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I've tried a few formats and ideas, but the syntax is kind of confusing. Help is appreciated, danke.

The function itself

KarlMarxsDog
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    I would start with learning about Python's [arithmetic operators](https://docs.python.org/3/library/stdtypes.html#numeric-types-int-float-complex) and the [`math`](https://docs.python.org/3/library/math.html) module. Otherwise, the main thing to keep in mind is that `sin^5(...)` is really `(sin(...))^5`. – chepner Jul 08 '21 at 13:59
  • Exactly what I was looking for, Chepner, thanks. – KarlMarxsDog Jul 08 '21 at 14:15

4 Answers4

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Only with python built-in functios:

import math
r = math.e**(math.cos(theta)) - 2 * math.cos(4 * theta) + math.sin(theta/12)**5

With Sympy(for symbolic computation):

from sympy import Symbol, cos, sin, E
t = Symbol('Θ')
E**(cos(t)) - 2 * cos(4 * t) + sin(t/12)**5

Result:

Hugo Amorim
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from math import exp, cos, sin

theta = 0.1  # Just as an example
r = exp(cos(theta)) - 2 * cos(4 * theta) + sin(theta / 12) ** 5
Pippet39
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A very straightforward, but naive approach would be to just use the math library, which is a standard library in python.

import math

def r(theta):
    return math.pow(math.e, math.cos(theta)) - 2 * math.cos(4*theta) + math.pow(math.sin(theta/12), 5)

better results are possible with libraries for scientific computing, like numpy or other members of the scipy ecosystem.

Simon B
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How about something like this?

import matplotlib.pyplot as plt
import numpy as np

theta = np.linspace(-2*np.pi, 2*np.pi, 10)
r = np.exp(np.cos(theta)) - 2*np.cos(4*theta) + np.sin(theta/12)**5

plt.plot(theta,r)
plt.savefig('parametric.jpg')

enter image description here

user32882
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