一个含参数三角函数方程组问题20

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+r4²cos(θ1-θ2)=(r3²+r4²-r1²-r2²)/2r1r2θ1-θ2=arcos[(r3²+r4²-r1²-r2²)/2r1r2]=φ(r4)θ1=θ2+φ代入r1*sinθ1-r2*sinθ2=r4r1(sinθ2cosφ+sinφcosθ2)-r2*sinθ2=r4(r1cosφ-r2)sinθ2+r1sinφcosθ2=r4令ψ(r4)=arcos[(r1cosφ-r2)/√{(r1cosφ-r2)²+(r1sinφ)²}/r4]cosψsinθ2+sinψcosθ2=1sin(θ2+ψ)=1θ2=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]