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

**rcs** · December 4th 2010, 04:07 AM

Can anybody

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

**Soroban** · December 4th 2010, 05:59 AM

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":

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

**TheCoffeeMachine** · December 4th 2010, 02:19 PM

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

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

$\displaystyle \sum_{0^{\circ}\le k \le 89^{\circ}}\cos\left({2k+1}\right) =\frac{\sin{90^{\circ}} \cdot \cos{90^{\circ}}}{\sin{1^{\circ}}} = 0.$

**Archie Meade** · December 4th 2010, 03:28 PM

Originally Posted by rcs

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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