$\sqrt7=?$由有理度数的三角函数表示

hbghlyj

哪些无理数可由有理度数的三角函数(可以使用多次不同角度)的有理函数表示出？
$\sqrt2=2\cos45°$
$\sqrt{2+\sqrt2}=2\cos22.5°$
$\underbrace{\sqrt{2+\cdots\sqrt{2+\sqrt{2+\sqrt2}}}}_{n重}=2\cos\left(\frac{45}{2^n}\right)°$
$\sqrt3=2\sin60°$
$\sqrt5=4\sin18°+1$
$\sqrt7=?$
$\sqrt17=-3+4\cos\left(\frac{360}{17}\right)°-152\cos^2\left(\frac{360}{17}\right)°+672\cos^4\left(\frac{360}{17}\right)°-1024\cos^6\left(\frac{360}{17}\right)°+512\cos^8\left(\frac{360}{17}\right)°$(参见这贴)
$\sqrt[3]2=?$

kuing

`\sqrt7` 的，昨天这帖正好用上，帖中我得出 `u^3-8u^2+16u-64/7=0` 的三根为 `u_k=\csc^2(k\pi/7)`, `k=1`, `2`, `3`，于是 `u_1u_2u_3=64/7`，即得 `\sqrt7=8\sin(\pi/7)\sin(2\pi/7)\sin(3\pi/7)`。

kuing

回复 2# kuing

呃，其实用 kuing.cjhb.site/forum.php?mod=viewthread&tid=2671 里的结论就好了

kuing

回复 2# kuing

哦，这帖 kuing.cjhb.site/forum.php?mod=viewthread&tid=3936 有
\[\sqrt7=2\left(\sin\frac{2\pi}7+\sin\frac{3\pi}7-\sin\frac\pi7\right)=\cot\frac\pi{14}-4\sin\frac\pi7.\]
还有这帖 kuing.cjhb.site/forum.php?mod=viewthread&tid=4795 也有很多东西，总之有大把这类恒等式……

hbghlyj

再考虑一下2的立方根
用三次单位根得到
$(\frac{-1+i\sqrt3}2)^{\frac13}+(\frac{-1-i\sqrt3}2)^{\frac13}=2\cos{\frac{\pi}9}$
$(\frac{-1+i\sqrt3}2)^{\frac13}-(\frac{-1-i\sqrt3}2)^{\frac13}=2i\sin{\frac{\pi}9}$

hbghlyj

hbghlyj 发表于 2019-12-27 15:27
$\sqrt[3]2=?$

顶一下

Czhang271828

hbghlyj 发表于 2023-4-29 19:40
$\sqrt[3]2=?$

不可能, 如果 $\sqrt[3]{2}$ 可以表示为有限个 $\{\sin p_i\pi, \cos q_i\pi\mid p_i,q_i\in \mathbb Q\}$ 的形式, 则取所有 $\{p_i,q_i\}$ 分母的最小公倍数 $n\in \mathbb N_+$, 得 $\sqrt[3]2\in \mathbb Q(e^{2\pi i/n},i)\subseteq \mathbb Q(e^{2\pi i/4n})$.

由于 $\mathbb Q(e^{2\pi i/n})$ 的 Galois 群总是交换群, $\mathbb Q(\sqrt[3]2)$ 的 Galois 群是 $S_3$, 从而 $\sqrt[3]2\notin \mathbb Q(e^{2\pi i/4n})$, 矛盾.

其实充要条件也做出来了, 无理数 $x$ 可被三角表示, 当且仅当 $\mathbb Q(x)$ 的 Galois 群是交换群.

Czhang271828

$(0^\circ,90^\circ)$ 的三角函数表, 原文链接.

#### 1°

$$
\begin{aligned}
\sin 1°&=-\frac{1}{4} (-1)^{1/4} \left(A_1 A_3(3) \left(2 B_2-i \sqrt{2} \sqrt[4]{5} B_4(2)\right)-A_3 A_1(3) \left(2 i B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\right)\\
\cos 1°&=+\frac{1}{4} (-1)^{3/4} \left(A_1 A_3(3) \left(2 B_2+i \sqrt{2} \sqrt[4]{5} B_4(2)\right)+A_3 A_1(3) \left(2 i B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\right)\\
\end{aligned}
$$



#### 2°

$$
\begin{aligned}
\sin 2°&=\frac{1}{4} \left(2 B_4 \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(-A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_2(2)\right)\\
\cos 2°&=\frac{1}{4} i \left(-2 B_4 \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_2(2)\right)\\
\end{aligned}
$$



