化简 arctan{1／[1+2cos(2π／9)]／[tan(7π／18)-2／(cot(π／9)+cot(2π／9))]}

天山草

化简一个三角表达式：



注： 结果是  \(\pi/18\)

fungarwai

化簡咗, 但未完成

\(t=\tan\dfrac{\pi}{9}\)
\(\cos\dfrac{2\pi}{9}=\dfrac{1-t^2}{1+t^2}\)
\(\tan\dfrac{7\pi}{18}=\tan\left(\dfrac{\pi}{2}-\dfrac{\pi}{9}\right)=\dfrac{1}{t}\)
\(\cot\dfrac{2\pi}{9}=\dfrac{1-t^2}{2t}\)

\(\dfrac{\dfrac{1}{1+2\cos\dfrac{2\pi}{9}}}
{\tan\dfrac{7\pi}{18}-\dfrac{2}{\cot\dfrac{\pi}{9}+\cot\dfrac{2\pi}{9}}}\)
\(=\dfrac{\dfrac{1}{1+2\cdot\dfrac{1-t^2}{1+t^2}}}
{\dfrac{1}{t}-\dfrac{2}{\dfrac{1}{t}+\dfrac{1-t^2}{2t}}}
=\dfrac{\dfrac{1+t^2}{3-t^2}}
{\dfrac{1}{t}-\dfrac{4t}{3-t^2}}
=\dfrac{t+t^3}{3-5t^2}\)

天山草

2# 楼演算正确。但未能得到结果。用软件化简可以：



只是我们不知道软件是咋算出来的？

王守恩

\(\frac{\frac{1}{1+2\cos40}}{\tan70-\frac{2}{\cot20+\cot40}}=\frac{\frac{1}{1+2\cos40}}{\frac{\sin70}{\cos70}-\frac{2}{\frac{\cos20}{\sin20}+\frac{\cos40}{\sin40}}}=\frac{\frac{1}{1+2\cos40}}{\frac{\cos20}{\sin20}-\frac{2\sin20\sin40}{\cos20\sin40+\cos40\sin20}}\)

\(=\frac{\frac{1}{2\cos^220+\cos40}}{\frac{\cos20}{\sin20}-\frac{2\sin40}{2\cos^220+\cos40}}=\frac{1}{\frac{\cos20(1+2\cos40)}{\sin20}-2\sin40}=\frac{\sin20/\cos20}{1+2\cos40-4\sin20\sin20}\)

\(=\frac{\sin20/\cos20}{3\cos^220-5\sin^220}=\tan10\)

最后一步是怎么来的？

fungarwai

\(\dfrac{t+t^3}{3-5t^2}
=\dfrac{\tan\dfrac{\pi}{9}+\tan^3\dfrac{\pi}{9}}{3-5\tan^2\dfrac{\pi}{9}}
=\dfrac{\sin\dfrac{\pi}{9}\cos^2\dfrac{\pi}{9}+\sin^3\dfrac{\pi}{9}}
{3\cos^3\dfrac{\pi}{9}-5\sin^2\dfrac{\pi}{9}\cos\dfrac{\pi}{9}}\)
\(=\dfrac{\sin\dfrac{\pi}{9}}
{3\cos^3\dfrac{\pi}{9}-5\sin^2\dfrac{\pi}{9}\cos\dfrac{\pi}{9}}\)

\(3\cos^3 \dfrac{\pi}{9}-5\sin^2 \dfrac{\pi}{9}\cos \dfrac{\pi}{9}
=8\cos^3 \dfrac{\pi}{9}-5\cos \dfrac{\pi}{9}\)
\(=6\cos \dfrac{\pi}{9}+2\cos \dfrac{\pi}{\color{red}{9}}-5\cos \dfrac{\pi}{9}
=1+\cos \dfrac{\pi}{9}=2\cos^2 \dfrac{\pi}{18}\)

\(\dfrac{\sin\dfrac{\pi}{9}}
{3\cos^3\dfrac{\pi}{9}-5\sin^2\dfrac{\pi}{9}\cos\dfrac{\pi}{9}}
=\dfrac{2\sin\dfrac{\pi}{18}\cos\dfrac{\pi}{18}}
{2\cos^2 \dfrac{\pi}{18}}=\tan\dfrac{\pi}{18}\)

---

\(=6\cos \dfrac{\pi}{9}+2\cos \dfrac{\pi}{\color{green}{3}}-5\cos \dfrac{\pi}{9}
=1+\cos \dfrac{\pi}{9}=2\cos^2 \dfrac{\pi}{18}\)

luyuanhong

楼上 fungarwai 的解答很好！已收藏。