# sum of cosine functions

• January 3rd 2011, 06:18 AM
vldo
sum of cosine functions
Find the sum of cos 1⁰ + cos 2⁰ + cos 23⁰ +…..+ cos 179⁰.

I know the sum would be zero but I can't show how to get it. Please help me to arrive an algebraic.
• January 3rd 2011, 06:32 AM
e^(i*pi)
$\displaystyle \cos(\theta+ \varphi) = 2 \cos\left( \dfrac{\theta + \varphi} {2} \right) \cos\left( \dfrac{\theta - \varphi}{2} \right)$

Take pairs from opposite ends such that $\alpha + \varphi = 180^{\circ}$

edit: Removed working because as Archie Meade pointed out in post 3 the difference is not 180. The sum is still 180 though and because each sum is 180 the answer is 0 since we're multiplying by 0.

Perhaps it would be better to follow the working in post 3 for a more complete solution
• January 3rd 2011, 07:53 AM
Quote:

Originally Posted by vldo
Find the sum of cos 1⁰ + cos 2⁰ + cos 23⁰ +…..+ cos 179⁰.

I know the sum would be zero but I can't show how to get it. Please help me to arrive an algebraic.

$cos(\pi-A)=cos{\pi}cos(-A)-sin{\pi}sin(-A)=-cos(-A)=-cosA$

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

$cos178^o=cos\left(180^o-2^o\right)=-cos2^o$

$\Rightarrow\ cos1^o+cos2^o.....+cos179^o=cos1^o+cos179^o+cos2^o +cos178^o+....+cos89^o+cos91^o+cos90^o$

$=cos1^o-cos1^o+cos2^o-cos2^o+......+cos89^o-cos89^o+cos90^o$

$=cos90^o$

@ e^(i*pi)

the difference of the angles is not 180 degrees.