任意四边形的一个面积公式，如何证明？

王守恩

王守恩 发表于 2021-10-24 07:40
还是这不用烧脑细胞的。

\(1，12\ 楼的图。\)谢谢 denglongshan！

有电脑罩着不会出错，用Simplify按钮手工可以慢慢摸索：这"1"是怎么来的？

王守恩

王守恩 发表于 2021-10-24 07:40
还是这不用烧脑细胞的。

\(1，12\ 楼的图。\)谢谢 denglongshan！

接10楼。Simplify
\(\bigg[\frac{\big(\frac{\sin(x)\sin(w)}{\sin(x+w)}+\frac{\sin(y)\sin(z)}{\sin(y+z)}+\frac{\sin(z)\sin(y)}{\sin(z+y)}+\frac{\sin(w)\sin(x)}{\sin(w+x)}\big)\big(\frac{\sin(x+y)}{1+\cos(x+y)}+\frac{1+\cos(y+z)}{\sin(y+z)}+\frac{\sin(z+w)}{1+\cos(z+w)}+\frac{1+\cos(w+x)}{\sin(w+x)}\big)\big(\frac{1+\cos(x+y)}{\sin(x+y)}+\frac{\sin(y+z)}{1+\cos(y+z)}+\frac{1+\cos(z+w)}{\sin(z+w)}+\frac{\sin(w+x)}{1+\cos(w+x)}\big)}{\big(\frac{\sin(x)}{\sin(x+w)}+\frac{\sin(y)}{\sin(y+z)}+\frac{\sin(z)}{\sin(z+y)}+\frac{\sin(w)}{\sin(w+x)}\big)^2\big(\frac{\sin(x+y)}{1+\cos(x+y)}+\frac{1+\cos(y+z)}{\sin(y+z)}+\frac{\sin(z+w)}{1+\cos(z+w)}+\frac{1+\cos(w+x)}{\sin(w+x)}\big)-\big(\frac{\sin(x)}{\sin(x+w)}-\frac{\sin(y)}{\sin(y+z)}+\frac{\sin(z)}{\sin(z+y)}-\frac{\sin(w)}{\sin(w+x)}\big)^2\big(\frac{1+\cos(x+y)}{\sin(x+y)}+\frac{\sin(y+z)}{1+\cos(y+z)}+\frac{1+\cos(z+w)}{\sin(z+w)}+\frac{\sin(w+x)}{1+\cos(w+x)}\big)}\bigg]\)
=1

denglongshan

王守恩 发表于 2021-10-26 07:51
接10楼。Simplify
\(\bigg[\frac{\big(\frac{\sin(x)\sin(w)}{\sin(x+w)}+\frac{\sin(y)\sin(z)}{\sin(y ...

完全解决，不用象你那么麻烦，十三行错误，多写了一个b，改为
2 (a d  Sin[A] + b c Sin[C1]); S3 = 2 (a b Sin[B] + d c Sin[D1]);即可

天山草