Trigonometry: Solving cos(2π/7)+cos(4π/7)+cos(6π/7) Problem

[terryds]

Homework Statement

What is $cos(\frac{2\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{6\pi}{7})$ ?

Homework Equations

Trigonometry

The Attempt at a Solution

[/B]
I remember that the answer is $-\frac{1}{2}$

But, I don't know how to do it since the angle is not nice. Any hint?

 

[.Scott]

I would consider: $cos(\frac{0\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{6\pi}{7})+cos(\frac{8\pi}{7})+cos(\frac{10\pi}{7})+cos(\frac{12\pi}{7})$
Which would trace either the x or y along a 7-sided polygon path.
Then I would notice that the terms can be reordered from 0, 1, 2, 3, 4, 5, 6 to (0), (1, 3, 5), (2, 4, 6).
And I would compare the terms in the (1, 3, 5) group with those in the (2, 4, 6) group.

With these insights, you should be able to do the math in your head.

 

[terryds]

I would consider: $cos(\frac{0\pi}{7})+cos(\frac{1\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{3\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{5\pi}{7})+cos(\frac{6\pi}{7})$
Which would trace either the x or y along a circular path.
Then I would notice that the terms can be reordered from 0, 1, 2, 3, 4, 5, 6 to (0), (1, 3, 5), (2, 4, 6).
And I would compare the terms in the (1, 3, 5) group with those in the (2, 4, 6) group.

Hmm
$=cos(\frac{1\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{3\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{5\pi}{7})+cos(\frac{6\pi}{7})+1\\ =cos(\frac{1\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{3\pi}{7})-cos(\frac{3\pi}{7})-cos(\frac{2\pi}{7})-cos(\frac{1\pi}{7})+1\\ \\ =1\\$

And then what?
I don't grasp the key what to do

 

[.Scott]

Here is a walk around a 7-point polygon:
0: $sin(2\pi\frac{0}{7}),cos(2\pi\frac{0}{7})$
1: $sin(2\pi\frac{1}{7}),cos(2\pi\frac{1}{7})$
2: $sin(2\pi\frac{2}{7}),cos(2\pi\frac{2}{7})$
3: $sin(2\pi\frac{3}{7}),cos(2\pi\frac{3}{7})$
4: $sin(2\pi\frac{4}{7}),cos(2\pi\frac{4}{7})$
5: $sin(2\pi\frac{5}{7}),cos(2\pi\frac{5}{7})$
6: $sin(2\pi\frac{6}{7}),cos(2\pi\frac{6}{7})$

Where would I end up after completing those 7 1-meter steps? In the X direction, all of those sin's are greater than or equal to 0, so they will add up to a net displacement.

 

[.Scott]

Hmm
$=cos(\frac{1\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{3\pi}{7})+cos(\frac{4\pi}{7})+cos(\frac{5\pi}{7})+cos(\frac{6\pi}{7})+1\\ =cos(\frac{1\pi}{7})+cos(\frac{2\pi}{7})+cos(\frac{3\pi}{7})-cos(\frac{3\pi}{7})-cos(\frac{2\pi}{7})-cos(\frac{1\pi}{7})+1\\ \\ =1\\$

And then what?
I don't grasp the key what to do

Actually, I'm wrong.
I was doing 2pi, not pi.

I will edit my previous post.

 

[terryds]

Actually, I'm wrong.
I was doing 2pi, not pi.
So we are only going half way around a 14-point polygon, not all the way around a 7-point.
But the arithmetic will be the same.

I will edit my previous post.

0_0

I'm confused. I don't even know what's wrong hahaha
Please help...

 

[.Scott]

Forget the other posts:

Here is a walk around a 7-point polygon:

0: $sin(2\pi(\frac{0}{7})),cos(2\pi(\frac{0}{7}))$
1: $sin(2\pi(\frac{1}{7})),cos(2\pi(\frac{1}{7}))$
2: $sin(2\pi(\frac{2}{7})),cos(2\pi(\frac{2}{7}))$
3: $sin(2\pi(\frac{3}{7})),cos(2\pi(\frac{3}{7}))$
4: $sin(2\pi(\frac{4}{7})),cos(2\pi(\frac{4}{7}))$
5: $sin(2\pi(\frac{5}{7})),cos(2\pi(\frac{5}{7}))$
6: $sin(2\pi(\frac{6}{7})),cos(2\pi(\frac{6}{7}))$

Terms 1, 2, and 3 are the same as the terms you are adding.

