Find $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ exactly by hand Show the exact value of $\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}$ is $\frac{-(1-\sqrt{17})}{4}$ (no calculator).
By and large these things require considerations of Euler's formula so I rewrote the expression as $$\frac{1}{2}(a+a^2+a^4+a^8)+\frac{1}{2}(a^{-1}+a^{-2}+a^{-4}+a^{-8})$$ where $a=e^\frac{2\pi i}{17}$. This looks really nice like you could sum it or something but it really just doesn't play nice when you try to. It is not geometric, so pretty much the only way forward is to find a function whose taylor series starts with $f=x+x^2+x^4+x^8+...$ or something workable to look like that, but this is not obvious at all either, and the next problem would be whether that function can be used to evaluate $f(e^\frac{2\pi i}{17})$ easily which at this point is hopelessly improbable. By now I have no where else to turn, it seems likely that the solution doesn't involve complex numbers.
 A: For anyone who stumbles across this problem, it is an utter waste of your time and you should go do something else. Nothing here can be motivated with honesty.

*

*Call the expression $\xi$ and expand $\xi ^2$ as a binomial

*Use Prosthaphaeresis Formulas and Cosine double angle and $\cos{(2\pi - x)}=\cos{x}$ and $\sum^{16}_{n=1} \cos{\frac{2n\pi}{17}}=0$ to get $\xi^2=2+\frac{\xi}{2}-\xi$

*Solve quadratic.

*Notice all but $\cos\frac{16\pi}{17}$ are positive in original expression but $\frac{2\pi}{17}<\frac{4\pi}{17}<\frac{\pi}{3}$ so certainly $\cos{\frac{2\pi}{17}+\cos\frac{4\pi}{17}}>1>-\cos{\frac{8\pi}{17}}$ and only one of the quadratic roots is positive.

A: Let
$$s_1=\cos\frac{2\pi}{17} + \cos\frac{4\pi}{17} + \cos\frac{8\pi}{17} + \cos\frac{16\pi}{17}
$$
$$s_2=\cos\frac{6\pi}{17} + \cos\frac{10\pi}{17} + \cos\frac{12\pi}{17} + \cos\frac{14\pi}{17}
$$
and note
$$s_1s_2=2(s_1+s_2) =2\cdot (-\frac12)=-1
$$
Thus, $s_1$ and $s_2$ satisfy $s^2+\frac12 s-1=0$, which yields $s_1= \frac{-1+\sqrt{17}}4$ and, as a by-product, $s_2= \frac{-1-\sqrt{17}}4$.
A: 1875 book by Reuschle. The technique was initiated by Gauss, some 30 years before Galois theory.
Indeed, on page 249 in Cox, Galois Theory, we find
$$ \sum_{w=1}^{16} \; (w|17) \; \zeta_{17}^w \; = \sqrt{17}.  $$
Letting $\eta_0$ refer to the quantity of interest, calling the sum of the others $\eta_1$ as in Reuschle below, we see $\eta_0 + \eta_1 +1=0, $ while $\eta_0 - \eta_1 = \sqrt{17}.$  So  $2\eta_0 +1 = \sqrt{17}.$
For any prime $p \equiv 1 \pmod 4$ we have
$$ \sum_{w=1}^{p-1} \; (w|p) \; \zeta_{p}^w = \sqrt{p}.  $$
while prime $q \equiv 3 \pmod 4$ gives
$$ \sum_{w=1}^{q-1} \; (w|q) \; \zeta_{q}^w = i\sqrt{q}.  $$
Cox refers to Ireland and Rosen for quadratic Gauss sums, chapter 6. The bits just preceding are I+R Theorem 1, on page 75.
Anyway, double your displayed quantity ( that is, drop the $1/2$), call it $\eta.$  Write out $\eta^2$ in terms of powers of $a,$ taking every exponent from $-8$ to $8$ as you began.   Add up $\eta^2 + \eta$ and reduce.
$$  \eta =   a + a^2 + a^4 + a^8 + a^9 + a^{13}   + a^{15}  + a^{16}   $$
If you write it with exponents from $0$ to $16,$  the result of $\eta^2 + \eta$   comes out to ( exponents $\pmod {17}$)
$$ \eta^2 + \eta = 4a^{16} + 4a^{15} + 4a^{14} + 4a^{13} + 4a^{12} + 4a^{11} + 4a^{10} + 4a^9 + 4a^8 + 4a^7 + 4a^6 + 4a^5 + 4a^4 + 4a^3 + 4a^2 + 4a + 8
$$
But this is ( we split the final $8$ into $4+4$)
$$ 4 \left(a^{16} + a^{15} + a^{14} + a^{13} + a^{12} + a^{11} + a^{10} + a^9 + a^8 + a^7 + a^6 + a^5 + a^4 + a^3 + a^2 + a + 1 \right) +4 = 4
$$
because
$$ a^{16} + a^{15} + a^{14} + a^{13} + a^{12} + a^{11} + a^{10} + a^9 + a^8 + a^7 + a^6 + a^5 + a^4 + a^3 + a^2 + a + 1 = 0$$
All together, as Reuschle tells us, $$  \eta^2 + \eta - 4 =0 $$

