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$\left\{\begin{array}{l}(\cos 2\alpha+\cos 2\beta {\small-1})m+(\sin 2\beta-\sin 2\alpha)n=\sin 2\alpha+\sin 2\beta-2\sin 2\gamma\\(\cos 2\alpha+\cos 2\gamma {\small-1})m+(\sin 2\alpha-\sin 2\gamma)n=2\sin 2\beta-(\sin 2\alpha+\sin 2\gamma) \end{array}\right.$"
$\quad$两式相减得~g;X@
$\quad 2\sin(\beta+\gamma)\sin(\gamma-\beta)m=(2\sin 2\alpha-\sin 2\beta-\sin 2\gamma)(n+1)\small\underset{\,}{\,}$:|#
$\quad$将$\;m=(n+1)\large\frac{2\sin 2\alpha-\sin 2\beta-\sin 2\gamma}{2\sin(\beta+\gamma)\sin(\gamma-\beta)}\,$代入原方程得整理得2G
$\small\left\{\begin{array}{l}m={\large\frac{\overset{^{\,}}{4}(2\sin 2x-\sin 2y-\sin 2z)\cos(y+z)\sin(y-z)}{\sin(2x-2y)+\sin(2x-2z)-\sin(2y+2z)-2\sin 2x+\sin 4x+\sin 2y+\sin 2z}}\\n={\large\frac{\sin 4y-2\sin(2y+2z)+\sin 4z}{\sin(2x-2y)+\sin(2x-2z)-\sin(2y+2z)-2\sin 2x+\sin 4x+\sin 2y+\sin 2z}}-1\end{array}\right.$\~6ZY
©Elinkage数学论坛 -- Elinkage极酷超级论坛 #5R
解二元线性方程组不难,但三角函数的化简较繁. 已知上面的式子还能简化.待更新...3ycV^H
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2020/06/11 07:26am 
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