<s>[INST] 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.
 [/INST] 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, $s_2= \frac{-1-\sqrt{17}}4$.
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