如何引导SymPy可靠消去三角函数表达式中的sin(th)因子以完成迭代简化计算

AI 小北

2026-4-14

我有这么一组表达式：

deque([-6*cos(th)**3 - 9*cos(th), (11*cos(th)**2 + 4)*sin(th), -6*sin(th)**2*cos(th), sin(th)**3])

接着我运行了一段迭代求导、相加再除以sin(th)的代码：

import sympy as sp
th = sp.symbols('th')
order = 4

for nu in range(order + 1, 2*order): # iterate order-1 more times to reach the constants
    q = 0
    for mu in range(1, nu): # Terms come from the previous derivative, so there are nu - 1 of them here.
        p = exprs.popleft()
        term = q + sp.diff(p, th)
        exprs.append(sp.cancel(term/sp.sin(th)))
        q = p
    exprs.append(sp.cancel(q/sp.sin(th)))
    print(nu, exprs)

但输出的结果一团糟：

5 deque([18*cos(th)**2 + 9, (-22*sin(th)**2*cos(th) + 5*cos(th)**3 - 5*cos(th))/sin(th), 6*sin(th)**2 - cos(th)**2 + 4, -3*sin(th)*cos(th), sin(th)**2])
6 deque([-36*cos(th), (22*sin(th)**4 - 19*sin(th)**2*cos(th)**2 + 14*sin(th)**2 - 5*cos(th)**4 + 5*cos(th)**2)/sin(th)**3, (-8*sin(th)**2*cos(th) + 5*cos(th)**3 - 5*cos(th))/sin(th)**2, (9*sin(th)**2 - 4*cos(th)**2 + 4)/sin(th), -cos(th), sin(th)])
7 deque([36, (24*sin(th)**4*cos(th) + 39*sin(th)**2*cos(th)**3 - 24*sin(th)**2*cos(th) + 15*cos(th)**5 - 15*cos(th)**3)/sin(th)**5, (30*sin(th)**4 - 34*sin(th)**2*cos(th)**2 + 19*sin(th)**2 - 15*cos(th)**4 + 15*cos(th)**2)/sin(th)**4, (9*sin(th)**2*cos(th) + 9*cos(th)**3 - 9*cos(th))/sin(th)**3, (10*sin(th)**2 - 4*cos(th)**2 + 4)/sin(th)**2, 0, 1])

而我预期的结果应该是随着迭代步骤逐渐简化，最终得到一组常数，正确的输出示例是这样的：

5 deque([9*cos(2*th) + 18, -27*sin(2*th)/2, 7*sin(th)**2 + 3, -3*sin(2*th)/2, sin(th)**2])
6 deque([-36*cos(th), 36*sin(th), -13*cos(th), 13*sin(th), -cos(th), sin(th)])
7 deque([36, 0, 49, 0, 14, 0, 1])

我试过添加各种.simplify()调用，在order = 4时还能凑效，但面对更复杂的表达式和更高阶的情况就不行了。我发现SymPy总是不能可靠地消去每一步中的sin(th)因子——尽管我知道理论上是可以消掉的。

我也试过用.trigsimp()和.simplify()，但它们有时候会生成更高频的项（比如sin(2th)），然后.cancel()就没法消去低频的sin(th)了；可如果完全不做简化，或者只在最后一步简化，SymPy的运行时间会变得极长，最终结果的复杂度也高得离谱。

下面是更高阶的表达式队列，我需要一个能处理所有这些情况的解决方案，而不只是阶数4的简单例子：

阶数5：

deque([-24*cos(th)**4 - 72*cos(th)**2 - 9, (50*cos(th)**3 + 55*cos(th))*sin(th), (-35*cos(th)**2 - 10)*sin(th)**2, 10*sin(th)**3*cos(th), -sin(th)**4])

阶数6：

deque([-120*cos(th)**5 - 600*cos(th)**3 - 225*cos(th), (274*cos(th)**4 + 607*cos(th)**2 + 64)*sin(th), (-225*cos(th)**3 - 195*cos(th))*sin(th)**2, (85*cos(th)**2 + 20)*sin(th)**3, -15*sin(th)**4*cos(th), sin(th)**5])

阶数7：

deque([-720*cos(th)**6 - 5400*cos(th)**4 - 4050*cos(th)**2 - 225, (1764*cos(th)**5 + 6552*cos(th)**3 + 2079*cos(th))*sin(th), (-1624*cos(th)**4 - 2842*cos(th)**2 - 259)*sin(th)**2, (735*cos(th)**3 + 525*cos(th))*sin(th)**3, (-175*cos(th)**2 - 35)*sin(th)**4, 21*sin(th)**5*cos(th), -sin(th)**6])

阶数8：

deque([-5040*cos(th)**7 - 52920*cos(th)**5 - 66150*cos(th)**3 - 11025*cos(th), (13068*cos(th)**6 + 73188*cos(th)**4 + 46575*cos(th)**2 + 2304)*sin(th), (-13132*cos(th)**5 - 38626*cos(th)**3 - 10612*cos(th))*sin(th)**2, (6769*cos(th)**4 + 9772*cos(th)**2 + 784)*sin(th)**3, (-1960*cos(th)**3 - 1190*cos(th))*sin(th)**4, (322*cos(th)**2 + 56)*sin(th)**5, -28*sin(th)**6*cos(th), sin(th)**7])

备注：内容来源于stack exchange，提问作者Pavel Komarov