Programming-how to find the angle

Question

I want to write a function in python that has as two arguments, the sine and the cosine of an angle and it returns the angle. Which formula do we use to calculate the angle knowing the sine and the cosine? If we could just use the inverse trigonometric functions we wouldn't have to define two arguments for the function. I don't have an idea how we can calculate the angle. Can you help please?

Answer 1

math.atan2(c, s)

will produce the angle, where $c$ is the known cosine and $s$ is the known sine. The value produced will lie between $-\pi$ and $\pi$; if you want it to be between $-180$ degrees and $180$ degrees, use

math.atan2(c, s) * 180.0/math.pi

Answer 2

If you are not able to use any inverse function, given the sine and the cosine of an angle, you can compute the tangent of the angle:

$$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$$

And then, compute the $\arctan$ using the power series:

$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - ...$$ or $$\arctan{x} = \sum_{n=0}^{\infty}(-1)^n\cdot\frac{x^{2n+1}}{2n+1}$$

In python code it would be something like this:

def get_angle(sinx, cosx):
    tanx = sinx / cosx
    arctanx = 0
   
    # More iterations means more accuracy
    for n in range(0, 20):
        arctanx += ((-1)**n)*((tanx**(2*n+1)) / (2*n+1))

    return arctanx

Answer 3

If math.atan2 (as mentioned in John's answer) isn't allowed, then you could use:

import cmath
    
def get_angle(sinx, cosx):
    return cmath.log(cosx + 1j * sinx).imag

This works because Euler's formula $e^{ix} = \cos(x) + i \sin(x)$ is equivalent to $\log(\cos(x) + i \sin(x)) = ix$.

Your instructor may still consider it “an inverse function”, though.

Answer 4

A non-“cheating” approach to use if the standard math.atan2 is not allowed:

import math

def atan2(y, x):
    # Use symmetry to restrict the result range to [0, pi/4]
    if y < 0:
        return -atan2(-y, x)
    if x < 0:
        return math.pi - atan2(y, -x)
    if y > x:
        return math.pi / 2 - atan2(x, y)
    # Normalize to unit circle to give c = cos(theta), s = sin(theta)
    r = math.hypot(x, y)
    c = x / r
    s = y / r
    # Use the small-angle identity sin(theta) = theta when applicable.
    if s <= 2e-8:
        return s
    # Otherwise, use half-angle identity
    tan_half = s / (1 + c)
    return 2.0 * atan2(tan_half, 1)

Note that there are no calls to trig functions nor inverse trig functions. Instead, the algorithm works by recursively bisecting the angle until the small-angle approximation can be used.