We know that all of the cosine terms will add up to 0, because we are returning to the starting point in our walk.

Term #0 is 1.
Terms 1, 2, and 3 are identical to terms 6, 5, and 4, respectively.

So:
Term0 + Terms(1 to 3) + Terms(4 to 6) = 0
= 1 + Terms(1 to 3) + Terms(1 to 3)
So:
2*Terms(1 to 3) = -1
Terms(1-3) = -0.5

Sorry for the confusion.
If you need more explanation, I will monitor this thread.

 

[terryds]

Forget the other posts:

Here is a walk around a 7-point polygon:

0: $sin(2\pi(\frac{0}{7})),cos(2\pi(\frac{0}{7}))$
1: $sin(2\pi(\frac{1}{7})),cos(2\pi(\frac{1}{7}))$
2: $sin(2\pi(\frac{2}{7})),cos(2\pi(\frac{2}{7}))$
3: $sin(2\pi(\frac{3}{7})),cos(2\pi(\frac{3}{7}))$
4: $sin(2\pi(\frac{4}{7})),cos(2\pi(\frac{4}{7}))$
5: $sin(2\pi(\frac{5}{7})),cos(2\pi(\frac{5}{7}))$
6: $sin(2\pi(\frac{6}{7})),cos(2\pi(\frac{6}{7}))$

Terms 1, 2, and 3 are the same as the terms you are adding.

We know that all of the cosine terms will add up to 0, because we are returning to the starting point in our walk.

Term #0 is 1.
Terms 1, 2, and 3 are identical to terms 6, 5, and 4, respectively.

So:
Term0 + Terms(1 to 3) + Terms(4 to 6) = 0
= 1 + Terms(1 to 3) + Terms(1 to 3)
So:
2*Terms(1 to 3) = -1
Terms(1-3) = -0.5

Sorry for the confusion.
If you need more explanation, I will monitor this thread.

Wow.. It's an interesting approach
Anyway, I don't think all the cosine terms will add to zero, it will be $1+2 cos(2 \pi( \pi/7 + 2\pi/7 + 3\pi/7))$
But, all the sine terms will add to zero, right ?

Since the sum of Y direction (cosine terms) is not zero, how come we return to the starting point ??

 

[.Scott]

Wow.. It's an interesting approach
Anyway, I don't think all the cosine terms will add to zero, it will be $1+2 cos(2 \pi( \pi/7 + 2\pi/7 + 3\pi/7))$
But, all the sine terms will add to zero, right ?

Since the sum of Y direction (cosine terms) is not zero, how come we return to the starting point ??

The sum of the sine terms from 0 to 6 will be zero, because you come back to the same X.
The sum of the cosine terms from 0 to 6 will be zero, because you come back to the same Y.
So:
$cos(2\pi\frac{0}{7})+cos(2\pi\frac{1}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{3}{7})+cos(2\pi\frac{4}{7})+cos(2\pi\frac{5}{7})+cos(2\pi\frac{6}{7}) = 0$
$1+cos(2\pi\frac{1}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{3}{7})+cos(2\pi\frac{4}{7})+cos(2\pi\frac{5}{7})+cos(2\pi\frac{6}{7}) = 0$
Now, since:
$cos(x)=cos(-x)=cos(2\pi-x),$
$cos(2\pi\frac{4}{7})=cos(-2\pi\frac{4}{7})=cos(2\pi-2\pi\frac{4}{7})=cos(2\pi\frac{3}{7})$
$cos(2\pi\frac{5}{7})=cos(-2\pi\frac{5}{7})=cos(2\pi-2\pi\frac{5}{7})=cos(2\pi\frac{2}{7})$
$cos(2\pi\frac{6}{7})=cos(-2\pi\frac{6}{7})=cos(2\pi-2\pi\frac{6}{7})=cos(2\pi\frac{1}{7})$
So:
$1+cos(2\pi\frac{1}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{3}{7})+cos(2\pi\frac{3}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{1}{7}) = 0$
$1+2cos(2\pi\frac{1}{7})+2cos(2\pi\frac{2}{7})+2cos(2\pi\frac{3}{7}) = 0$
$2cos(2\pi\frac{1}{7})+2cos(2\pi\frac{2}{7})+2cos(2\pi\frac{3}{7}) = -1$
$cos(2\pi\frac{1}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{3}{7}) = -\frac{1}{2}$
And:
$cos(\pi\frac{2}{7})+cos(\pi\frac{4}{7})+cos(\pi\frac{6}{7}) = -\frac{1}{2}$
QED

