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Math Help - cos 1 deg. + cos 3 deg. + cos 5 deg.+ ... + cos 177 deg. + cos 179 deg.

  1. #1
    rcs
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    cos 1 deg. + cos 3 deg. + cos 5 deg.+ ... + cos 177 deg. + cos 179 deg.

    Can anybody

    Find the sum of
    cos 1 deg. + cos 3 deg. + cos 5 deg.+ ... + cos 177 deg. + cos 179 deg.
    Last edited by mr fantastic; December 4th 2010 at 01:10 PM. Reason: Re-titled.
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    Hello, rcs!

    \text{Find the sum: }\;\cos 1^o + \cos 3^o + \cos5^o + \hdots + \cos 177^o + \cos179^o

    Recall the sum-to-product identity:

    . . \cos A + \cos B \;=\;2\cos\left(\frac{A+B}{2}\right)\cos\left(\fra  c{A-B}{2}\right)


    Then take the pairs of cosines "from each end":

    . . \begin{array}{cccccccccc} \cos1 + \cos 179 &=& 2\cos90\cos89 &=& 0 \\ \cos3 + \cos177 &=& 2\cos90\cos87 &=& 0 \\ \cos5 + \cos175 &=& 2\cos90\cos85 &=& 0 \\ \vdots && \vdots && \vdots \\ \cos89 + \cos91 &=& 2\cos(90)\cos1 &=& 0 \\ \\[-4mm] \hline \\[-4mm] && \text{Total:} && 0 \end{array}

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    \displaystyle \sum_{0\le k \le n}\cos\left({\varphi+k\alpha}\right) =\frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)} \cdot \cos{(\varphi + \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}

    Putting \alpha = 2, \varphi = 1 and ranging k from 0^{\circ} to 89^{\circ}:

    \displaystyle \sum_{0^{\circ}\le k \le 89^{\circ}}\cos\left({2k+1}\right) =\frac{\sin{90^{\circ}} \cdot \cos{90^{\circ}}}{\sin{1^{\circ}}} = 0.
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    Quote Originally Posted by rcs View Post
    Can anybody

    Find the sum of
    cos 1 deg. + cos 3 deg. + cos 5 deg.+ ... + cos 177 deg. + cos 179 deg.
    cosA gives the horizontal co-ordinate of a point on the "unit-circle" at an angle A from the origin.

    Hence,

    cosA=-cos\left(180^o-A\right)

    Therefore, as there are 90 terms in the sum (an even number)...

    cos1^o=-cos\left(180^o-1^o\right)\Rightarrow\ cos1^0=-cos179^o

    cos3^o=-cos\left(180^o-3^o\right)\Rightarrow\ cos3^o=-cos177^o

    ...........

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