$\sqrt{58+\sqrt{2}}$用有理数度的sin/cos写出？

hbghlyj

$\sqrt{58+\sqrt{2}}$
$=2 \sqrt{29} \cos \left(\dfrac{1}{2} \tan ^{-1}(41)\right)$
$=\dfrac{\sqrt{1-41 i}+\sqrt{1+41 i}}{\sqrt[4]{2}}$
$=\dfrac{1}{2}(\sqrt{-20-21 i}+5+i 2) \sqrt{1+i} 2^{3 / 4}$
$=(5+2 i) \sqrt[8]{-1}+\dfrac{(3-7 i) \sqrt[8]{-1}}{\sqrt{2}}$
用有理数度的sin/cos写出（不含根式）？

Sage

1. R.<x> = CyclotomicField(16)['x']
2. f = x^4 - 116*x^2 + 3362
3. f.factor()

Copy the Code

$$\small(x - 5 \zeta_{16}^{7} + 2 \zeta_{16}^{5} - 2 \zeta_{16}^{3} + 5 \zeta_{16}) \cdot (x - 2 \zeta_{16}^{7} - 5 \zeta_{16}^{5} + 5 \zeta_{16}^{3} + 2 \zeta_{16}) \cdot (x + 2 \zeta_{16}^{7} + 5 \zeta_{16}^{5} - 5 \zeta_{16}^{3} - 2 \zeta_{16}) \cdot (x + 5 \zeta_{16}^{7} - 2 \zeta_{16}^{5} + 2 \zeta_{16}^{3} - 5 \zeta_{16})$$

hbghlyj

找到了：
\begin{align*}\sqrt{58+\sqrt{2}}&=\frac{1}{2}(7 \sqrt{2}-4) \csc \left(\frac{\pi}{8}\right)\\
&=\color{blue}{\frac{7\sin\left(\frac\pi4\right)-2}{\sin\left(\frac{\pi}{8}\right)}}
\end{align*}
验证：wolframalpha.com/input?i=%287+sin%28%CF%80%2F … +csc%28%CF%80%2F8%29
右边用到的角是$\frac{\pi}{8}$。它是怎么得出的？怎样从$\sqrt{58+\sqrt{2}}$看出$\frac\pi8$？

hbghlyj

$\sqrt{n+\sqrt{n}},n=m^2+1,n\inZ^+$
$f(X)=X^4-2 n X^2+n(n-1)$.


$\sqrt{10+\sqrt{10}}$
$=\sqrt[4]{\frac{5}{2}}(\sqrt{1-3 i}+\sqrt{1+3 i})$
$=\frac{1}{2} \sqrt[4]{5} 2^{3 / 4}(\sqrt{2+i}+\sqrt{-1-2 i}) \sqrt{1+i}$
$=2 \sqrt{5} \cos \left(\frac{1}{2} \tan ^{-1}(3)\right)$


如何用有理数度的sin/cos写出（不含根式）？