#### 3°

$$
\begin{aligned}
\sin 3°&=-\frac{A_2 B_4}{\sqrt{2}}+\frac{1}{2} i \sqrt[4]{5} A_4 B_2(2)\\
\cos 3°&=\frac{A_4 B_4}{\sqrt{2}}+\frac{1}{2} i \sqrt[4]{5} A_2 B_2(2)\\
\end{aligned}
$$



#### 4°

$$
\begin{aligned}
\sin 4°&=\frac{1}{2} i B_2 \left(A_1(3)-A_3(3)\right)-\frac{\sqrt[4]{5} \left(A_1(3)+A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\cos 4°&=-\frac{1}{2} B_2 \left(A_1(3)+A_3(3)\right)-\frac{i \sqrt[4]{5} \left(A_1(3)-A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 5°

$$
\begin{aligned}
\sin 5°&=\frac{1}{2} (-1)^{1/4} \left(A_1(3)-i A_3(3)\right)\\
\cos 5°&=\frac{1}{2} (-1)^{1/4} \left(-i A_1(3)+A_3(3)\right)\\
\end{aligned}
$$



#### 6°

$$
\begin{aligned}
\sin 6°&=\frac{1}{4} \left(2 B_2+\sqrt[4]{5} \sqrt{6} B_4(2)\right)\\
\cos 6°&=\frac{1}{4} \left(-2 \sqrt{3} B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\\
\end{aligned}
$$



#### 7°

$$
\begin{aligned}
\sin 7°&=\frac{(-1+i) A_3\left(\frac{3}{4}\right) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1+i) A_1\left(\frac{3}{4}\right) \left(2 B_4+\sqrt{2} \sqrt[4]{5}
B_2(2)\right)}{4 \sqrt{2}}\\
\cos 7°&=\frac{(1+i) A_3\left(\frac{3}{4}\right) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1-i) A_1\left(\frac{3}{4}\right) \left(2 B_4+\sqrt{2} \sqrt[4]{5}
B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 8°

$$
\begin{aligned}
\sin 8°&=\frac{1}{4} i \left(-2 B_4 \left(-A_3 A_1(3)+A_1 A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_3 A_1(3)+A_1 A_3(3)\right) B_2(2)\right)\\
\cos 8°&=\frac{1}{2} B_4 \left(A_3 A_1(3)+A_1 A_3(3)\right)-\frac{\sqrt[4]{5} \left(-A_3 A_1(3)+A_1 A_3(3)\right) B_2(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 9°

$$
\begin{aligned}
\sin 9°&=-\frac{B_2}{\sqrt{2}}-\frac{1}{2} \sqrt[4]{5} B_4(2)\\
\cos 9°&=-\frac{B_2}{\sqrt{2}}+\frac{1}{2} \sqrt[4]{5} B_4(2)\\
\end{aligned}
$$



#### 10°

$$
\begin{aligned}
\sin 10°&=\frac{1}{2} \left(A_3 A_1(3)+A_1 A_3(3)\right)\\
\cos 10°&=-\frac{1}{2} i \left(-A_3 A_1(3)+A_1 A_3(3)\right)\\
\end{aligned}
$$



#### 11°

$$
\begin{aligned}
\sin 11°&=\frac{(-2+2 i) B_2 \left(A_1\left(\frac{3}{4}\right)+i A_3\left(\frac{3}{4}\right)\right)-2 (-5)^{1/4} \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{4 \sqrt{2}}\\
\cos 11°&=\frac{(1+i) B_2 \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)+(-5)^{1/4} \left(i A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 12°

$$
\begin{aligned}
\sin 12°&=-\frac{1}{2} \sqrt{3} B_4-\frac{i \sqrt[4]{5} B_2(2)}{2 \sqrt{2}}\\
\cos 12°&=\frac{1}{4} \left(2 B_4-i \sqrt[4]{5} \sqrt{6} B_2(2)\right)\\
\end{aligned}
$$



#### 13°

$$
\begin{aligned}
\sin 13°&=\frac{(-1+i) A_1(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1+i) A_3(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\cos 13°&=\frac{(-1-i) A_1(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1-i) A_3(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 14°

$$
\begin{aligned}
\sin 14°&=-\frac{1}{2} B_2 \left(A_1(3)+A_3(3)\right)+\frac{i \sqrt[4]{5} \left(A_1(3)-A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\cos 14°&=\frac{1}{4} \left(2 i B_2 \left(A_1(3)-A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1(3)+A_3(3)\right) B_4(2)\right)\\
\end{aligned}
$$



#### 15°

$$
\begin{aligned}
\sin 15°&=\frac{A_4}{\sqrt{2}}\\
\cos 15°&=-\frac{A_2}{\sqrt{2}}\\
\end{aligned}
$$



