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Math Help - Prove that :-

  1. #1
    Newbie Sanjana Das's Avatar
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    Question Prove that :-

    sin^25^\circ+ sin^210^\circ+..........+sin^285^\circ+sin^290^\ci  rc=9 \frac{1}{2}

    I have been solving for a long time but in my answer my LHS is not equal to RHS....therefore pls help me....
    Last edited by Sanjana Das; September 30th 2009 at 10:35 AM.
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  2. #2
    Newbie Sanjana Das's Avatar
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    do not solve plss its a totally wrong quest.
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  3. #3
    Newbie Sanjana Das's Avatar
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    corrected question


    sin^25^\circ+ sin^210^\circ+..........+sin^285^\circ+sin^290^\ci  rc=9 \frac{1}{2}

    I have been solving for a long time but in my answer my LHS is not equal to RHS....therefore pls help me....
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  4. #4
    Newbie Sanjana Das's Avatar
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    No need to solve i have solved............
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  5. #5
    Senior Member DeMath's Avatar
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    You have

    \sum\limits_{n = 1}^{18} {{{\sin }^2}\left( {{5^ \circ }n} \right)}  = {\sin ^2}{5^ \circ } + {\sin ^2}{10^ \circ } +  \ldots  + {\sin ^2}{85^ \circ } + {\sin ^2}{90^ \circ } = \frac{{19}}{2}.

    Now simplify the left side by formula \sin^2\theta  = \frac{1-\cos2\theta}{2} left side and transform

    \sum\limits_{n = 1}^{18} {{{\sin }^2}\left( {{5^ \circ }n} \right)}  = \sum\limits_{n = 1}^{18} {{{\sin }^2}\frac{{n\pi }}{{36}}}  = \frac{1}{2}\sum\limits_{n = 1}^{18} {\left( {1 - \cos \frac{{n\pi }}{{18}}}\right)} = 9 - \frac{1}{2}\sum\limits_{n = 1}^{18} {\cos \frac{{n\pi }}{{18}}}=

    = 9 - \frac{1}{2}\left( {\sum\limits_{n = 1}^{17} {\cos \frac{{n\pi}}{{18}}}  + \cos \pi } \right) = \frac{{19}}{2} - \frac{1}{2} \sum\limits_{n = 1}^{17} {\cos \frac{{n\pi }}{{18}}}.

    Now estimate the sum \sum\limits_{n = 1}^{17}{\cos\frac{n\pi}{18}}.
    If n=9, then \cos\frac{9\pi}{18}=\cos\frac{\pi }{2}=0.
    Since the cosine function in the interval \left[\frac{\pi}{18};\,\frac{\pi}{2} \right) positive and negative in \left(\frac{\pi}{2};\,\frac{17\pi}{18}\right], it follows that \sum\limits_{n = 1}^{17} {\cos\frac{n\pi}{18}}=0,
    ie you have
    \sum\limits_{n = 1}^{18} {{{\sin }^2}\left( {{5^ \circ }n} \right)}  = \frac{{19}}{2} - \frac{1}{2}\underbrace {\sum\limits_{n = 1}^{17} {\cos \frac{n\pi}{18}}}_0 = \frac{19}{2}.

    Thus, the initial equality is proved.
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