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Prove that:
cos(2pi/7)+cos(4pi/7)+cos(8pi/7)= - 1/2
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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.
Solved.