#### 16°

$$
\begin{aligned}
\sin 16°&=-\frac{1}{2} i B_2 \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)-\frac{\sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\cos 16°&=\frac{1}{2} B_2 \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)-\frac{i \sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 17°

$$
\begin{aligned}
\sin 17°&=\frac{(1+i) A_3 A_1(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1-i) A_1 A_3(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\cos 17°&=\frac{(1-i) A_3 A_1(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1+i) A_1 A_3(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 18°

$$
\begin{aligned}
\sin 18°&=B_4\\
\cos 18°&=-\frac{i \sqrt[4]{5} B_2(2)}{\sqrt{2}}\\
\end{aligned}
$$



#### 19°

$$
\begin{aligned}
\sin 19°&=\frac{1}{4} (-1)^{1/4} \left(A_3 A_1(3) \left(-2 B_2-i \sqrt{2} \sqrt[4]{5} B_4(2)\right)+A_1 A_3(3) \left(2 i B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\right)\\
\cos 19°&=\frac{1}{4} (-1)^{1/4} \left(A_1 A_3(3) \left(2 B_2-i \sqrt{2} \sqrt[4]{5} B_4(2)\right)+A_3 A_1(3) \left(-2 i B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\right)\\
\end{aligned}
$$



#### 20°

$$
\begin{aligned}
\sin 20°&=-\frac{1}{2} i \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)\\
\cos 20°&=\frac{1}{2} \left(-A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)\\
\end{aligned}
$$



#### 21°

$$
\begin{aligned}
\sin 21°&=\frac{A_4 B_2}{\sqrt{2}}-\frac{1}{2} \sqrt[4]{5} A_2 B_4(2)\\
\cos 21°&=\frac{A_2 B_2}{\sqrt{2}}+\frac{1}{2} \sqrt[4]{5} A_4 B_4(2)\\
\end{aligned}
$$



#### 22°

$$
\begin{aligned}
\sin 22°&=-\frac{1}{2} B_4 \left(A_1(3)+A_3(3)\right)-\frac{\sqrt[4]{5} \left(A_1(3)-A_3(3)\right) B_2(2)}{2 \sqrt{2}}\\
\cos 22°&=-\frac{1}{4} i \left(2 B_4 \left(A_1(3)-A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1(3)+A_3(3)\right) B_2(2)\right)\\
\end{aligned}
$$



#### 23°

$$
\begin{aligned}
\sin 23°&=\frac{(-1-i) A_3(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1-i) A_1(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\cos 23°&=\frac{(1-i) A_3(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1+i) A_1(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 24°

$$
\begin{aligned}
\sin 24°&=\frac{1}{4} \left(-2 \sqrt{3} B_2-\sqrt{2} \sqrt[4]{5} B_4(2)\right)\\
\cos 24°&=\frac{1}{4} \left(-2 B_2+\sqrt[4]{5} \sqrt{6} B_4(2)\right)\\
\end{aligned}
$$



#### 25°

$$
\begin{aligned}
\sin 25°&=\frac{1}{2} (-1)^{3/4} \left(A_1\left(\frac{3}{4}\right)+i A_3\left(\frac{3}{4}\right)\right)\\
\cos 25°&=-\frac{1}{2} (-1)^{1/4} \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)\\
\end{aligned}
$$



#### 26°

$$
\begin{aligned}
\sin 26°&=\frac{1}{2} B_2 \left(A_3 A_1(3)+A_1 A_3(3)\right)-\frac{i \sqrt[4]{5} \left(-A_3 A_1(3)+A_1 A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\cos 26°&=\frac{1}{4} \left(2 i B_2 \left(-A_3 A_1(3)+A_1 A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_3 A_1(3)+A_1 A_3(3)\right) B_4(2)\right)\\
\end{aligned}
$$



#### 27°

$$
\begin{aligned}
\sin 27°&=-\frac{B_4}{\sqrt{2}}-\frac{1}{2} i \sqrt[4]{5} B_2(2)\\
\cos 27°&=\frac{B_4}{\sqrt{2}}-\frac{1}{2} i \sqrt[4]{5} B_2(2)\\
\end{aligned}
$$



#### 28°

$$
\begin{aligned}
\sin 28°&=-\frac{1}{4} i \left(2 B_4 \left(-A_3 A_1(3)+A_1 A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_3 A_1(3)+A_1 A_3(3)\right) B_2(2)\right)\\
\cos 28°&=\frac{1}{4} \left(-2 B_4 \left(A_3 A_1(3)+A_1 A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_3 A_1(3)-A_1 A_3(3)\right) B_2(2)\right)\\
\end{aligned}
$$



