甘志国的“三角函数关系求解”问题

数学星空

1.设变量\(\alpha,\beta,\gamma\)满足\(0 \leqslant  \alpha \lt  \beta \lt  \gamma \lt  2\pi\) ,且

     \[x=\frac{\cos(\alpha)\sin(\frac{\gamma-\beta}{2})+\cos(\beta)\sin(\frac{\gamma-\alpha}{2})+\cos(\gamma)\sin(\frac{\beta-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

     \[y=\frac{\sin(\alpha)\sin(\frac{\gamma-\beta}{2})+\sin(\beta)\sin(\frac{\gamma-\alpha}{2})+\sin(\gamma)\sin(\frac{\beta-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

     \[z=\frac{2\sin(\frac{\beta-\alpha}{2})\sin(\frac{\gamma-\beta}{2})\sin(\frac{\gamma-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

       猜测变量x,y,z之间存在不依赖变量\(\alpha,\beta,\gamma\)的等量关系式（可能是二次式）？



2.设变量\(\alpha,\beta,\gamma\)满足\(0 \leqslant  \alpha \lt  \beta \lt  \gamma \lt  2\pi\) ,且

     \[x=\frac{\cos(\alpha)\sin(\frac{\beta-\gamma}{2})+\cos(\beta)\sin(\frac{\gamma-\alpha}{2})+\cos(\gamma)\sin(\frac{\beta-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

     \[y=\frac{\sin(\alpha)\sin(\frac{\beta-\gamma}{2})+\sin(\beta)\sin(\frac{\gamma-\alpha}{2})+\sin(\gamma)\sin(\frac{\beta-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

     \[z=\frac{2\sin(\frac{\beta-\alpha}{2})\sin(\frac{\gamma-\beta}{2})\sin(\frac{\gamma-\alpha}{2})}{\sin(\frac{\beta-\gamma}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

       猜测变量x,y,z之间存在不依赖变量\(\alpha,\beta,\gamma\)的等量关系式（可能是二次式）？


3.设变量\(\alpha,\beta,\gamma\)满足\(0 \leqslant  \alpha \lt  \beta \lt  \gamma \lt  2\pi\) ,且

     \[x=\frac{\cos(\alpha)\sin(\frac{\gamma-\beta}{2})+\cos(\beta)\sin(\frac{\alpha-\gamma}{2})+\cos(\gamma)\sin(\frac{\beta-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

     \[y=\frac{\sin(\alpha)\sin(\frac{\gamma-\beta}{2})+\sin(\beta)\sin(\frac{\alpha-\gamma}{2})+\sin(\gamma)\sin(\frac{\beta-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

     \[z=\frac{2\sin(\frac{\beta-\alpha}{2})\sin(\frac{\gamma-\beta}{2})\sin(\frac{\alpha-\gamma}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

       猜测变量x,y,z之间存在不依赖变量\(\alpha,\beta,\gamma\)的等量关系式（可能是二次式）？


4.设变量\(\alpha,\beta,\gamma\)满足\(0 \leqslant  \alpha \lt  \beta \lt  \gamma \lt  2\pi \),且

     \[x=\frac{\cos(\alpha)\sin(\frac{\gamma-\beta}{2})+\cos(\beta)\sin(\frac{\gamma-\alpha}{2})+\cos(\gamma)\sin(\frac{\alpha-\beta}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\alpha-\beta}{2})}\]

     \[y=\frac{\sin(\alpha)\sin(\frac{\gamma-\beta}{2})+\sin(\beta)\sin(\frac{\gamma-\alpha}{2})+\sin(\gamma)\sin(\frac{\alpha-\beta}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\alpha-\beta}{2})}\]

     \[z=\frac{2\sin(\frac{\alpha-\beta}{2})\sin(\frac{\gamma-\beta}{2})\sin(\frac{\gamma-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\alpha-\beta}{2})}\]

       猜测变量x,y,z之间存在不依赖变量\(\alpha,\beta,\gamma\)的等量关系式（可能是二次式）？

注：甘志国猜测2,3,4满足同一个关系式。

转自：http://www.cdmath.org/Article/ShowArticle.asp?ArticleID=973

数学星空

有趣的是：

将\(\alpha=\frac{\pi}{2},\frac{\pi}{3},\frac{\pi}{4},\frac{\pi}{5}\)的函数图像叠合在一起，很像‘’燕子‘’的形状：

数学星空

取\(\alpha=0\),消元结果:

