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.
 A: 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
$$
\fbox{$x^4 - x^2 + \frac{1}{8}$}.
$$
A: 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 $\mathbb{Q}$.
A: 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:
$$16 \cos^4{\frac{\pi}{9}}-8 \cos^3{\frac{\pi}{9}}-12 \cos^2{\frac{\pi}{9}}+4 \cos{\frac{\pi}{9}}+1=0 $$
A: $\frac{1}{\sqrt{2}}=\cos(\pi/4)=1-2\sin^2(\sin\pi/8)$ 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 $\cos(3\theta)$
A: 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.
