Prove that cos (2π/13)+ cos (6π/13) + cos (8π/13)= (sqrt(13)-1)/4

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- Jun 8th 2011, 07:45 AMhooketrigonometry proof
Prove that cos (2π/13)+ cos (6π/13) + cos (8π/13)= (sqrt(13)-1)/4

- Jun 8th 2011, 01:20 PMOpalg
Outline proof: For , let denote the angle . The points are equally spaced around the unit circle, so their centre of mass is at the origin, and it follows that

Since and , it follows that

Next,

.

But (with similar results for and ). Also, (with similar results for the other two products and ).

Putting all that together, you find that

, and

Now let and It follows from the above calculations that and Eliminate T between those equations to get , a quadratic with solutions Now all that remains for you to do is to figure out why the square root must have the plus sign. - Jun 9th 2011, 08:00 AMhooke
Thank you so much Opalg, as for the last piece of parcel you left for me, do i consider the quadrants cos A1, cos A2 and cos A3 belong to? I don't see a way of doing this without a calculator.

- Jun 9th 2011, 08:53 AMOpalg
The quantities S and T are the two roots of the quadratic equation. It should be clear from a diagram of the 13 points on the unit circle that the three cosines making up the sum T are smaller than their counterparts in S. Specifically, , and . Therefore T<S and so S must correspond to the larger root of the quadratic.

- Jun 10th 2011, 08:05 PMhooke
Thanks Opalg.