I know if $x=e^{\frac{2\pi i}{17}}$ then $x^{17}=1$ and $\Re(x)=\cos\left(\frac{2\pi}{17}\right)$.
But how do I form a polynomial which has root $\cos\left(\frac{2\pi}{17}\right)$.
I know you can consider de Moivre's theorem and expand the LHS using binomial theorem but that will take a long time.
Probably the best way is to just show that the sum of two algebraic numbers is algebraic. This is not obvious, but if you look at it just right it's much easier than it seems at first.
Regard $\Bbb C$ as a vector space over $\Bbb Q$. Any linear-algebra concepts below refer to $\Bbb Q$-linear subspaces of $\Bbb C$.
Lemma 0. The number $a\in \Bbb C$ is algebraic if and only if the span of $1,a,a^2\dots$ is finite-dimensional.
Proof: Easy exercise. QED.
Lemma 1. Suppose $A,B\subset\Bbb C$ are subspaces, and let $C$ be the span of the $xy$ with $x\in A$, $y\in B$. If $A$ is spanned by $a_1,\dots,a_n$ and $B$ is spanned by $b_1,\dots,b_m$ then $C$ is spanned by $a_jb_k$, $1\le j\le n$, $1\le k\le m$ (so in particular $C$ is finite dimensional).
Proof: Easy exercise. QED.
Theorem. If $a,b\in\Bbb C$ are algebraic then $a+b$ is algebraic.
Proof. Let $A$ be the span of the powers of $a$ and $B$ the span of the powers of $b$. Let $C$ be as in Lemma 1. Now there exist $n$ and $m$ such that $A$ is spanned by $1,a,\dots, a^n$ and $B$ is spanned by $1,b,\dots,b^m$. So Lemma 1 shows that $C$ is finite dimensional.
But every power of $a+b$ lies in $C$. So the span of the powers of $a+b$ is finite dimensional, and hence Lemma 0 shows that $a+b$ is algebraic. QED.
This shows that your number is algebraic since $\cos(t)=(e^{it}+e^{-it})/2$.
Edit. One could use the argument above to find $P$ with $P(a+b)=0$, given $p(a)=0$ and $q(b)=0$. Any power of $A$ can be written explicitly as a linear combination of $1,a,\dots a^n$, and similarly for $b$. So any $a^jb^k$ can be written explicitly as a linear combination of $a^jb^k$ with $0\le j\le n$ and $0\le k\le m$. Hence the same is true of any power of $a+b$. So write down the powers of $a+b$ as such linear combinations, one by one, and check the vectors of coefficients for linear dependence. Eventually a dependence relation appears, and that gives you $P$ with $P(a+b)=0$.
By adding equalities $$\cos(n+1)x=\cos nx\cos x-\sin nx\sin x\\ \cos(n-1)x=\cos nx\cos x+\sin nx\sin x$$ we get an equality $$\cos(n+1)x+\cos(n-1)x=2\cos nx\cos x$$ If we now define, by induction, Chebyshev polynomials $T_0(y)=1,T_1(y)=y,T_{n+1}(y)=2yT_n(y)-T_{n-1}(y)$ then it follows, by taking $y=\cos x$, that $$T_n(\cos x)=\cos nx$$ It follows that $\cos\frac{2\pi}{17}$ is a root of $T_{17}(x)-\cos 2\pi=T_{17}(x)-1$.
Chebyshev polynomials are a bit tedious to calculate by hand, but thanks to the recurrence relation this can be done in quite a short amount of time. You can draw a Pascal-triangle like table containing their coefficients, which would make it even faster.
The number $x=cos\left(\frac{2\pi}{17}\right)$ is a root of the polynomial $$\sum_{k=0}^{8} \binom{17}{2k+1}x^{2k+1}\cdot i^{16-2k}\cdot (1-x^2)^{8-k}=1$$.
Let $c=\cos\left(\frac{2\pi}{17}\right)$ and $s=\sin\left(\frac{2\pi}{17}\right)$.
Then
$ 1=\Re(1)=\Re ((c+s\, i)^{17})=c^{17}-136 c^{15} s^2+2380 c^{13} s^4-12376 c^{11} s^6+24310 c^9 s^8-19448 c^7 s^{10}+6188 c^5 s^{12}-680 c^3 s^{14}+17 c s^{16} $
Note that $s$ appears only with even powers. Now replace $s^2=1-c^2$ to get a polynomial in $c$.
Consider that
$$\sin{\left ( \frac{9 \pi}{17} \right )} = \sin{\left ( \frac{8 \pi}{17} \right )} $$
or, letting $y = \frac{\pi}{17}$,
$$3 \sin{3 y} - 4 \sin^3{3 y} = 2 \sin{4 y} \cos{4 y} $$
or
$$9 \sin{y} - 120 \sin^3{y} + 432 \sin^5{y} - 576 \sin^7{y} + 256 \sin^9{y} \\= 8 \sin{y} \cos{y} (2 \cos^2{y}-1) [2 (2 \cos^2{y}-1)^2-1] $$
Note that $\sin{y}$ cancels on both sides. Using $\sin^2{y}=1-\cos^2{y}$, we get an $8$th degree polynomial in $\cos{y}$, from which $\cos{2 y} = 2 \cos^2{y}-1$ may be determined.
