计算下面三角式

ejsoon

笨笨 发表于 2025-1-9 14:38
就是链接打不开啊

可以在此看到兩題的解，已下載單頁面。

https://quanquan.space/viewtopic.php?p=6541

或許這兩題的解法可作參考。

yigo

知道答案是\(\sqrt{7}\)，那就来凑吧。
令\(x=\cos(\frac{\pi}{7})\)，由\(\cos(\frac{3\pi}{7})=-\cos(\frac{4\pi}{7})\)可得
\(8x^3-4x^2-4x+1=0\)
令\(u=\frac{2\sin(\frac{\pi}{7})+\sin(\frac{2\pi}{7})}{\cos(\frac{2\pi}{7})}=\frac{2\sqrt{1-x^2}(1+x)}{2x^2-1}＞0\)，则
\(u^2=\frac{4(1-x^2)(1+x)^2}{4x^4-4x^2+1}\),
利用前面的方程将\(x^4\)降幂到\(x^3\)，可得\(u^2=7\),故\(u=\sqrt{7}\).

wayne

楼主的答案是$\sqrt{7}$,式子并不具备对称性,那就只能死算了.
既然是死算, 咱也来一个, 可不能输了气势.
.

王守恩

\(\D\frac{2\sin(\pi/7)+\sin(2\pi/7)}{\cos(2\pi/7)}\)

=\(\D\frac{2\sin(\pi/7)\big(1+\cos(\pi/7)\big)}{\cos(2\pi/7)}\)

=\(\D\frac{4\sin(\pi/7)\cos(\pi/14)\cos(\pi/14)}{\cos(2\pi/7)}\)

=\(\D\frac{4\sin(\pi/7)\cos(\pi/14)\sin(6\pi/14)}{\sin(3\pi/14)}\)

=\(\D8\sin(\pi/7)\cos(\pi/14)\cos(3\pi/14)\)

=\(\D8\sin(2\pi/14)\sin(4\pi/14)(\sin(6\pi/14)\)

=\(\D8\big(2\sin(\pi/14)\sin(6\pi/14)\big)\big(2\sin(2\pi/14)\sin(5\pi/14)\big)\big(2\sin(3\pi/14)\sin(4\pi/14)\big)\)

=\(\D(2\sin(\pi/14))(2\sin(2\pi/14))(2\sin(3\pi/14))(2\sin(4\pi/14))(2\sin(5\pi/14))(2\sin(6\pi/14))\)

=\(\D\sqrt{\ 7\ }\)

详见：小题大作> 求证：\(\D(2\sin(\pi/4))^2=2\)

yigo

13#求解

令\(t=\cos{x}\),方程\(\cos{2x}=\cos{5x}\)化为：

\(16t^5-20t^3-2t^2+5t+1=0\)

即：\((t-1)(2t+1)(8t^3+4t^2-4t-1)=0\)

\(t_1=\cos(\frac{2\pi}{7}),t_2=\cos(\frac{4\pi}{7}),t_3=\cos(\frac{6\pi}{7})\)是\(8t^3+4t^2-4t-1=0\)的三个根。

令\(t_1=s_1^3,t_2=s_2^3,t_3=s_3^3\),

则：\(s_1^3+s_2^3+s_3^3=-\frac{1}{2},s_1^3s_2^3+s_2^3s_3^3+s_3^3s_1^3=-\frac{1}{2},s_1s_2s_3=\frac{1}{2}\)

令\(u=s_1+s_2+s_3,v=s_1s_2+s_2s_3+s_3s_1\),题目即求\(u\)

利用恒等式:\(x^3+y^3+z^3-3xyz=(x+y+z)^3-3(xy+yz+zx)(x+y+z)\)

有\(u^3-3uv=-2,v^3-\frac{3}{2}uv=-\frac{5}{4}\)

消去\(v\)得：\((u^3+2)^3-\frac{1}{2}(3u)^3(u^3+2)+\frac{5}{4}(3u)^3=0\)

软件求得：\(u^3=\frac{5-3\sqrt[3]{7}}{2}\),即：\(u=\sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}\)