#### 29°

$$
\begin{aligned}
\sin 29°&=\frac{1}{4} (-1)^{1/4} \left(2 B_2 \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(-i A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_4(2)\right)\\
\cos 29°&=\frac{(2+2 i) B_2 \left(-i A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)-2 (-5)^{1/4} \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 30°

$$
\begin{aligned}
\sin 30°&=\frac{1}{2}\\
\cos 30°&=\frac{\sqrt{3}}{2}\\
\end{aligned}
$$



#### 31°

$$
\begin{aligned}
\sin 31°&=\frac{1}{4} (-1)^{1/4} \left(2 B_2 \left(A_1(3)-i A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(-i A_1(3)+A_3(3)\right) B_4(2)\right)\\
\cos 31°&=\frac{(-1+i) B_2 \left(A_1(3)+i A_3(3)\right)+(-5)^{1/4} \left(A_1(3)-i A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 32°

$$
\begin{aligned}
\sin 32°&=-\frac{1}{4} i \left(-2 B_4 \left(A_1(3)-A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1(3)+A_3(3)\right) B_2(2)\right)\\
\cos 32°&=\frac{1}{4} \left(2 B_4 \left(A_1(3)+A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(-A_1(3)+A_3(3)\right) B_2(2)\right)\\
\end{aligned}
$$



#### 33°

$$
\begin{aligned}
\sin 33°&=-\frac{A_2 B_4}{\sqrt{2}}-\frac{1}{2} i \sqrt[4]{5} A_4 B_2(2)\\
\cos 33°&=-\frac{A_4 B_4}{\sqrt{2}}+\frac{1}{2} i \sqrt[4]{5} A_2 B_2(2)\\
\end{aligned}
$$



#### 34°

$$
\begin{aligned}
\sin 34°&=\frac{1}{2} B_2 \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)+\frac{i \sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\cos 34°&=\frac{1}{2} i B_2 \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)-\frac{\sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 35°

$$
\begin{aligned}
\sin 35°&=\frac{(-1+i) A_3 A_1(3)-(1+i) A_1 A_3(3)}{2 \sqrt{2}}\\
\cos 35°&=\frac{(1+i) A_3 A_1(3)+(1-i) A_1 A_3(3)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 36°

$$
\begin{aligned}
\sin 36°&=\frac{\sqrt[4]{5}}{\sqrt{2}}B_4(2)\\
\cos 36°&=-B_2\\
\end{aligned}
$$



#### 37°

$$
\begin{aligned}
\sin 37°&=\frac{(-1-i) A_1 A_3(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1-i) A_3 A_1(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\cos 37°&=\frac{(-1+i) A_1 A_3(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1+i) A_3 A_1(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 38°

$$
\begin{aligned}
\sin 38°&=\frac{1}{4} \left(-2 B_4 \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(-A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_2(2)\right)\\
\cos 38°&=\frac{1}{4} i \left(2 B_4 \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_2(2)\right)\\
\end{aligned}
$$



#### 39°

$$
\begin{aligned}
\sin 39°&=\frac{A_2 B_2}{\sqrt{2}}-\frac{1}{2} \sqrt[4]{5} A_4 B_4(2)\\
\cos 39°&=-\frac{A_4 B_2}{\sqrt{2}}-\frac{1}{2} \sqrt[4]{5} A_2 B_4(2)\\
\end{aligned}
$$



#### 40°

$$
\begin{aligned}
\sin 40°&=-\frac{1}{2} i \left(A_1(3)-A_3(3)\right)\\
\cos 40°&=\frac{1}{2} \left(A_1(3)+A_3(3)\right)\\
\end{aligned}
$$



#### 41°

$$
\begin{aligned}
\sin 41°&=\frac{(-1-i) B_2 \left(A_1(3)-i A_3(3)\right)+(-5)^{1/4} \left(-i A_1(3)+A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\cos 41°&=\frac{(-2+2 i) B_2 \left(A_1(3)+i A_3(3)\right)-2 (-5)^{1/4} \left(A_1(3)-i A_3(3)\right) B_4(2)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 42°

$$
\begin{aligned}
\sin 42°&=-\frac{1}{4} i \left(-2 i B_4+\sqrt[4]{5} \sqrt{6} B_2(2)\right)\\
\cos 42°&=\frac{1}{4} \left(2 \sqrt{3} B_4-i \sqrt{2} \sqrt[4]{5} B_2(2)\right)\\
\end{aligned}
$$



