Why is sin(180) not zero when using python and numpy?

The number π cannot be represented exactly as a floating-point number. So, np.radians(180) doesn't give you π, it gives you 3.1415926535897931.

And sin(3.1415926535897931) is in fact something like 1.22e-16.

So, how do you deal with this?

You have to work out, or at least guess at, appropriate absolute and/or relative error bounds, and then instead of x == y, you write:

abs(y - x) < abs_bounds and abs(y-x) < rel_bounds * y

(This also means that you have to organize your computation so that the relative error is larger relative to y than to x. In your case, because y is the constant 0, that's trivial—just do it backward.)

Numpy provides a function that does this for you across a whole array, allclose:

np.allclose(x, y, rel_bounds, abs_bounds)

(This actually checks abs(y - x) < abs_ bounds + rel_bounds * y), but that's almost always sufficient, and you can easily reorganize your code when it's not.)

In your case:

np.allclose(0, np.sin(np.radians(180)), rel_bounds, abs_bounds)

So, how do you know what the right bounds are? There's no way to teach you enough error analysis in an SO answer. Propagation of uncertainty at Wikipedia gives a high-level overview. If you really have no clue, you can use the defaults, which are 1e-5 relative and 1e-8 absolute.

Faced same problem,

import numpy as npprint(np.cos(math.radians(90)))>> 6.123233995736766e-17

and tried this,

print(np.around(np.cos(math.radians(90)), decimals=5))>> 0

Worked in my case. I set decimal 5 not lose too many information. As you can think of round function get rid of after 5 digit values.

One solution is to switch to sympy when calculating sin's and cos's, then to switch back to numpy using sp.N(...) function:

>>> # Numpy not exactly zero>>> import numpy as np>>> value = np.cos(np.pi/2)6.123233995736766e-17# Sympy workaround>>> import sympy as sp>>> def scos(x): return sp.N(sp.cos(x))>>> def ssin(x): return sp.N(sp.sin(x))>>> value = scos(sp.pi/2)0

just remember to use sp.pi instead of sp.np when using scos and ssin functions.