据我所知，在Python sin(pi)和cos(pi/2)不会产生0，但我用的矩阵进行计算，我需要使用这些值。
我使用SymPy，并在第一的价值观sin(pi)和cos(pi/2)是有点讨厌。经过一些乘法后，它们开始妨碍您。有没有办法使这些值0在整个模块中等于？如何在表达式中间更改它？
我将使用此矩阵作为示例：
A = Matrix([
[(-sin(theta1)*sin(theta2)*cos(alpha2) + cos(theta1)*cos(theta2))*cos(theta3) + (-sin(theta1)*cos(alpha2)*cos(theta2) - sin(theta2)*cos(theta1))*sin(theta3)*cos(alpha3) + sin(alpha2)*sin(alpha3)*sin(theta1)*sin(theta3), -(-sin(theta1)*sin(theta2)*cos(alpha2) + cos(theta1)*cos(theta2))*sin(theta3) + (-sin(theta1)*cos(alpha2)*cos(theta2) - sin(theta2)*cos(theta1))*cos(alpha3)*cos(theta3) + sin(alpha2)*sin(alpha3)*sin(theta1)*cos(theta3), -(-sin(theta1)*cos(alpha2)*cos(theta2) - sin(theta2)*cos(theta1))*sin(alpha3) + sin(alpha2)*sin(theta1)*cos(alpha3), a3*(-sin(theta1)*sin(theta2)*cos(alpha2) + cos(theta1)*cos(theta2)) + d2*sin(alpha2)*sin(theta1) - d3*(-sin(theta1)*cos(alpha2)*cos(theta2) - sin(theta2)*cos(theta1))*sin(alpha3) + d3*sin(alpha2)*sin(theta1)*cos(alpha3)],
[(-sin(theta1)*sin(theta2) + cos(alpha2)*cos(theta1)*cos(theta2))*sin(theta3)*cos(alpha3) + (sin(theta1)*cos(theta2) + sin(theta2)*cos(alpha2)*cos(theta1))*cos(theta3) - sin(alpha2)*sin(alpha3)*sin(theta3)*cos(theta1), (-sin(theta1)*sin(theta2) + cos(alpha2)*cos(theta1)*cos(theta2))*cos(alpha3)*cos(theta3) - (sin(theta1)*cos(theta2) + sin(theta2)*cos(alpha2)*cos(theta1))*sin(theta3) - sin(alpha2)*sin(alpha3)*cos(theta1)*cos(theta3), -(-sin(theta1)*sin(theta2) + cos(alpha2)*cos(theta1)*cos(theta2))*sin(alpha3) - sin(alpha2)*cos(alpha3)*cos(theta1), a3*(sin(theta1)*cos(theta2) + sin(theta2)*cos(alpha2)*cos(theta1)) - d2*sin(alpha2)*cos(theta1) …