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{}，但是我用矩阵进行计算，我需要使用这些值。在

我使用SymPy，起初sin(pi)和{}的值有点烦人。经过几次乘法之后，它们开始碍事了。有没有办法使这些值在整个模块中等于0？ 我怎样才能改变中间的表情？在

我以这个矩阵为例：

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]])

我用SymPy代替值

^{pr2}$

我会在中间放很多6.12323399573677e-17。在

Answer 1

你总是可以做一个函数！有点像

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)

Answer 2

使用SymPy中的符号pi，而不是math或NumPy模块中的数字pi。这就是你可能正在做的：

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

你应该这样做：

^{pr2}$