow
    \Leftrightarrow \cos 41\cos 51=\sin 15\sin 73+\sin 97\sin 73-\sin 49\sin 73\Leftrightarrow
    \Leftrightarrow \cos 92+\cos 10=\cos 58-\cos 88+\cos 24-\cos 170-\cos 24+\cos 122\Leftrightarrow
    \Leftrightarrow \cos 92+\cos 10=-\cos 88-\cos 170
    which is true because \cos 92=\cos (180-88)=-\cos 88
    and \cos 10=\cos (180-170)=-\cos 170
    Dang!

    -Dan
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