Can we use Vieta's formula in solving Trigonometric equations? **The value of $$\sec\frac{\pi}{11}-\sec\frac{2\pi}{11}+\sec\frac{3\pi}{11}-\sec\frac{4\pi}{11}+\sec\frac{5\pi}{11}$$
is ...
My Approach
I used the fact that $$\sec (\pi-x)=-\sec x$$ to simplify the equation to $$\sec\frac{\pi}{11}+\sec\frac{3\pi}{11}+\sec\frac{5\pi}{11}+\sec\frac{7\pi}{11}+\sec\frac{9\pi}{11}$$
Now I tried to devise an equation whose roots are $$\sec\frac{\pi}{11}, \sec\frac{3\pi}{11}, \sec\frac{5\pi}{11}, \sec\frac{7\pi}{11}, \sec\frac{9\pi}{11}$$
Afterwards, I found that the equation $$\cos \frac{11x}{2}=0 $$ satisfy the condition. But the equation has infinite number of roots, so my plan to use Vieta's formula to calculate the required sum did not work.
Please suggest how to proceed in this problem or share any other method.
 A: Observe that $\cos(2n+1)\pi=-1$ for any integer $n$
If $11x=(2n+1)\pi,\cos6x=-\cos5x$
With $\cos x=c,$
$$\cos6x=2\cos^23x-1=2(4c^3-3c)^2-1=?$$
Using Prosthaphaeresis Formulas,
$$\cos5x+\cos x=2\cos3x\cos2x=2(4c^3-3c)(2c^2-1)$$  to find
$$0=\cos6x+\cos5x=32c^6+16c^5-48c^4-20c^3+18c^2+5c-1$$
So, the roots of $$32c^6+16c^5-48c^4-20c^3+18c^2+5c-1=0$$ are $c_n=\cos\dfrac{(2n+1)\pi}{11},0\le n\le 5$
But $c_5=-1$
So, the roots of $$32c^5-16c^4-32c^3+12c^2+6c-1=0$$ are $c_n=\cos\dfrac{(2n+1)\pi}{11},0\le n\le 4$
Set $\dfrac1c=s$ to find the roots of $$s^5-6s^4+\cdots-32=0$$ to be  $s_n=\sec\dfrac{(2n+1)\pi}{11},0\le n\le 4$
Can you take it from here?
A: Let $\alpha = \dfrac{(2n+1)\pi}{11}$, which is the form of angles we are interested. Then $6\alpha = (2n+1)\pi - 5\alpha$ which implies $\cos 6\alpha +\cos 5\alpha = 0$. Now, expressing this in terms of $\cos \alpha$, we observe
$$32\cos^6\alpha + 16\cos^5\alpha- 48\cos^4\alpha - 20\cos^3\alpha + 18\cos^2\alpha + 5\cos\alpha - 1 = 0$$
From this, we conclude that the equation
$$32x^6 + 16x^5-48x^4-20x^3+18x^2+5x-1 = 0$$
has the roots of exactly same angles that you want (with $\cos$) and $1$, so divide by $(x-1)$ and substitute $1/x$ to get $\sec$ and use Vieta's formula.
Note: We partitioned $11\alpha$ to $5\alpha$ and $6\alpha$ since we needed a polynomial with (possibly exactly) $5$ roots (and $\lfloor 11 / 2 \rfloor = 5$).
