Minimal polynomials of $\sin(\pi/8)$ and $\cos(\pi/9)$ Is there an "easy" way to find the minimal polynomials of $\sin(\pi/8)$ and $\cos(\pi/9)$ without the help of any computer programme?
If I knew $\sin(\pi/8)=\frac{\sqrt{2-\sqrt2}}{2}$ then it would obviously be easy to find it, but how would i evaluate $\sin(\pi/8)$ in the first place?
And $\cos(\pi/9)$ does not seem to have a nice alternate form. So I see no point to start.

This is an old question, but it doesn't seem to have a correct answer with full details.
Let $\omega = e^{i\pi/9}$ and $\zeta = e^{i\pi/8}$, so $\cos(\pi/9) = \mathrm{Re}(\omega)$ and $\sin(\pi/8) = \mathrm{Im}(\zeta)$, and note that $\omega$ is a primitive 18th root of unity and $\zeta$ is a primitive 16th root of unity.  The conjugates of $\cos(\pi/9)$ must be the real parts of the other primitive 18th roots of unity, namely $\cos(5\pi/9)$ and $\cos(7\pi/9)$, and therefore $\cos(\pi/9)$ has degree three.  Similarly $\sin(\pi/8)$ has degree $4$, with conjugates $-{\sin(\pi/8)}$ and $\pm{\sin(3\pi/8)}$.
To find the minimum polynomial for $\cos(\pi/9)$, we start by finding the minimum polynomial for $\omega=e^{i\pi/9}$.  The roots of this polynomial must be the primitive 18th roots of unity, and with a little thought we can see that the desired cyclotomic polynomial is
$$
\frac{(x^{18}-1)(x^3-1)}{(x^9-1)(x^6-1)} \;=\; \frac{x^9+1}{x^3+1} \;=\; x^6 - x^3 + 1.
$$
Thus $\omega^6 - \omega^3 + 1 = 0$.  Dividing through by $\omega^3$ gives the relation
$$
\omega^3 + \omega^{-3} - 1 \;=\; 0.
$$
Let $\alpha = 2\cos(\pi/9)$.  Then
$$
\alpha \;=\; \omega + \omega^{-1}
\qquad\text{and}\qquad
\alpha^3 \;=\; \omega^3 + 3\omega + 3\omega^{-1} + \omega^{-3}
$$
so
$$
\alpha^3 - 3\alpha - 1 \;=\; 0.
$$
Then the minimum polynomial for $\cos(\pi/9)$ is $(2x)^3 - 3(2x) - 1$, or equivalently
$$
\fbox{$x^3 \,-\, \frac{3}{4} x  \,-\, \frac{1}{8}$}.
$$
We can do something similar for $\sin(\pi/8)$.  First, since $\zeta = e^{i\pi/8}$ is a primitive 16th root of unity, the minimum polynomial for $\zeta$ is
$$
\frac{x^{16}-1}{x^8-1} \;=\; x^8 + 1.
$$
Thus $\zeta^8 + 1 = 0$.  Dividing through by $\zeta^4$ gives the equation
$$
\zeta^4 + \zeta^{-4} \;=\; 0.
$$
Let $\beta = 2i\sin(\pi/8)$.  Then $\beta = \zeta - \zeta^{-1}$, so
$$
\beta^2 \;=\; \zeta^2 - 2 + \zeta^{-2}
\qquad\text{and}\qquad
\beta^4 \;=\; \zeta^4 - 4\zeta^2 + 6 - 4\zeta^{-2} + \zeta^{-4}.
$$
Then
$$
\beta^4 + 4\beta^2 + 2 \;=\; 0
$$
so the minimum polynomial for $\sin(\pi/8)$ is $(2ix)^4 + 4(2ix)^2 + 2$, or equivalently
$$
\fbox{$x^4 - x^2 + \frac{1}{8}$}.
$$
