一个含参数三角函数方程组问题
r2*cosθ2-r1*cosθ1=r3r1*sinθ1-r2*sinθ2=r4其中r1、r2、r3为已知量,r4为输入变量,θ1、θ2为输出量,想求出θ1与r4、θ2与r4的两个单独的关系式,两个未知数两个方程感觉应该可以求解吧
2式平方和可得到θ1-θ2,可是较难得到θ1+θ2的简洁表示
r2²+r1²-2r1r2(cosθ1cosθ1+sinθ1sinθ1)=r3²+r4²
cos(θ1-θ2)=(r3²+r4²-r1²-r2²)/2r1r2
θ1-θ2=arcos[(r3²+r4²-r1²-r2²)/2r1r2]=φ(r4)
θ1=θ2+φ代入r1*sinθ1-r2*sinθ2=r4
r1(sinθ2cosφ+sinφcosθ2)-r2*sinθ2=r4
(r1cosφ-r2)sinθ2+r1sinφcosθ2=r4
令ψ(r4)=arcos[(r1cosφ-r2)/√{(r1cosφ-r2)²+(r1sinφ)²}/r4]
cosψsinθ2+sinψcosθ2=1
sin(θ2+ψ)=1
θ2=2kπ-½π-ψ(r4)
θ1=2kπ-½π-ψ(r4)+φ(r4)
φ(r4)=arcos[(r3²+r4²-r1²-r2²)/2r1r2]
ψ(r4)=arcos[(r1cosφ-r2)/√{(r1cosφ-r2)²+(r1sinφ)²}/r4]
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