3 ms·
That's because of the equality relationship between 2π radians and the length of the unit circle perimeter. If one is working with a sine taking in turns, one c
by shadowgovt 4y ago
That's because of the equality relationship between 2π radians and the length of the unit circle perimeter. If one is working with a sine taking in turns, one can just adjust by saying sin(q) ≈ 2π * q for small q.