\(x^6+4x^5z+3x^4y^2+4x^4z^2+8x^3y^2z+3x^2y^4+8x^2y^2z^2+4xy^4z+y^6+4y^4z^2-2x^5-4x^4z-4x^3y^2-8x^2y^2z-2xy^4-4y^4z-5x^4-24x^3z-6x^2y^2-24x^2z^2-24xy^2z-y^4-24y^2z^2+
20x^3+56x^2z+12xy^2+32xz^2+24y^2z-25x^2-44xz-5y^2-12z^2+14x+12z-3=0\)

看来\(x,y,z\)的关系式没这么简单，更不可能是二次关系啦！


取\(\alpha=\frac{\pi}{3}\),消元结果:

\(x^{12}+4x^{11}z+6x^{10}y^2+12x^{10}z^2+20x^9y^2z+16x^9z^3+15x^8y^4+44x^8y^2z^2+16x^8z^4+40x^7y^4z+64x^7y^2z^3+20x^6y^6+56x^6y^4z^2+64x^6y^2z^4+40x^5y^6z+96x^5y^4z^3+15x^4y^8+
24x^4y^6z^2+96x^4y^4z^4+20x^3y^8z+64x^3y^6z^3+6x^2y^{10}-4x^2y^8z^2+64x^2y^6z^4+4xy^{10}z+16xy^8z^3+y^{12}-4y^{10}z^2+16y^8z^4-2x^{11}-12x^{10}z-10x^9y^2-24x^9z^2-44x^8y^2z-
32x^8z^3-20x^7y^4-96x^7y^2z^2-56x^6y^4z-128x^6y^2z^3-20x^5y^6-144x^5y^4z^2-24x^4y^6z-192x^4y^4z^3-10x^3y^8-96x^3y^6z^2+4x^2y^8z-128x^2y^6z^3-2xy^{10}-24xy^8z^2+4y^{10}z-32y^8z^3-3x^{10}-
24x^9z-23x^8y^2-96x^8z^2-80x^7y^2z-192x^7z^3-62x^6y^4-208x^6y^2z^2-192x^6z^4-96x^5y^4z-576x^5y^2z^3-78x^4y^6-48x^4y^4z^2-576x^4y^2z^4-48x^3y^6z-576x^3y^4z^3-47x^2y^8+144x^2y^6z^2-
576x^2y^4z^4-8xy^8z-192xy^6z^3-11y^{10}+80y^8z^2-192y^6z^4+18x^9+132x^8z+80x^7y^2+360x^7z^2+376x^6y^2z+512x^6z^3+132x^5y^4+1016x^5y^2z^2+128x^5z^4+304x^4y^4z+1408x^4y^2z^3+96x^3y^6+952x^3y^4z^2+
256x^3y^2z^4+8x^2y^6z+1280x^2y^4z^3+26xy^8+296xy^6z^2+128xy^4z^4-52y^8z+384y^6z^3-30x^8-156x^7z-92x^6y^2-240x^6z^2-484x^5y^2z+224x^5z^3-64x^4y^4-928x^4y^2z^2+480x^4z^4-500x^3y^4z+448x^3y^2z^3+
28x^2y^6-1136x^2y^4z^2+960x^2y^2z^4-172xy^6z+224xy^4z^3+30y^8-448y^6z^2+480y^4z^4+6x^7-120x^6z+26x^5y^2-864x^5z^2+56x^4y^2z-1728x^4z^3+2x^3y^4-1216x^3y^2z^2-768x^3z^4+408x^2y^4z-2688x^2y^2z^3-18xy^6-
352xy^4z^2-768xy^2z^4+232y^6z-960y^4z^3+69x^6+684x^5z+43x^4y^2+2184x^4z^2+584x^3y^2z+2368x^3z^3-9x^2y^4+2208x^2y^2z^2+832x^2z^4+156xy^4z+1344xy^2z^3-47y^6+792y^4z^2-192y^2z^4-
162x^5-1140x^4z-28x^3y^2-2520x^3z^2-512x^2y^2z-2048x^2z^3-26xy^4-664xy^2z^2-384xz^4-300y^4z+384y^2z^3+210x^4+1128x^3z-32x^2y^2+1860x^2z^2-8xy^2z+912xz^3+46y^4-348y^2z^2+144z^4-182x^3-740x^2z+
50xy^2-792xz^2+156y^2z-288z^3+109x^2+300xz-27y^2+216z^2-42x-72z+9=0\)

