# Sum of Cosines

## 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.

tiny-tim
Homework Helper
Hi Dragonfall! (have a pi: π and a sigma: ∑ and try using the X2 icon just above the Reply box )

Use one of the standard trigonometric identities for cos2 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.

tiny-tim
Homework Helper
Actually, half that! 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? Unfortunately, if I draw a diagram, the Y values cancel out. The X values, which are cos, do not.

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tiny-tim
Homework Helper
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. SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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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