Python: Calculate sine/cosine with a precision of up to 1 million digits

Question

Question is pretty self-explanatory. I\'ve seen a couple of examples for pi but not for trigo functions. Maybe one could use a Taylor series as done here but I\'m not entire

Answer 1

import math
x = .5
def sin(x):
    sum = 0
    for a in range(0,50): #this number (50) to be changed for more accurate results
        sum+=(math.pow(-1,a))/(math.factorial(2*a+1))*(math.pow(x,2*a+1))
    return sum

ans = sin(x)
print(str.format('{0:.15f}', ans)) #change the 15 for more decimal places

Here is an example of implementing the Taylor series using python as you suggested above. Changing to cos wouldn't be too hard after that.

EDIT:

Added in the formatting of the last line in order to actual print out more decimal places.

Answer 2

Try this

import math
from decimal import *


def sin_taylor(x, decimals):
    p = 0
    getcontext().prec = decimals
    for n in range(decimals):
        p += Decimal(((-1)**n)*(x**(2*n+1)))/(Decimal(math.factorial(2*n+1)))
    return p


def cos_taylor(x, decimals):
    p = 0
    getcontext().prec = decimals
    for n in range(decimals):
        p += Decimal(((-1)**n)*(x**(2*n)))/(Decimal(math.factorial(2*n)))
    return p

if __name__ == "__main__":
    ang = 0.1
    decimals = 1000000
    print 'sin:', sin_taylor(ang, decimals)
    print 'cos:', cos_taylor(ang, decimals)

Answer 3

mpmath is the way:

from mpmath import mp
precision = 1000000
mp.dps = precision
mp.cos(0.1)

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If unable to install mpmath or any other module you could try polynomial approximation as suggested.

where Rn is the Lagrange Remainder

Note that Rn grows fast as soon as x moves away from the center x₀, so be careful using Maclaurin series (Taylor series centered in 0) when trying to calculate sin(x) or cos(x) of arbitrary x.