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Math Help - Problem #20: Sum of Sines

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    Forum Admin topsquark's Avatar
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    Problem #20: Sum of Sines

    We haven't done a Trigonometry problem in a while, so here we go!

    Evaluate: sin^4 \left ( \frac{\pi}{8} \right ) + sin ^4 \left ( \frac{3 \pi}{8} \right ) + sin ^4 \left ( \frac{5 \pi}{8} \right ) + sin ^4 \left ( \frac{7 \pi}{8} \right )

    -Dan
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    Re: Problem #20: Sum of Sines

    Do you know any half angle formulas? Do you know what sin(\pi/2)? So what are sin(\pi/4) and sin(\pi/8)? Once you have that the rest is easy.
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    Re: Problem #20: Sum of Sines

    That serves as a good hint.

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    Forum Admin topsquark's Avatar
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    Re: Problem #20: Sum of Sines

    No final takers? Here's my solution.

    Spoiler:
    I prefer a slightly different approach than that given by HallsofIvy:

    cos(2 \theta ) = 1 - 2~sin^2( \theta ) \implies sin^4( \theta) = \left ( \frac{ 1 - 2~cos( 2 \theta )}{2} \right )^2

    So for \theta = \frac{\pi}{8} we get
    sin^4 \left ( \frac{\pi}{8} \right ) = \left ( \frac{ 1 - 2~cos \left (  2 \cdot \frac{\pi}{8} \right ) }{2} \right )^2 = \left ( \frac{ 1 - 2~cos \left ( \frac{\pi}{4} \right ) }{2} \right )^2 = \frac{3}{8} - \frac{\sqrt{2}}{4}

    Similarly
    sin^4 \left ( \frac{3 \pi}{8} \right ) = \frac{3}{8} + \frac{\sqrt{2}}{4}

    sin^4 \left ( \frac{5 \pi}{8} \right ) = \frac{3}{8} + \frac{\sqrt{2}}{4}

    sin^4 \left ( \frac{7 \pi}{8} \right ) = \frac{3}{8} - \frac{\sqrt{2}}{4}

    Adding these gives 3/2.


    -Dan
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