TITLE: Exact values of $\cos(2\pi/7)$ and $\sin(2\pi/7)$
QUESTION [20 upvotes]: What are the exact values of $\cos(2\pi/7)$ and $\sin(2\pi/7)$ and how do I work it out?
I know that $\cos(2\pi/7)$ and $\sin(2\pi/7)$ are the real and imaginary parts of $e^{2\pi i/7}$ but I am not sure if that helps me...

REPLY [3 votes]: case = {7},
first, set $\zeta^k_7$ as seventh roots of unity, using both Euler and De Moiver Formula,
$\zeta^{k}_{7}=\text{e}^{\frac{2ki\pi}{7}}=\cos\frac{2k\pi}{7} \pm i\sin\frac{2k\pi}{7}$
hence
$\zeta^{-k}_{7}=\text{e}^{\frac{2ki\pi}{7}}=\cos\frac{2k\pi}{7} \mp i\sin\frac{2k\pi}{7}$
hence
$$\zeta^{k}_{7}+\zeta^{-k}_{7}=2\cos\frac{2k\pi}{7}$$
$$\zeta^{k}_{7}-\zeta^{-k}_{7}=2i\sin\frac{2k\pi}{7}$$
$$\zeta^{2k}_{7}-2+\zeta^{-2k}_{7}=-4\sin\frac{2k\pi}{7}$$
$$-4\sin\frac{2k\pi}{7}=2\cos\frac{4k\pi}{7}-2$$
$$2i\sin\frac{2k\pi}{7}=\sqrt{2\cos\frac{4k\pi}{7}-2}$$
and some certain properties of $7^{th}$ roots of unity,
$$\zeta^{7}_{7}=1$$
$$\zeta^{k}_{7}=\zeta^{k\text{mod}7}_{7}$$
$$\zeta^{-k}_{7}=\zeta^{7-k}_{7}$$
and sum of all $n^{th}$ (where $n$ is more than $1$) roots of unity is $0$, so we can write
$$\zeta^{1}_{7}+\zeta^{2}_{7}+\zeta^{3}_{7}+\zeta^{4}_{7}+\zeta^{5}_{7}+\zeta^{6}_{7}+1=0$$
move 1
$$\zeta^{1}_{7}+\zeta^{2}_{7}+\zeta^{3}_{7}+\zeta^{4}_{7}+\zeta^{5}_{7}+\zeta^{6}_{7}=-1$$
hence
$$\zeta^{1}_{7}+\zeta^{2}_{7}+\zeta^{3}_{7}+\zeta^{-3}_{7}+\zeta^{-2}_{7}+\zeta^{-1}_{7}=-1$$
set
$$t_{k}=\zeta^{k}_{7}+\zeta^{-k}_{7}=2\cos\frac{2k\pi}{7}$$
this simplifies
$$t_1+t_2+t_3=-1$$
multiply $t_1$ with $t_2$ , remember $\zeta^{k}_{7}=\zeta^{k\text{mod}7}_{7}$
$$t_{1}=\left(\zeta^{1}_{7}+\zeta^{6}_{7}\right)$$
$$t_{2}=\left(\zeta^{2}_{7}+\zeta^{5}_{7}\right)$$
$$t_{1} t_{2}=\zeta^{3}_{7}+\zeta^{6}_{7}+\zeta^{1}_{7}+\zeta^{4}_{7}$$
$$t_{1} t_{2}=t_{1}+t_{3}$$
similarly
$$t_{1} t_{3}=t_{3}+t_{2}$$
$$t_{2} t_{3}=t_{2}+t_{1}$$
so
$$t_{1} t_{2}+t_{1} t_{3}+t_{2} t_{3}=-2$$
then, find $t_{1} t_{2} t_{3}$
$$t_{1} t_{2}=t_{1}+t_{3}$$
since $t_1+t_2+t_3=-1$
$$t_{1} t_{2}=-1-t_{2}$$
$$t_{1} t_{2} t_{3}=\left(-1-t_{2}\right)t_{3}$$
$$t_{1} t_{2} t_{3}=-t_{3}-t_{2}t_{3}$$
$$t_{1} t_{2} t_{3}=-t_{3}-t_{2}-t_{1}$$
$$t_{1} t_{2} t_{3}=1$$
then,
$$t_1 + t_2 + t_3 = -1$$
$$t_2 + t_3 = -1 -t_1$$
$$t_1 t_2 t_3 = 1$$
$$t_2 t_3 = \frac {1}{t_1} = t^{-1}_1$$
$$t_1 t_2 + t_1 t_3 + t_2 t_3 +2 = 0$$
$$t_1 ( t_2 + t_3 ) + t_2 t_3 +2 = 0$$
$$t_1 ( -1 -t_1 ) + t^{-1}_1 +2 = 0$$
$$-t^2_1 -t_1 + 2 + t^{-1}_1 = 0$$