这说明：\(x,y,z\)之间的关系并不是确定的

hujunhua

如果存在方程f(x,y,z)=0，代入后可得到f(x(α,β,γ),y(α,β,γ),z(α,β,γ))=0, 即存在约束g(α,β,γ)=0。但是甘志国给出的条件中，α,β,γ是独立变量，不存在约束关系。

hujunhua

以第1个问题为例，可以设
`x=\cos(\alpha)\sin(\frac{\gamma-\beta}{2})+\cos(\beta)\sin(\frac{\gamma-\alpha}{2})+\cos(\gamma)\sin(\frac{\beta-\alpha}{2})`
`y=\sin(\alpha)\sin(\frac{\gamma-\beta}{2})+\sin(\beta)\sin(\frac{\gamma-\alpha}{2})+\sin(\gamma)\sin(\frac{\beta-\alpha}{2})`
`z=2\sin(\frac{\beta-\alpha}{2})\sin(\frac{\gamma-\beta}{2})\sin(\frac{\gamma-\alpha}{2})`
`w=\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})`
从以上四式中消元可以得到`f(x,y,z,w)=0`, 如果甘志国的猜想成立，`f(x,y,z,w)=0`应该可以化为`f(x/w,y/w,z/w,1)=0`. 也就是说，如果`f(x,y,z,w)`是一个多项式的话，应该是一个齐次式。

数学星空

对于第1问：



若\(\alpha+\beta+\gamma=\pi\),我们可以得到：

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复制代码

若\(\alpha+\beta+\gamma=2\pi\),我们可以得到：

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数学星空

很郁闷啊，猛然发现以前计算的式子写错了，将（1）的条件写成了

1.设变量\(\alpha,\beta,\gamma\)满足\(0 \leqslant  \alpha \lt  \beta \lt  \gamma \lt  2\pi\) ,且

     \[x=\frac{\cos(\alpha)\sin(\frac{\gamma-\beta}{2})+\cos(\beta)\sin(\frac{\gamma-\alpha}{2})+\cos(\gamma)\sin(\frac{\beta-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

     \[y=\frac{\sin(\alpha)\sin(\frac{\gamma-\beta}{2})+\sin(\beta)\sin(\frac{\gamma-\alpha}{2})+\sin(\gamma)\sin(\frac{\beta-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

     \[z=\frac{2\sin(\frac{\beta-\alpha}{2})\sin(\frac{\beta-\alpha}{2})\sin(\frac{\gamma-\alpha}{2})}{\sin(\frac{\gamma-\beta}{2})+\sin(\frac{\gamma-\alpha}{2})+\sin(\frac{\beta-\alpha}{2})}\]

       猜测变量x,y,z之间存在不依赖变量\(\alpha,\beta,\gamma\)的等量关系式（可能是二次式）？

数学星空

重新将（1）~（4）全部消元计算得到了甘志国想要的答案：

\[x^2+y^2+2z=1\]


将(1)~(4)代入上式很容易验证正确性，具体可见附件

三角函数式关系化简结果001.pdf (498.66 KB, 下载次数: 3)

2014-8-19 00:20 上传

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chyanog

Mathematica code:

1. {x==(Cos[\[Alpha]] Sin[(\[Gamma]-\[Beta])/2]+Cos[\[Beta]] Sin[(\[Gamma]-\[Alpha])/2]+Cos[\[Gamma]] Sin[(\[Beta]-\[Alpha])/2])/(Sin[(\[Gamma]-\[Beta])/2]+Sin[(\[Gamma]-\[Alpha])/2]+Sin[(\[Beta]-\[Alpha])/2]),y==(Sin[\[Alpha]] Sin[(\[Gamma]-\[Beta])/2]+Sin[\[Beta]] Sin[(\[Gamma]-\[Alpha])/2]+Sin[\[Gamma]] Sin[(\[Beta]-\[Alpha])/2])/(Sin[(\[Gamma]-\[Beta])/2]+Sin[(\[Gamma]-\[Alpha])/2]+Sin[(\[Beta]-\[Alpha])/2]),z==(2 Sin[(\[Beta]-\[Alpha])/2] Sin[(\[Gamma]-\[Beta])/2] Sin[(\[Gamma]-\[Alpha])/2])/(Sin[(\[Gamma]-\[Beta])/2]+Sin[(\[Gamma]-\[Alpha])/2]+Sin[(\[Beta]-\[Alpha])/2])}//TrigFactor
   
2. %/.{\[Alpha]->4ArcTan[a1],\[Beta]->4ArcTan[a2],\[Gamma]->4ArcTan[a3]}//TrigExpand//Simplify
   
3. GroebnerBasis[%,{x,y,z},{a1,a2,a3},MonomialOrder->EliminationOrder]//AbsoluteTiming

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