#### 43°

$$
\begin{aligned}
\sin 43°&=\frac{(1+i) A_1\left(\frac{3}{4}\right) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1-i) A_3\left(\frac{3}{4}\right) \left(2 B_4+\sqrt{2} \sqrt[4]{5}
B_2(2)\right)}{4 \sqrt{2}}\\
\cos 43°&=\frac{(-1+i) A_1\left(\frac{3}{4}\right) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1+i) A_3\left(\frac{3}{4}\right) \left(2 B_4+\sqrt{2} \sqrt[4]{5}
B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 44°

$$
\begin{aligned}
\sin 44°&=\frac{1}{2} i B_2 \left(-A_3 A_1(3)+A_1 A_3(3)\right)-\frac{\sqrt[4]{5} \left(A_3 A_1(3)+A_1 A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\cos 44°&=-\frac{1}{2} B_2 \left(A_3 A_1(3)+A_1 A_3(3)\right)-\frac{i \sqrt[4]{5} \left(-A_3 A_1(3)+A_1 A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 45°

$$
\begin{aligned}
\sin 45°&=\frac{1}{\sqrt{2}}\\
\cos 45°&=\frac{1}{\sqrt{2}}\\
\end{aligned}
$$



#### 46°

$$
\begin{aligned}
\sin 46°&=-\frac{1}{2} B_2 \left(A_3 A_1(3)+A_1 A_3(3)\right)-\frac{i \sqrt[4]{5} \left(-A_3 A_1(3)+A_1 A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\cos 46°&=\frac{1}{2} i B_2 \left(-A_3 A_1(3)+A_1 A_3(3)\right)-\frac{\sqrt[4]{5} \left(A_3 A_1(3)+A_1 A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 47°

$$
\begin{aligned}
\sin 47°&=\frac{(-1+i) A_1\left(\frac{3}{4}\right) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1+i) A_3\left(\frac{3}{4}\right) \left(2 B_4+\sqrt{2} \sqrt[4]{5}
B_2(2)\right)}{4 \sqrt{2}}\\
\cos 47°&=\frac{(1+i) A_1\left(\frac{3}{4}\right) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1-i) A_3\left(\frac{3}{4}\right) \left(2 B_4+\sqrt{2} \sqrt[4]{5}
B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 48°

$$
\begin{aligned}
\sin 48°&=\frac{1}{4} \left(2 \sqrt{3} B_4-i \sqrt{2} \sqrt[4]{5} B_2(2)\right)\\
\cos 48°&=-\frac{1}{4} i \left(-2 i B_4+\sqrt[4]{5} \sqrt{6} B_2(2)\right)\\
\end{aligned}
$$



#### 49°

$$
\begin{aligned}
\sin 49°&=\frac{(-2+2 i) B_2 \left(A_1(3)+i A_3(3)\right)-2 (-5)^{1/4} \left(A_1(3)-i A_3(3)\right) B_4(2)}{4 \sqrt{2}}\\
\cos 49°&=\frac{(-1-i) B_2 \left(A_1(3)-i A_3(3)\right)+(-5)^{1/4} \left(-i A_1(3)+A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 50°

$$
\begin{aligned}
\sin 50°&=\frac{1}{2} \left(A_1(3)+A_3(3)\right)\\
\cos 50°&=-\frac{1}{2} i \left(A_1(3)-A_3(3)\right)\\
\end{aligned}
$$



#### 51°

$$
\begin{aligned}
\sin 51°&=-\frac{A_4 B_2}{\sqrt{2}}-\frac{1}{2} \sqrt[4]{5} A_2 B_4(2)\\
\cos 51°&=\frac{A_2 B_2}{\sqrt{2}}-\frac{1}{2} \sqrt[4]{5} A_4 B_4(2)\\
\end{aligned}
$$



#### 52°

$$
\begin{aligned}
\sin 52°&=\frac{1}{4} i \left(2 B_4 \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_2(2)\right)\\
\cos 52°&=\frac{1}{4} \left(-2 B_4 \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(-A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_2(2)\right)\\
\end{aligned}
$$



#### 53°

$$
\begin{aligned}
\sin 53°&=\frac{(-1+i) A_1 A_3(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1+i) A_3 A_1(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\cos 53°&=\frac{(-1-i) A_1 A_3(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1-i) A_3 A_1(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 54°

$$
\begin{aligned}
\sin 54°&=-B_2\\
\cos 54°&=\frac{\sqrt[4]{5} B_4(2)}{\sqrt{2}}\\
\end{aligned}
$$



