Minimal polynomials of $\sin(\pi/8)$ and $\cos(\pi/9)$

Solution 1:

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 $$ \fbox{$x^4 - x^2 + \frac{1}{8}$}. $$

Solution 2:

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 $\mathbb{Q}$.

Solution 3:

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:

$$16 \cos^4{\frac{\pi}{9}}-8 \cos^3{\frac{\pi}{9}}-12 \cos^2{\frac{\pi}{9}}+4 \cos{\frac{\pi}{9}}+1=0 $$