Python: Find the degree of the Angle [0:360] from cos(a) and sin(a) values

user_13183396 0
Viewed 4066

I would like to find the degree between 0 to 360 of my angle. I have a DataFrame with 2 columns: cos and sin values.

df['cos'] = vector values between 0 and 1

df['sin'] = vector values between 0 and 1
trigonometry python
4 Answers

I gues you mean something like:

import math

angle = math.degrees(math.acos(df['cos']))

To really stay in [0, 360] you will have to check for negative cos and adapt the code like:

import math
a_acos = math.acos(df['cos'])
if df['sin'] < 0:
   angle = math.degrees(-a_acos) % 360
else: 
   angle = math.degrees(a_acos)
user_106982610

Don't mess with sign check. You need both cos and sin

import math
for i in range(360):
   angle = i * math.pi / 180
   cs = math.cos(angle)
   sn = math.sin(angle)
   angle2 = math.atan2(sn, cs)  # ALWAYS USE THIS
   angle2 *= 180 / math.pi
   if angle2 < 0: angle2 += 360
   print(angle2)
user_14384650

You can use the numpy module, which contains trigonometric functions that work on vectors, such as arcsin and arccos, which take sin and cos values and return the angle. You can use the degrees function to convert from radians to degrees.

user_48522900

The best way of doing this is by using the np.angle function which returns the 'angle' associated to a complex number. For a bit of theory, any complex number z has a magnitude r and an angle theta, and is given by

z = r*cos(theta) + 1j * r*sin(theta)

np.angle takes a complex number as input and returns the angle in radians from $-pi$ to $pi$ (corresponding to -180 to 180 in degrees). Which means that what you're looking for is essentially this

angle_negpi_to_pi = np.angle(df['cos'] + 1j*df['sin'])
angle = ((angle_negpi_to_pi + 2*np.pi) % (2*np.pi)) * 180/np.pi
user_31407500
Related