Is there an "easy" way to find the minimal polynomials of $\sin(\pi/8)$ and $\cos(\pi/9)$ without the help of any computer program?
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.
On
$$\frac{1}{\sqrt{2}}=\cos(\pi/4)=1-2\sin^2(\pi/8)$$ or we have $$1/2=1-4\sin^2(\pi/8)+4\sin^4(\pi/8),$$ so the minimal polynomial is $8x^4-8x^2+1$.
Similarly for $\cos(\pi/9)$ use the fact that $\cos(\pi/3)=1/2$ and the formula for $\cos(3\theta)$.
On
Theoretically all fractional trig expressing sin( pi* q) where q is a fraction can be expressed in an algebraic form. This is because if q = a/b where a and b are integers. We can recursively determine what sin(az) and sin(bz) equals. This enables us to determine what sin(ax) and sin(x/b) equals. So sin(pi*q) can be expanded to an Algebraic expression of sin(pi/b) which too can be expanded to an algebraic expression over sin(pi) = 0.
To answer your question though, you don't need that much work. Sin(f*pi/8) can be expanded into terms involving sin(f*pi/4) which can be then reduced and solved for x to find minimal polynomial.
On
For $\cos{(\pi/9)}$, use the fact that
$$\sin{\frac{4 \pi}{9}} = \sin{\frac{5 \pi}{9}}$$
Use the formula
$$\sin{5 x} = \sin^5{x} - 10 \cos^2{x} \sin^3{x} + 5 \cos^4{x} \sin{x}$$
the double-angle formula for sine (twice applied), and some algebra to obtain the following 4th degree equation:
$$16 \cos^4{\frac{\pi}{9}}-8 \cos^3{\frac{\pi}{9}}-12 \cos^2{\frac{\pi}{9}}+4 \cos{\frac{\pi}{9}}+1=0 $$
We know that $\zeta = \cos(\pi / 8) + \mathbf{i} \sin(\pi / 8)$ is a primitive eighth root of unity. Therefore its minimal polynomial is
$$ f_\zeta(x) = \frac{x^8 - 1}{x^4 - 1} = x^4 + 1 $$
However, $\cos(\pi / 8) = (1/2)(\zeta + \bar{\zeta}) = (1/2)(\zeta + 1/\zeta)$.
From this information, it's easy to show that the minimal polynomial of $\cos(\pi / 8)$ is quadratic, and using the fact that $\zeta^2 + \zeta^{-2} = 0$, we can find a linear combination of the powers of $\cos(\pi / 8)$ that sum to zero without much trouble.
Finding the minimal polynomial of $\sin(\pi/8)$ over $\mathbb{Q}(i)$ can be done the same way. Once you have that, you can use it to determine the minimal polynomial over $\mathbb{Q}$.