3 ms·
If we use a degrees version of sin and cos (call them sind and cosd), then we cannot have e on the left side without a conversion factor. (iπx/180)
by kazinator 2y ago
If we use a degrees version of sin and cos (call them sind and cosd), then we cannot have e on the left side without a conversion factor.
(iπx/180)
e = cosd x + i sind x
ix
π/180
-> e = cosd x + i sind x
π/180
-> let f = e
ix
f = cosd x + i sind x
Probem is, f doesn't have nice properties like:
d x x
- f /= f
dx
There is something uniquely special about the unit circle, and about using the unscaled distance around the unit circle as the measure of the angle.
- kazinator 2y agoRadians have the property that if we step x by some tiny amount δ, then the cos/sin coordinates will move by that same distance around the unit circle: |[cos(x+δ) + i sin(x+δ)] - [cos(x) + i sin(x)]| = δ This is also related to how we can estimate sin(x) = x for small values next to zero, if using radians. In radians, the derivative sin'(x) is cos(x), and cos'(x) is -sin(x). Derviation just shifts the waveform left by ninety degrees. In units other than radians, we get wacky constant terms that change at each step. That's related to how e^x is its own derivative.