2 ms·

You don't need to treat "[cos(x)]^2 + [sin(x)]^2 = 1" separately either. e^(iθ) times e^(-iθ) = e^(iθ-iθ) = e^0 = 1 = (cosθ+isinθ)(cosθ-isinθ) = cos²θ + sin²θ

• josephcsible 6y ago
  Sure, this takes "[cos(x)]^2 + [sin(x)]^2 = 1" off the list, but doesn't it require you to put "cos(-x) = cos(x)" and "sin(-x) = -sin(x)" on it, to get "(cosθ-isinθ)" from "e^(-iθ)"?
  • evanb 6y ago
    Only if you forget about how complex conjugation works. (e^-ix) = (e^ix)* = cos x - i sin x
  • sriku 6y ago
    As indicated in the other reply, think of the minus as changing the sign of i instead of x.