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Math Help - prove that

  1. #1
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    prove that

    \cos(\pi/14)\cos(3\pi/14)\cos(5\pi/14)=\sqrt7/8
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  2. #2
    Member Henderson's Avatar
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    Use the Product-to-Sum Formula:

    \cos(A)\cos(B) = \frac{1}{2}(\cos(A-B) + \cos(A+B)).

    Use it first on the first two, then use that result with the third.
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  3. #3
    Senior Member JaneBennet's Avatar
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    Hendersonís suggestion will lead to \cos{\frac{\pi}{14}}+\cos{\frac{3\pi}{14}}-\cos{\frac{5\pi}{14}}=\frac{\sqrt{7}}{2}.

    Also, note that

    \color{white}.\quad. \cos{\frac{5\pi}{14}}\left(\cos{\frac{3\pi}{14}}+\  cos{\frac{\pi}{14}}\right)

    =\ \cos{\frac{5\pi}{14}}\left(2\cos{\frac{\pi}{14}}\c  os{\frac{2\pi}{14}}\right)

    =\ \cos{\frac{\pi}{14}}\left(2\cos{\frac{5\pi}{14}}\c  os{\frac{2\pi}{14}}\right)

    =\ \cos{\frac{\pi}{14}}\left(\cos{\frac{3\pi}{14}}+\c  os{\frac{\pi}{2}}\right)

    =\ \cos{\frac{\pi}{14}}\cos{\frac{3\pi}{14}}

    i.e. \cos{\frac{\pi}{14}}\cos{\frac{3\pi}{14}}-\cos{\frac{3\pi}{14}}\cos{\frac{5\pi}{14}}-\cos{\frac{5\pi}{14}}\cos{\frac{\pi}{14}}=0.

    So if you can prove that \cos{\frac{\pi}{14}},\cos{\frac{3\pi}{14}},-\!\cos{\frac{5\pi}{14}} are roots of the equation 8x^3-4\sqrt{7}x^2+\sqrt{7}=0, youíre done.
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  4. #4
    Senior Member JaneBennet's Avatar
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    I got it! I got it!

    Use the identity


    \cos{7\theta}\ =\ 64\cos^7{\theta}-112\cos^5{\theta}+56\cos^3{\theta}-7\cos{\theta}


    Substitute \theta=\frac{\pi}{7},\frac{3\pi}{7},\frac{5\pi}{7} into that. Thus youíll find that \cos{\frac{\pi}{7}},\cos{\frac{3\pi}{7}},\cos{\fra  c{5\pi}{7}} are the roots of the equation


    0\ =\ 64x^7-112x^5+56x^3-7x+1\ =\ (x+1)(8x^3-4x^2-4x+1)^2


    Obviously \cos{\frac{\pi}{7}},\cos{\frac{3\pi}{7}},\cos{\fra  c{5\pi}{7}}\ne-1; therefore they are the roots of 8x^3-4x^2-4x+1=0.

    Now \cos{\frac{\pi}{7}}=2\cos^2{\frac{\pi}{14}}-1, etc. Hence the roots of the equation


    8(2x^2-1)^3-4(2x^2-1)^2-4(2x^2-1)+1=0


    are \pm\cos{\frac{\pi}{14}},\pm\cos{\frac{3\pi}{14}},\  pm\cos{\frac{5\pi}{14}}. All you need from the sextic equation are the leading coefficient, which is 64, and the constant term, which is -7. Now multiply all the six roots together and you have your answer!

    Oh, I am so happy!
    Last edited by JaneBennet; March 18th 2008 at 02:32 AM.
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  5. #5
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    Consider the equation, z^{13}+z^{12}+...+z+1 = 0. The solutions are \zeta, \zeta^2, .... , \zeta^{13} where \zeta = \cos \frac{\pi }{14}+i\sin \frac{\pi}{14}.
    Thus, we can factorize, 1+z+z^2+...+z^{13} = \prod_{n=1}^{13}(z - \zeta^n).

    Let z=1 and we get, 14 = \prod_{n=1}^{13} = (1 - \zeta^n). This implies, 14 = |14| = \left| \prod_{n=1}^{13} (1-\zeta^n) \right| = \prod_{n=1}^{13} \left| 1 - \zeta^n \right|.

    Note, \left| 1 - \zeta^n\right|^2 = \left| 1 - \cos \frac{2\pi n}{14} - i\sin \frac{2\pi n}{14} \right|^2 = \left( 1-\cos \frac{2\pi n}{14} \right)^2 + \sin^2 \frac{2\pi n}{14} =   1 - 2\cos \frac{2\pi n}{14} + \cos^2 \frac{2\pi n}{14} + \sin^2 \frac{2\pi n}{14} = 2\left(1 - \cos \frac{2\pi n}{14}\right) = 4\sin ^2 \frac{\pi n}{14}.
    Which means, |1-\zeta^n | = \sqrt{4\sin^2 \frac{\pi n}{14} } = 2\sin \frac{\pi n}{14}.

    Thus, we established that, 14 = \prod_{n=1}^{13} 2\sin \frac{\pi n}{14} \implies \prod_{n=1}^{13}\sin \frac{\pi n}{14} = \frac{14}{2^{13}}.

    In the product, \sin \frac{\pi }{14} \cdot ... \cdot \sin \frac{5\pi}{14}\cdot\sin \frac{6\pi }{14}\cdot \sin \frac{7\pi}{14}\cdot ... \cdot \sin \frac{13\pi}{14} = \frac{14}{2^{13}}
    Look at the sines from 8\pi/14,...,13\pi/14.
    Use the fact that \sin \frac{8\pi}{14} = \sin \left( \pi - \frac{8\pi}{14} \right) = \sin \frac{6\pi}{14}, \sin \frac{9\pi}{14} = \sin \left( \pi - \frac{9\pi}{14} \right) = \sin \frac{5\pi}{14}, ... , \sin \frac{13\pi}{14} = \sin \frac{\pi}{14}.

    This means, we can write this product as,
    \sin^2 \frac{\pi}{14}\cdot \sin^2 \frac{2\pi}{14} \cdot ... \cdot \sin ^2\frac{6\pi}{14} \sin \frac{7\pi}{14} = \frac{14}{2^{13}}=\frac{7}{2^{12}}.

    This means, (note \sin \frac{7\pi}{14} = 1),
    \sin \frac{\pi}{14} \sin \frac{2\pi}{14} \cdot ... \cdot \sin \frac{6\pi}{14} = \frac{\sqrt{7}}{2^6}.

    Now, use fact that, \sin A\sin B = \frac{1}{2}\left( \cos (A-B) - \cos (A+B) \right).

    Apply the above identity to the opposite ends and work inside-out.
    \sin \frac{\pi}{14} \sin \frac{6\pi}{14} = \frac{1}{2}\cos \frac{5\pi}{14}
    \sin \frac{2\pi}{14} \sin\frac{5\pi}{14} = \frac{1}{2}\cos \frac{3\pi}{14}
    \sin \frac{3\pi}{14}\sin \frac{4\pi}{13} = \frac{1}{2}\cos \frac{\pi}{14}.

    Thus,
    \cos \frac{\pi}{14} \cos \frac{3\pi}{14} \cos \frac{5\pi}{14} = \frac{\sqrt{7}}{2^3}.

    Generalize! If n is odd then: \boxed{ \cos \frac{\pi}{2n} \cos \frac{3\pi}{2n} \cdot ... \cdot \cos \frac{(n-2)\pi}{2n} = \frac{\sqrt{n}}{2^{(n-1)/2}} }.
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