Proving the Sum of Cosines: A Homework Challenge

Dragonfall
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Homework Statement



\sum_{k=0}^{4}\cos^2\left({\frac{2\pi k}{5}\right) = 5/2

The Attempt at a Solution



I have to prove that. I don't know how. Maple gives

1+2*cos((2/5)*Pi)^2+2*cos((1/5)*Pi)^2

evaluated numerically it is equal to 2.499999. But I don't remember the trick to doing this algebraically.
 
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Hi Dragonfall! :smile:

(have a pi: π and a sigma: ∑ and try using the X2 icon just above the Reply box :wink:)

Use one of the standard trigonometric identities for cos2 :wink:
 
Using the half angle formula I get the terms as

1+\cos\left(\frac{4k\pi}{5}\right)

There does not seem to be a way to cancel them out or write them as nice, rational numbers.
 
Actually, half that! :wink:

ok, the "1" neatly gives the correct answer on its own, so all you need to do is to prove that the five cosines all add to zero …

have you tried drawing a diagram of them? :smile:
 
Unfortunately, if I draw a diagram, the Y values cancel out. The X values, which are cos, do not.
 
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Dragonfall said:
Unfortunately, if I draw a diagram, the Y values cancel out. The X values, which are cos, do not.

ok, maybe just looking at the diagram doesn't do it,

but you can see they're evenly spaced, so you're adding ∑ cos(nπ/5) …

that's the real part of ∑ en(2πi/5), = ∑ (e(2πi/5))n,

which is easy algebra. :wink:
 
Dragonfall said:

Homework Statement



\sum_{k=0}^{4}\cos^2\left({\frac{2\pi k}{5}\right) = 5/2

The Attempt at a Solution



I have to prove that. I don't know how. Maple gives
1+2*cos((2/5)*Pi)^2+2*cos((1/5)*Pi)^2
I believe you mean:

1+2\cos^{2}((2/5)\pi)+2\cos^{2}((1/5)\pi)
evaluated numerically it is equal to 2.499999. But I don't remember the trick to doing this algebraically.
For the following see: Weisstein, Eric W. "Trigonometry Angles--Pi/5." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/TrigonometryAnglesPi5.html

\cos\left(\frac{\pi}{5}\right)=\frac{1}{4}\left(\sqrt{5}+1\right)

\cos\left(\frac{2\pi}{5}\right)=\frac{1}{4}\left(\sqrt{5}-1\right)
 
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