Proving the Sum of Cosines: A Homework Challenge

[Dragonfall]

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]

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

 

[Dragonfall]

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]

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? ​

 

[Dragonfall]

Unfortunately, if I draw a diagram, the Y values cancel out. The X values, which are cos, do not.

 

[tiny-tim]

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 ∑ e^n(2πi/5), = ∑ (e^(2πi/5))ⁿ,

which is easy algebra.

 

[SammyS]

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

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

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