如何使 sin(pi) 和 cos(pi/2) 为零？

Question

我知道在 Python 中 sin(pi) 和 cos(pi/2) 不会产生 0，但我正在使用矩阵进行计算，我需要使用这些值。

我正在使用 SymPy，起初 sin(pi) 和 cos(pi/2) 的值有点烦人。在一些乘法之后，它们开始妨碍。有没有办法让整个模块中的这些值等于0？ 如何在表达式中间更改它？

我将以这个矩阵为例：

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) - d3*(-sin(theta1)*sin(theta2) + cos(alpha2)*cos(theta1)*cos(theta2))*sin(alpha3) - d3*sin(alpha2)*cos(alpha3)*cos(theta1)],
[sin(alpha2)*sin(theta2)*cos(theta3) + sin(alpha2)*sin(theta3)*cos(alpha3)*cos(theta2) + sin(alpha3)*sin(theta3)*cos(alpha2), -sin(alpha2)*sin(theta2)*sin(theta3) + sin(alpha2)*cos(alpha3)*cos(theta2)*cos(theta3) + sin(alpha3)*cos(alpha2)*cos(theta3), -sin(alpha2)*sin(alpha3)*cos(theta2) + cos(alpha2)*cos(alpha3),a3*sin(alpha2)*sin(theta2) + d2*cos(alpha2) - d3*sin(alpha2)*sin(alpha3)*cos(theta2) + d3*cos(alpha2)*cos(alpha3)],
[0,0,0,1]])

使用 SymPy 我将替换该值

substitution = A.subs(alpha2, (-pi/2))

我中间会有很多6.12323399573677e-17。

Answer 1

使用 SymPy 中的符号 pi，而不是数学或 NumPy 模块中的数字 pi。这就是你可能正在做的事情：

from sympy import sin, cos
from math import pi
print([sin(pi), cos(pi/2)])   # [1.22464679914735e-16, 6.12323399573677e-17]

而你应该这样做：

from sympy import sin, cos, pi
print([sin(pi), cos(pi/2)])  #  [0, 0]

Answer 2

你总是可以做一个函数！像

from math import sin as oldsin

def sin(x):
    if x % pi == 0: 
        #if x is an integer mult of pi, like pi, 2pi, -7pi, etc.
        return 0
    else:
        return oldsin(x)