SOLUTION: Prove that: cos(2π/7)+cos(4π/7)+cos(8π/7)= - 1/2

Imagine the unit circle on a coordinate plane with the center O at the origin of the coordinate system.
Imagine that the regular 7-sided polygon ABCDEFG with the center at the point O 
is inscribed into this circle in a way that one its vertex is located at the point A = (1,0).

Then the other 6 vertices are the points

#2, B: (cos( 2pi/7), sin( 2pi/7))
#3, C: (cos( 4pi/7), sin( 4pi/7))
#4, D: (cos( 6pi/7), sin( 6pi/7))
#5, E: (cos( 8pi/7), sin( 8pi/7))
#6, F: (cos(10pi/7), sin(10pi/7))
#7, G: (cos(12pi/7), sin(12pi/7))

listed in the anticlockwise order.


1.  First of all, it is clear that the x-coordinates of the points D and E are the same: cos( 6pi/7) = cos( 8pi/7).

    So, we can consider the equal sum  cos(2pi/7)+cos(4pi/7)+cos(6pi/7)  instead  of  cos(2pi/7)+cos(4pi/7)+cos(8pi/7).


2.  From symmetry, it is clear that the sum of vectors  OA + OB + OC + OD + OE + OF + OG  is equal to zero:

    OA + OB + OC + OD + OE + OF + OG = 0.

    It is so obvious that I will omit the proof of this fact.

    It gives us the equality

    
    1 + cos(2pi/7)+cos(4pi/7)+cos(6pi/7)+cos(8pi/7)+cos(10pi/7)+cos(12pi/7) = 0.     (1)


    Notice that "1", the first addend, is x-coordinate of the point A.


3.  From symmetry, it is also clear that the x-coordinates of the pairs of the points B and G; C and F; D and E are the same:

    cos(2pi/7) = cos(12pi/7);  cos(4pi/7) = cos(10pi/7);  cos(6pi/7) = cos(8pi/7).


    Therefore, we can re-write the sum (1) in the form

    2*(cos(2pi/7)+cos(4pi/7)+cos(6pi/7)) = -1.


4.  Now all that we need to do is to divide both sides of (2) by 2, and we will get 

       cos(2pi/7)+cos(4pi/7)+cos(6pi/7) = ,

    which is the same as 

       cos(2pi/7)+cos(4pi/7)+cos(8pi/7) = , 

    as we noted above.


5.  Proved.