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Let $\zeta=\cos(\frac{2\pi}{16})+i\sin(\frac{2\pi}{16})$ be a 16th root of unity, so that it is a primitive root of unity. I need to explicitly express this number in terms of radicals: $a+ib$, where $a$ and $b$ are real numbers.

My approach:

We know that $\zeta^1+\zeta^2+...+\zeta^{16} = 0$. We can now compute $\zeta$ in even powers to see that they all cancel out and so we have:

$$\zeta^2+...+\zeta^{16}=0$$ Dividing by $\zeta^2$ we get:

$$ 1+\frac{1}{\zeta^2}+\zeta^2+\zeta^4+\zeta^6+\zeta^8+...+\zeta^{14}=0 $$ which we can rewrite as $$ 1+ \frac{1}{\zeta^2}+(\zeta^2+\zeta^4+\zeta^6)+\zeta^6(\zeta^2+\zeta^4+\zeta^6)=1+ \frac{1}{\zeta^2}+(1+\zeta^6)(\zeta^2+\zeta^4+\zeta^6)$$

Well, I don't think this is useful anyway. Would appreciate some hints. I'm not sure how to approach this problem in a right way.

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Given that $\cos \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$, and that $\cos (\frac{a}{2})=\pm \sqrt{\frac{1+\cos a}{2}}$ and $\sin (\frac{a}{2})=\pm \sqrt{\frac{1-\cos a}{2}}$, you find that $\cos \frac{\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$ and $\sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$. That is,

$$\cos(\frac{2\pi}{16})+i\sin(\frac{2\pi}{16})=\frac{\sqrt{2+\sqrt{2}}}{2}+i\cdot\frac{\sqrt{2-\sqrt{2}}}{2}$$

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To express $z$ as $a + ib$ where $a$ and $b$ are radicals, you'll just need to find the value of $\cos {2π\over 16}$ and $\sin {2π\over 16}$. The fact that $2π\over 16$ lies in the first quadrant makes the work easier. You can use the identity:

$$ 2\cos x = \sqrt{2 + 2\cos 2x}. $$

Repetitive application of this identity for $x = {2π\over 16} = {π\over 8}$ gets you to:

$$ \cos{π\over 8} = \frac12 \sqrt{2 + 2\cos {π\over 4}} $$

Simplification is up to you. For $\sin {π\over 8}$, use the identity:

$$ 2\sin x = \sqrt{2 - 2\cos 2x}. $$

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