2015/112) show that $\cos2A + \cos6A + \cos8A = \dfrac{\sqrt{13)}-1}{4}$ where $A = \dfrac{pi}{13}$

Let
$x = \cos2A + \cos6A + \cos8A \cdots(1)$
by seeing that $\sqrt{13}$ on right
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 $26 A = 2\pi$
$\cos 14 A = \cos 12 A$ as $26 A = 2\pi$
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 $\cos 2A + \cos 4A + \cos 6A + \cos 8A + \cos 10 A + \cos 12A = \dfrac{-1}{2}$

So $\cos 4A + \cos 12 A+ \cos 10 A = (\dfrac{- 1}{2}-x)$

So $2x^2 = 3 + 2x + 3(\dfrac{-1}{2}- x)$
Or $4x^2 = 6 + 4x -3 – 6x$
Or $4x^2 + 2x -3 = 0$

This has one positive solution $\dfrac{\sqrt{13)}-1}{4}$ 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 $\dfrac{\sqrt{13)}-1}{4}$


I have solved at http://in.answers.yahoo.com/question/index;_ylt=ApwbfmdwbadqGq5DHY.KZsiRHQx.;_ylv=3?qid=20120607073236AAFCjFo