# Cosine and sine of 2Pi/5 problem

1. Nov 19, 2015

### cdummie

We have a complex number ω = cos(2π/5) +isin(2π/5) and we have two complex numbers a and b, such that:

a= ω + ω4 and b= ω2 + ω3. I have to prove that a + b = -1 and a*b= -1. Then, based on that, determine cos(2π/5) and sin(2π/5).

I've tried to solve this using trigonometry. First a+b, i got:

cos(2Pi/5) + cos(4Pi/5) + cos(6Pi/5) + cos(8Pi/5) + i[ sin(2Pi/5) + sin(4Pi/5) + sin(6Pi/5) + sin(8Pi/5)]

then i tried expressing cos(6Pi/5) as cos(2Pi/5 + 5Pi/5) = cos(2Pi/5)cos(4Pi/5) + sin(2Pi/5)sin(4Pi/5)

and cos(4Pi/5) as cos2(2Pi/5) - sin2(2Pi/5). But that wasn't helpful. Knowing that a + b equals -1 that means that this whole expression should end up as cosPi + isinPi since that is equal to -1. But i just can't get there.

2. Nov 19, 2015

### BvU

Are you familiar with the Euler formula ?
Mark your $\omega^n$ terms in a and b on a unit circle in the complex plane and see what that brings you....

3. Nov 19, 2015

### cdummie

I know Euler's formula in this example i'd have: $ω=e^\frac{i2π}{5}$ or for a + b i would have: $a+b = e^\frac{i2π}{5} + e^\frac{i4π}{5} + e^\frac{i6π}{5} + e^\frac{i8π}{5}$ but i don't know how this helps...

When i draw a unit circle i don't see much, i mean i see the points that represent corresponding complex numbers but i don't see how it gets me closer to the solution.

4. Nov 19, 2015

### BvU

Do you see the same thing I see ? Addition of complex numbers is like vector addition in the x,y plane.
Can you post your sketch ?

Last edited: Nov 20, 2015
5. Nov 20, 2015

### cdummie

Here is the sketch:

But, i can't see what will i get exactly by just looking at this picture (i'd like you to explain it to me though), and i'd like to know if there's any algebraic way to solve this?

6. Nov 20, 2015

### BvU

So where are a and b in the picture ?

7. Nov 20, 2015

### cdummie

a would be somewhere in the middle of the positive part of x axis in the circle (even though it seems a little bit closer to zero, but this is just a sketch) and b is on the negative part of the x axis in the circle little bit closer to (-1,0)

8. Nov 20, 2015

### Samy_A

I'm confused now. I do get a+b=-1 with the same results for a and b as you.
Mark44 edit: I made a mistake in my calculations, so the text that Samy_A is quoting is in error. I have also deleted my reply that is quoted above.

Last edited by a moderator: Nov 20, 2015
9. Nov 20, 2015

### Staff: Mentor

Oops, my mistake. I'll go back and edit my earlier post.

10. Nov 20, 2015

### Staff: Mentor

Have you calculated $ω^5, ω^6, ω^7$ to calculate $a \cdot b$? And are you familiar with polynomial divisions?
There is an interesting generalization of the 3rd binomial formula $(x-1)(x+1) = x^2 -1$ for $x^n -1$.

11. Nov 20, 2015

### BvU

Check b: it should be outside the circle (a.b is supposed to give -1, remember ?)

And: what is $\omega^5$ ?

Last edited: Nov 21, 2015
12. Nov 20, 2015

### WWGD

Maybe you can use the fact that if $w$ is an n-th root of unity, then EDIT $1+ w+w^2+...+w^n =0$ , changing the proof a bit to adapt it to your situation.

13. Nov 21, 2015

### cdummie

Oh i see now, b IS outside, but it's somewhere outside of circle, just like a is somewhere inside circle, that won't help if i don't know where they are exactly.
$\omega^5 = 1$

14. Nov 23, 2015

### cdummie

So, what can i do next?

15. Nov 23, 2015

### Staff: Mentor

a + b = ω + ω4 + ω2 + ω3
ω and ω4 are the two complex numbers on the right side of the drawing. They have the same x-coordinate and their y-coordinates are opposites. Similarly, ω2 and ω3 are on the left side of the drawing. Their x-coordinates are equal and their y-coordinates are opposite in value.

Ordinary trig can be used to show that a + b = -1.

For the product ab, it's helpful to know that ω, ω2, ω34, and ω5 are all equally spaced around the unit circle, and that higher powers simply retrace certain lower powers. By that I mean that ω6 = ω, ω7 = ω2, and so on.

16. Nov 23, 2015

### BvU

It's difficult to help you without giving away too much. So far we've established that a and b are real and that 0<a<1 and -2<b<-1.

You have also seen that $\omega^5 = 1$ which makes ab = a + b $\ \ \$ (clear ?)

Unfortunately, Mark's 'ordinary trig' doesn't help me other than that I can confirm a + b = -1 numerically. But the exercised wanted us to prove it.

For that we can now do something useful with $0 = \omega^5 - 1\ \ \$ -- see hint by fresh in post #10

lovely exercise !
--

17. Nov 23, 2015

### Staff: Mentor

Yes, you're correct. I just used a calculator to get a value, but didn't show it algebraically.
Yes!

To expand slightly on the hint from fresh_42 that BvU mentions, the idea is related to these facts:
$(x - 1)(x + 1) = x^2 - 1$
$(x - 1)(x^2 + x + 1) = x^3 - 1$
$(x - 1)(x^3 + + x^2 + x + 1) = x^4 - 1$
and so on. This idea can be helpful in evaluating a*b.

Last edited: Nov 23, 2015
18. Nov 23, 2015

### ehild

a + b = ω + ω4 + ω2 + ω3.
You can add 1 to it and rearrange the terms: a+b+1= 1+ω+ω234. It is a the sum of the first five elements of a geometric progression. You certainly know the formula of the sum.
If you expand ab+1 and use that ω5 = 1, you get the same sum.

Last edited: Nov 23, 2015
19. Nov 23, 2015

### BvU

Didn't we want to leave something for cd to discover ?

20. Nov 23, 2015