#### 55°

$$
\begin{aligned}
\sin 55°&=\frac{(1+i) A_3 A_1(3)+(1-i) A_1 A_3(3)}{2 \sqrt{2}}\\
\cos 55°&=\frac{(-1+i) A_3 A_1(3)-(1+i) A_1 A_3(3)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 56°

$$
\begin{aligned}
\sin 56°&=\frac{1}{2} i B_2 \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)-\frac{\sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\cos 56°&=\frac{1}{2} B_2 \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)+\frac{i \sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 57°

$$
\begin{aligned}
\sin 57°&=-\frac{A_4 B_4}{\sqrt{2}}+\frac{1}{2} i \sqrt[4]{5} A_2 B_2(2)\\
\cos 57°&=-\frac{A_2 B_4}{\sqrt{2}}-\frac{1}{2} i \sqrt[4]{5} A_4 B_2(2)\\
\end{aligned}
$$



#### 58°

$$
\begin{aligned}
\sin 58°&=\frac{1}{4} \left(2 B_4 \left(A_1(3)+A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(-A_1(3)+A_3(3)\right) B_2(2)\right)\\
\cos 58°&=-\frac{1}{4} i \left(-2 B_4 \left(A_1(3)-A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1(3)+A_3(3)\right) B_2(2)\right)\\
\end{aligned}
$$



#### 59°

$$
\begin{aligned}
\sin 59°&=\frac{(-1+i) B_2 \left(A_1(3)+i A_3(3)\right)+(-5)^{1/4} \left(A_1(3)-i A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\cos 59°&=\frac{1}{4} (-1)^{1/4} \left(2 B_2 \left(A_1(3)-i A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(-i A_1(3)+A_3(3)\right) B_4(2)\right)\\
\end{aligned}
$$



#### 60°

$$
\begin{aligned}
\sin 60°&=\frac{\sqrt{3}}{2}\\
\cos 60°&=\frac{1}{2}\\
\end{aligned}
$$



#### 61°

$$
\begin{aligned}
\sin 61°&=\frac{(2+2 i) B_2 \left(-i A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)-2 (-5)^{1/4} \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{4 \sqrt{2}}\\
\cos 61°&=\frac{1}{4} (-1)^{1/4} \left(2 B_2 \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(-i A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_4(2)\right)\\
\end{aligned}
$$



#### 62°

$$
\begin{aligned}
\sin 62°&=\frac{1}{4} \left(-2 B_4 \left(A_3 A_1(3)+A_1 A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_3 A_1(3)-A_1 A_3(3)\right) B_2(2)\right)\\
\cos 62°&=-\frac{1}{4} i \left(2 B_4 \left(-A_3 A_1(3)+A_1 A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_3 A_1(3)+A_1 A_3(3)\right) B_2(2)\right)\\
\end{aligned}
$$



#### 63°

$$
\begin{aligned}
\sin 63°&=\frac{B_4}{\sqrt{2}}-\frac{1}{2} i \sqrt[4]{5} B_2(2)\\
\cos 63°&=-\frac{B_4}{\sqrt{2}}-\frac{1}{2} i \sqrt[4]{5} B_2(2)\\
\end{aligned}
$$



#### 64°

$$
\begin{aligned}
\sin 64°&=\frac{1}{4} \left(2 i B_2 \left(-A_3 A_1(3)+A_1 A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_3 A_1(3)+A_1 A_3(3)\right) B_4(2)\right)\\
\cos 64°&=\frac{1}{2} B_2 \left(A_3 A_1(3)+A_1 A_3(3)\right)-\frac{i \sqrt[4]{5} \left(-A_3 A_1(3)+A_1 A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 65°

$$
\begin{aligned}
\sin 65°&=-\frac{1}{2} (-1)^{1/4} \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)\\
\cos 65°&=\frac{1}{2} (-1)^{3/4} \left(A_1\left(\frac{3}{4}\right)+i A_3\left(\frac{3}{4}\right)\right)\\
\end{aligned}
$$



#### 66°

$$
\begin{aligned}
\sin 66°&=\frac{1}{4} \left(-2 B_2+\sqrt[4]{5} \sqrt{6} B_4(2)\right)\\
\cos 66°&=\frac{1}{4} \left(-2 \sqrt{3} B_2-\sqrt{2} \sqrt[4]{5} B_4(2)\right)\\
\end{aligned}
$$



