Finding Cosine Using Python

I must write a function that computes and returns the cosine of an angle using the first 10 terms of the following series: cosx = 1 - (x**2)/2! + (x**4)/4! - (x**6)/6!.... I can't

Solution 1:

Maybe something like this:

#! /usr/bin/python3.2

def cos(a):
    d =1c=1for i inrange(2,20,2):
        d *= i *(i -1)sign=-1if i % 4else1
        print ('adding {} * a ** {} / {}'.format (sign, i, d))c+=sign* a ** i / d
        print ('cosine is now {}'.format (c))returnccos(1.0)

Basically d (as in Denominator) is your accumulator.

Solution 2:

Note: If you are using Python2.x, you should use

from __future__ import division

as the first line of the file

---

One way is to alternate the sign like this

def cos(angle):sign=-1
    cosx =1for i inrange(2,20,2):
         cosx +=sign*(x**i)/(i)sign=-sign

You'll still need to get the factorial part correct

Here is a simpler version that calculates each term based on the previous one.

defcos(x):
    res = 0
    term = 1for i inrange(1, 20, 2):
        res += term
        term *= -x * x/ i /(i + 1)
    return res

Solution 3:

Why not make a function for factorial? (I added an optional parameter for repetitions):

EDIT:

as per a comment, here's a "learning" factorial function that will make sure not to recalculate any values (granted this will only work up to 199!, I'm assuming there will never be more than 99 repetitions with the cos function since it will probably throw an overflow error around 90 already):

facts = [0]*200
facts[0] = 1deffactorial(x): 
    if(facts[x] != 0): return facts[x]
    return x * factorial(x-1)

defcos(x, reps=10):
    final_val = 1
    neg = -1for n inrange(2, reps*2, 2):
        final_val += neg*(x**n)/factorial(n)
        neg *= -1return final_val

print cos(3.14/2)

Solution 4:

For calculating series, use the power of list comprehensions.

In [1]: a = range(1,11)

In [2]: a
Out[2]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

The mul function multiplies the elements of the list it is given:

In [3]: def mul(r):
   ...:     rv =1
   ...:     for j in r:
   ...:         rv *= j
   ...:     return rv
   ...: 

Now you can generate the factorials:

In [7]: b = [mul(a[:i]) for i in a]

In [8]: b
Out[8]: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]

To pick the right elements from the lists;

In[10]: a[1::2]Out[10]: [2, 4, 6, 8, 10]In[11]: b[1::2]Out[11]: [2, 24, 720, 40320, 3628800]

To generate the plusses and minuses:

In[12]: [(-1)**k for k in a[:5]]
Out[12]: [-1, 1, -1, 1, -1]

Combine all the elements;

In [14]: data = zip(a[1::2], b[1::2], [(-1)**k for k in a[:5]])

Then do the calculation for a certain x;

In [21]: x = 12

In [22]: sum([s*x**t/float(b) for t, b, s in data])
Out[22]: -9753.737142857142