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

rcs

Jul 2010
487
4
iligan city. Philippines
Can anybody

Find the sum of
cos 1 deg. + cos 3 deg. + cos 5 deg.+ ... + cos 177 deg. + cos 179 deg.
 
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Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, rcs!

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

Recall the sum-to-product identity:

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


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

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

 
Mar 2010
715
381
\(\displaystyle \displaystyle \sum_{0\le k \le n}\cos\left({\varphi+k\alpha}\right) =\frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)} \cdot \cos{(\varphi + \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}\)

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

\(\displaystyle \displaystyle \sum_{0^{\circ}\le k \le 89^{\circ}}\cos\left({2k+1}\right) =\frac{\sin{90^{\circ}} \cdot \cos{90^{\circ}}}{\sin{1^{\circ}}} = 0.\)
 
Dec 2009
3,120
1,342
Can anybody

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

Hence,

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

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

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

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

...........

 
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