Python trigonometric calculations in degrees

Question

I am converting a MATLAB program to Python for a project. I am facing some major issues with converting MATLAB's sind() syntax to Python. I'm using

numpy.sin(numpy.radians())

but some of the results in Python compared to Matlab are displaying tremendous variations. Is there an easier way to tell Python in degrees instead of radians?

Answer 1

In Octave, sind is:

function y = sind (x)
  ...
  I = x / 180;
  y = sin (I .* pi);
  y(I == fix (I) & isfinite (I)) = 0;
endfunction

np.radians (np.deg2rad) is, according to a note x * pi / 180.

So for most values np.sin(np.radians(x)) should be the same as sind(x). There might be some variation off at the limits of floating point precision. Also I'm not sure what that last line is supposed to be doing.

In [327]: np.sin(np.radians(0))
Out[327]: 0.0
In [328]: np.sin(np.radians(90))
Out[328]: 1.0
In [329]: np.sin(np.radians(180))
Out[329]: 1.2246467991473532e-16

>> sind(0)
ans = 0
>> sind(90)
ans =  1
>> sind(180)
ans = 0

The Octave docs add: Returns zero for elements where 'X/180' is an integer.

So yes there may be differences, but 1e-16 is not a tremendous difference.

I can replicate sind with:

def sind(x):
    I = x/180.
    y = np.sin(I * np.pi)
    mask = (I == np.trunc(I)) & np.isfinite(I)
    y[mask] = 0
    return y

In [356]: x=np.array([0,90,180,359,360])
In [357]: np.sin(np.radians(x))
Out[357]: 
array([  0.00000000e+00,   1.00000000e+00,   1.22464680e-16,
        -1.74524064e-02,  -2.44929360e-16])
In [358]: sind(x)
Out[358]: array([ 0.        ,  1.        ,  0.        , -0.01745241,  0.        ])

>> x=[0,90,180,359, 360]
x =
     0    90   180   359   360
>> sind(x)
ans =
   0.00000   1.00000   0.00000  -0.01745   0.00000