How can I simplify iterative trigonometric derivatives in SymPy without introducing higher-frequency terms?

Rewriting to exp and doing the simplifications seems to work in this case:

PYTHONCopied
from collections import deque
from sympy import *
import sympy as sp
th = sp.symbols('th')
order = 4
exprs = deque([i.simplify() for i in [-6*cos(th)**3 - 9*cos(th), (11*cos(th)**2 + 4)*sin(th), -6*sin(th)**2*cos(th), sin(th)**3]])

def simpler(e):
    return e.rewrite(exp).simplify().rewrite(cos).cancel()

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(simpler(term/sp.sin(th)))
        q = p
    this = simpler(q/sp.sin(th))
    exprs.append(this)
    print(nu,exprs)

outputs

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

If you just want to get the constants at the end, don't simplify until the end. Here is a complete working example:

PYTHONCopied
from collections import deque
import sympy as sp
th = sp.symbols('th')
E = {4:deque([-6*cos(th)**3 - 9*cos(th), (11*cos(th)**2 + 4)*sin(th), -6*sin(th)**2*cos(th), sin(th)**3]),
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]),
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]),
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]),
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])}
from time import time
for order in E:
 t=time()
 exprs = E[order]
 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)))
 t1=time()-t
 t=time()
 print(order,round(t1,2),[i.rewrite(sp.exp).simplify() for i in exprs],round(time()-t,2))

4 0.06 [36, 0, 49, 0, 14, 0, 1] 0.16
5 0.13 [-576, 0, -820, 0, -273, 0, -30, 0, -1] 0.4
6 0.29 [14400, 0, 21076, 0, 7645, 0, 1023, 0, 55, 0, 1] 0.68
7 0.59 [-518400, 0, -773136, 0, -296296, 0, -44473, 0, -3003, 0, -91, 0, -1] 1.14
8 0.97 [25401600, 0, 38402064, 0, 15291640, 0, 2475473, 0, 191620, 0, 7462, 0, 140, 0, 1] 1.7
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