#### 67°

$$
\begin{aligned}
\sin 67°&=\frac{(1-i) A_3(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1+i) A_1(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\cos 67°&=\frac{(-1-i) A_3(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1-i) A_1(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 68°

$$
\begin{aligned}
\sin 68°&=-\frac{1}{4} i \left(2 B_4 \left(A_1(3)-A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1(3)+A_3(3)\right) B_2(2)\right)\\
\cos 68°&=-\frac{1}{2} B_4 \left(A_1(3)+A_3(3)\right)-\frac{\sqrt[4]{5} \left(A_1(3)-A_3(3)\right) B_2(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 69°

$$
\begin{aligned}
\sin 69°&=\frac{A_2 B_2}{\sqrt{2}}+\frac{1}{2} \sqrt[4]{5} A_4 B_4(2)\\
\cos 69°&=\frac{A_4 B_2}{\sqrt{2}}-\frac{1}{2} \sqrt[4]{5} A_2 B_4(2)\\
\end{aligned}
$$



#### 70°

$$
\begin{aligned}
\sin 70°&=\frac{1}{2} \left(-A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)\\
\cos 70°&=-\frac{1}{2} i \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)\\
\end{aligned}
$$



#### 71°

$$
\begin{aligned}
\sin 71°&=\frac{1}{4} (-1)^{1/4} \left(A_1 A_3(3) \left(2 B_2-i \sqrt{2} \sqrt[4]{5} B_4(2)\right)+A_3 A_1(3) \left(-2 i B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\right)\\
\cos 71°&=\frac{1}{4} (-1)^{1/4} \left(A_3 A_1(3) \left(-2 B_2-i \sqrt{2} \sqrt[4]{5} B_4(2)\right)+A_1 A_3(3) \left(2 i B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\right)\\
\end{aligned}
$$



#### 72°

$$
\begin{aligned}
\sin 72°&=-\frac{i \sqrt[4]{5} B_2(2)}{\sqrt{2}}\\
\cos 72°&=B_4\\
\end{aligned}
$$



#### 73°

$$
\begin{aligned}
\sin 73°&=\frac{(1-i) A_3 A_1(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1+i) A_1 A_3(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\cos 73°&=\frac{(1+i) A_3 A_1(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1-i) A_1 A_3(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 74°

$$
\begin{aligned}
\sin 74°&=\frac{1}{2} B_2 \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)-\frac{i \sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\cos 74°&=-\frac{1}{2} i B_2 \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)-\frac{\sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 75°

$$
\begin{aligned}
\sin 75°&=-\frac{A_2}{\sqrt{2}}\\
\cos 75°&=\frac{A_4}{\sqrt{2}}\\
\end{aligned}
$$



#### 76°

$$
\begin{aligned}
\sin 76°&=\frac{1}{4} \left(2 i B_2 \left(A_1(3)-A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1(3)+A_3(3)\right) B_4(2)\right)\\
\cos 76°&=-\frac{1}{2} B_2 \left(A_1(3)+A_3(3)\right)+\frac{i \sqrt[4]{5} \left(A_1(3)-A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 77°

$$
\begin{aligned}
\sin 77°&=\frac{(-1-i) A_1(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1-i) A_3(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\cos 77°&=\frac{(-1+i) A_1(3) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1+i) A_3(3) \left(2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 78°

$$
\begin{aligned}
\sin 78°&=\frac{1}{4} \left(2 B_4-i \sqrt[4]{5} \sqrt{6} B_2(2)\right)\\
\cos 78°&=-\frac{1}{2} \sqrt{3} B_4-\frac{i \sqrt[4]{5} B_2(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 79°

$$
\begin{aligned}
\sin 79°&=\frac{(1+i) B_2 \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)+(-5)^{1/4} \left(i A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{2 \sqrt{2}}\\
\cos 79°&=\frac{(-2+2 i) B_2 \left(A_1\left(\frac{3}{4}\right)+i A_3\left(\frac{3}{4}\right)\right)-2 (-5)^{1/4} \left(A_1\left(\frac{3}{4}\right)-i A_3\left(\frac{3}{4}\right)\right)
B_4(2)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 80°

$$
\begin{aligned}
\sin 80°&=-\frac{1}{2} i \left(-A_3 A_1(3)+A_1 A_3(3)\right)\\
\cos 80°&=\frac{1}{2} \left(A_3 A_1(3)+A_1 A_3(3)\right)\\
\end{aligned}
$$



#### 81°

$$
\begin{aligned}
\sin 81°&=-\frac{B_2}{\sqrt{2}}+\frac{1}{2} \sqrt[4]{5} B_4(2)\\
\cos 81°&=-\frac{B_2}{\sqrt{2}}-\frac{1}{2} \sqrt[4]{5} B_4(2)\\
\end{aligned}
$$



