matlab 求解含有三角函数的多变量方程组

Question

Answer 1

你好，利用matlab的符号工具箱，可以解出a和b，有两对解：

syms x y L1 L2 L0 a b
eq = [x-L2*cos(a+b)+L0-L1*cos(a);
    y - L2*sin(a+b) - L1*sin(a)];
[a,b] = solve(eq, [a,b])

解的结果为

a =
 
 2*atan((((8*L1^2*L2*y)/((L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2)*(L0^2 + 2*L0*L1 + 2*L0*x + L1^2 + 2*L1*x - L2^2 + x^2 + y^2)) + (4*L1*L2*(-(L0^2 + 2*L0*x - L1^2 - 2*L1*L2 - L2^2 + x^2 + y^2)*(L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2))/((L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2)*(L0^2 + 2*L0*L1 + 2*L0*x + L1^2 + 2*L1*x - L2^2 + x^2 + y^2)))*(L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))/(4*L1*L2))
 2*atan((((8*L1^2*L2*y)/((L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2)*(L0^2 + 2*L0*L1 + 2*L0*x + L1^2 + 2*L1*x - L2^2 + x^2 + y^2)) - (4*L1*L2*(-(L0^2 + 2*L0*x - L1^2 - 2*L1*L2 - L2^2 + x^2 + y^2)*(L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2))/((L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2)*(L0^2 + 2*L0*L1 + 2*L0*x + L1^2 + 2*L1*x - L2^2 + x^2 + y^2)))*(L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))/(4*L1*L2))
 
 
b =
 
 -2*atan((-(L0^2 + 2*L0*x - L1^2 - 2*L1*L2 - L2^2 + x^2 + y^2)*(L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2)/(L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))
  2*atan((-(L0^2 + 2*L0*x - L1^2 - 2*L1*L2 - L2^2 + x^2 + y^2)*(L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))^(1/2)/(L0^2 + 2*L0*x - L1^2 + 2*L1*L2 - L2^2 + x^2 + y^2))

用公式打印可以得到解的具体表达式

Answer 2

你好，用你的代码跑出来这个结果
a =

Empty sym: 0-by-1

b =

Empty sym: 0-by-1
请问这是什么问题，谢谢

Answer 3

Matlab版本问题，我的已经2020a版本了，我不知道你更新了没有

Answer 4

我是19b版本，那我更新试下，谢谢