$$t^3_1 + t^2_1 -2t_1 -1 = 0$$
this statement also applies for other $t_k$
$$t^3+t^2-2-1=0$$
back to the statement that we've set
$$t_{k}=2\cos\frac{2k\pi}{7}$$
the roots of the cubic polynomial
$$t^3+t^2-2-1=0$$
have the roots cosine, applying Lagrange Resolvent ( if you want a hard way ) or Tartaglia Formula,
$$2\cos\frac{2\pi}{7}=\frac{-1+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}$$
$$2\cos\frac{4\pi}{7}=\frac{-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}$$
$$2\cos\frac{6\pi}{7}=\frac{-1+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}$$
and, notice
$t_{k}=\zeta^{k}_{7}+\zeta^{-k}_{7}$
$t_{-k}=\zeta^{-k}_{7}+\zeta^{k}_{7}$
so
$t_{1}=t_{1}
t_{2}=t_{2}
t_{3}=t_{3}
t_{4}=t_{3}
t_{5}=t_{2}
t_{6}=t_{1}$
$$\zeta^{2k}_{7}-2+\zeta^{-2k}_{7}=-4\sin\frac{2k\pi}{7}$$
so
$$t_{2k}-2=-4\sin\frac{2k\pi}{7}$$
$$\sqrt{t_{2k}-2}=2i\sin\frac{2k\pi}{7}$$
$$\sqrt{t_{2}-2}=2i\sin\frac{2\pi}{7}$$
$$\sqrt{t_{3}-2}=2i\sin\frac{4\pi}{7}$$
$$\sqrt{t_{1}-2}=2i\sin\frac{6\pi}{7}$$
substitute each $t_{k}$
$$\sqrt{t_{2}-2}=\sqrt{\frac{-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}-2}$$
$$\sqrt{t_{2}-2}=\sqrt{\frac{-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{3}}$$
$$\sqrt{t_{2}-2}=\sqrt{\frac{3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}{9}}$$
$$\sqrt{t_{2}-2}=\frac{\sqrt{3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$
$$2i\sin\frac{2\pi}{7}=\frac{\sqrt{3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$
so
$$2\sin\frac{2\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$
$$2\sin\frac{4\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$
$$2\sin\frac{6\pi}{7}=\frac{\sqrt{-3\left(-7+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{3}$$
now, we are done with
$$\cos\frac{2\pi}{7}=\frac{-1+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{6}$$
$$\cos\frac{4\pi}{7}=\frac{-1+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{6}$$
$$\cos\frac{6\pi}{7}=\frac{-1+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}}{6}$$
$$\sin\frac{2\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{6}$$
$$\sin\frac{4\pi}{7}=\frac{\sqrt{-3\left(-7+\frac{-1+\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\frac{-1-\sqrt{-3}}{2}\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{6}$$
$$\sin\frac{6\pi}{7}=\frac{\sqrt{-3\left(-7+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}+\sqrt[3]{\frac{7}{2}\left(1+3\sqrt{-3}\right)}\right)}}{6}$$