Is there any particular method to deal with sine, cosine and other ratios of pi/7, pi/13, pi/17 etc?

There are some methods which use in http://momentum21.forumeiros.com/t6-prove-soma-de-...

but these are so involved that I do not see a pattern here

I set

x = cos2A + cos6A + cos8A … 1

by seeing that sqrt13 on right

and square both sides of (1) to get

x^2 = cos^2 2A + cos^2 6A + cos^2 8A + 2 cos 2A cos 6A + 2 cos 2A cos 8A + 2 cos 6A cos 8A

multiply by 2 to get

2 x^2 = 2 cos^2 2A + 2 cos^2 6A + 2 cos^2 8A + 2(2 cos 2A cos 6A + 2 cos 2A cos 8A + 2 cos 6A cos 8A)

= cos 4A + 1 + cos 12 A + 1 + cos 16 A + 1 + 2( cos 8A + cos 4 A + cos 10 A + cos 6A + cos 14 A + cos 2 A)

= 3 + cos 4A + cos 12 A + cos 16 A + 2 ( cos 8A + cos 6A + cos 2A + cos 4A + cos 10 A + cos 14A)

Now cos 16 A = cos 10 A as 26 A = 2pi

Cos 14 A = cos 12 A as 26 A = 2pi

So we continue

= 3 + cos 4A + cos 12 A + cos 10 A + 2(x+ cos 4A + cos 10 A + cos 12 A)

= 3 + 2x + 3 (cos 4A + cos 12 A+ cos 10 A)

Now cos 2A + cos 4A + cos 6A + cos 8A + cos 10 A + cos 12A = -1/2

So cos 4A + cos 12 A+ cos 10 A = (- 1/2-x)

So 2x^2 = 3 + 2x + 3(-1/2- x)

Or 4x^2 = 6 + 4x -3 – 6x

Or 4x^2 + 2x -3 = 0

This has one positive solution (sqrt(13)-1)/ 4 and one negative solution

As cos2A + cos6A + cos8A = cos 2A + cos 6A – cos 5A and cos6 A > 0 and cos 2A > cos 5A so this is > 0

So this is (sqrt(13)-1)/ 4

Note: I squared as sqrt(13) was in RHS and I was expected to get a quadratic equation in x. This was guaranted. Cannot be sure of the method if the expression was to be evaluated.

2) we could have put x = (sqrt(13)-1)/ 4 and form an equation

And the 4x +1 = sqrt(13)

Square both sides to get 16x^2 + 8x + 1 = 13

Or 16x^2 + 8x- 12 = 0

Or 4x^2 + 2x -3 = 0

and see that cos2A + cos6A + cos8A satisfies the above equation

3) I do not see pattern, each to be treated on merit

the link mentioned is not accessible. It may kindly be provided