In my early posts I was seeing 2pi where there was only pi. We are still going around a full circle (or 7-sided polygon). I just confused myself over exactly how. So ignore my first couple of posts.

 

[terryds]

The sum of the sine terms from 0 to 6 will be zero, because you come back to the same X.
The sum of the cosine terms from 0 to 6 will be zero, because you come back to the same Y.
So:
$cos(2\pi\frac{0}{7})+cos(2\pi\frac{1}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{3}{7})+cos(2\pi\frac{4}{7})+cos(2\pi\frac{5}{7})+cos(2\pi\frac{6}{7}) = 0$
$1+cos(2\pi\frac{1}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{3}{7})+cos(2\pi\frac{4}{7})+cos(2\pi\frac{5}{7})+cos(2\pi\frac{6}{7}) = 0$
Now, since:
$cos(x)=cos(-x)=cos(2\pi-x),$
$cos(2\pi\frac{4}{7})=cos(-2\pi\frac{4}{7})=cos(2\pi-2\pi\frac{4}{7})=cos(2\pi\frac{3}{7})$
$cos(2\pi\frac{5}{7})=cos(-2\pi\frac{5}{7})=cos(2\pi-2\pi\frac{5}{7})=cos(2\pi\frac{2}{7})$
$cos(2\pi\frac{6}{7})=cos(-2\pi\frac{6}{7})=cos(2\pi-2\pi\frac{6}{7})=cos(2\pi\frac{1}{7})$
So:
$1+cos(2\pi\frac{1}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{3}{7})+cos(2\pi\frac{3}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{1}{7}) = 0$
$1+2cos(2\pi\frac{1}{7})+2cos(2\pi\frac{2}{7})+2cos(2\pi\frac{3}{7}) = 0$
$2cos(2\pi\frac{1}{7})+2cos(2\pi\frac{2}{7})+2cos(2\pi\frac{3}{7}) = -1$
$cos(2\pi\frac{1}{7})+cos(2\pi\frac{2}{7})+cos(2\pi\frac{3}{7}) = -\frac{1}{2}$
And:
$cos(\pi\frac{2}{7})+cos(\pi\frac{4}{7})+cos(\pi\frac{6}{7}) = -\frac{1}{2}$
QED

In my early posts I was seeing 2pi where there was only pi. We are still going around a full circle (or 7-sided polygon). I just confused myself over exactly how. So ignore my first couple of posts.

Actually, I'm rather confused when you say

The sum of the sine terms from 0 to 6 will be zero, because you come back to the same X.
The sum of the cosine terms from 0 to 6 will be zero, because you come back to the same Y.

How to know that these vectors will add up and return to the starting point again? How to know that all of they will form a closed shape?
What if those vectors doesn't return to the starting point?
Any explanation or theorem about this?

 

[ehild]

Actually, I'm rather confused when you say

How to know that these vectors will add up and return to the starting point again? How to know that all of they will form a closed shape?
What if those vectors doesn't return to the starting point?
Any explanation or theorem about this?

There are seven vectors in the first figure, labelled by 1,2,,,7. They are all of the same magnitude and enclosing the same angle of 360/7 degrees with each other. When you add them, vector 2 to vector 1, then vector 3 to the sum, and so on, the endpoints of the consecutive sums are on a circle, and the central angle of the red triangles is also 360/7 degrees. At every addition, the whole angle at the centre increases by 360/7 degrees, and when add vector 7, it becomes 7*360/7= 360 degrees, the head of vector 7 returning to the tail of vector 1.

You can also consider the cosine terms to add as the real parts of the complex numbers e^{k(2π/7 i)} . The terms from k=0 to 6 make a geometric sequence with first terms z1=1 and quotient q= e^{(2π/7) i} . If you add the 7 terms, you get zero.