#### 82°

$$
\begin{aligned}
\sin 82°&=\frac{1}{2} B_4 \left(A_3 A_1(3)+A_1 A_3(3)\right)-\frac{\sqrt[4]{5} \left(-A_3 A_1(3)+A_1 A_3(3)\right) B_2(2)}{2 \sqrt{2}}\\
\cos 82°&=\frac{1}{4} i \left(-2 B_4 \left(-A_3 A_1(3)+A_1 A_3(3)\right)+\sqrt{2} \sqrt[4]{5} \left(A_3 A_1(3)+A_1 A_3(3)\right) B_2(2)\right)\\
\end{aligned}
$$



#### 83°

$$
\begin{aligned}
\sin 83°&=\frac{(1+i) A_3\left(\frac{3}{4}\right) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)-(1-i) A_1\left(\frac{3}{4}\right) \left(2 B_4+\sqrt{2} \sqrt[4]{5}
B_2(2)\right)}{4 \sqrt{2}}\\
\cos 83°&=\frac{(-1+i) A_3\left(\frac{3}{4}\right) \left(-2 B_4+\sqrt{2} \sqrt[4]{5} B_2(2)\right)+(1+i) A_1\left(\frac{3}{4}\right) \left(2 B_4+\sqrt{2} \sqrt[4]{5}
B_2(2)\right)}{4 \sqrt{2}}\\
\end{aligned}
$$



#### 84°

$$
\begin{aligned}
\sin 84°&=\frac{1}{4} \left(-2 \sqrt{3} B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\\
\cos 84°&=\frac{1}{4} \left(2 B_2+\sqrt[4]{5} \sqrt{6} B_4(2)\right)\\
\end{aligned}
$$



#### 85°

$$
\begin{aligned}
\sin 85°&=\frac{1}{2} (-1)^{1/4} \left(-i A_1(3)+A_3(3)\right)\\
\cos 85°&=\frac{1}{2} (-1)^{1/4} \left(A_1(3)-i A_3(3)\right)\\
\end{aligned}
$$



#### 86°

$$
\begin{aligned}
\sin 86°&=-\frac{1}{2} B_2 \left(A_1(3)+A_3(3)\right)-\frac{i \sqrt[4]{5} \left(A_1(3)-A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\cos 86°&=\frac{1}{2} i B_2 \left(A_1(3)-A_3(3)\right)-\frac{\sqrt[4]{5} \left(A_1(3)+A_3(3)\right) B_4(2)}{2 \sqrt{2}}\\
\end{aligned}
$$



#### 87°

$$
\begin{aligned}
\sin 87°&=\frac{A_4 B_4}{\sqrt{2}}+\frac{1}{2} i \sqrt[4]{5} A_2 B_2(2)\\
\cos 87°&=-\frac{A_2 B_4}{\sqrt{2}}+\frac{1}{2} i \sqrt[4]{5} A_4 B_2(2)\\
\end{aligned}
$$



#### 88°

$$
\begin{aligned}
\sin 88°&=\frac{1}{4} i \left(-2 B_4 \left(A_1\left(\frac{3}{4}\right)-A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_2(2)\right)\\
\cos 88°&=\frac{1}{4} \left(2 B_4 \left(A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)+\sqrt{2} \sqrt[4]{5} \left(-A_1\left(\frac{3}{4}\right)+A_3\left(\frac{3}{4}\right)\right)
B_2(2)\right)\\
\end{aligned}
$$



#### 89°

$$
\begin{aligned}
\sin 89°&=\frac{1}{4} (-1)^{3/4} \left(A_1 A_3(3) \left(2 B_2+i \sqrt{2} \sqrt[4]{5} B_4(2)\right)+A_3 A_1(3) \left(2 i B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\right)\\
\cos 89°&=\frac{1}{4} (-1)^{1/4} \left(A_1 A_3(3) \left(-2 B_2-i \sqrt{2} \sqrt[4]{5} B_4(2)\right)+A_3 A_1(3) \left(2 i B_2+\sqrt{2} \sqrt[4]{5} B_4(2)\right)\right)\\
\end{aligned}
$$



#### 90°

$$
\begin{aligned}
\sin 90°&=1\\
\cos 90°&=0\\
\end{aligned}
$$

hbghlyj

MathOverflow When is sin(r \pi) expressible in radicals for r rational?

hbghlyj

Czhang271828 发表于 2023-4-30 06:00
当且仅当 $\mathbb Q(x)$ 的 Galois 群是交换群.

$\Rightarrow$是因为：$\operatorname {Gal} (\mathbf {Q} (\zeta _{n})/\mathbf {Q} )\cong(\mathbf {Z} /n\mathbf {Z} )^{\times }$ 是交换群.
$\Leftarrow$是因为：The Kronecker–Weber theorem states that every finite abelian extension of Q in C is contained in Q(ζ_n) for some n.