The product of $1-e^{ix}$ and $1-e^{-ix}$

Question

I came across a question which needs to simplify $(1-e^{ix})(1-e^{-ix})$ (It's a small part of mathematics in a physics question). I tried to simplify it but I obtained two answers instead after using two different approaches.

Approach 1: Expanding the terms

$(1-e^{ix})(1-e^{-ix})$

$=2 - (e^{ix} + e^{-ix})$

$=2(1-\cos x)$

$=4 \sin^2(\frac{x}{2})$

Approach 2:

$(1-e^{ix})(1-e^{-ix})$

$= e^{\frac{ix}{2}}(e^{-\frac{ix}{2}}-e^{\frac{ix}{2}})e^{-\frac{ix}{2}}(e^{\frac{ix}{2}}-e^{-\frac{ix}{2}})$

$=-(e^{\frac{ix}{2}}-e^{-\frac{ix}{2}})^2$ [the two exponents cancel each other and the sign of the first bracket is flipped so that both bracket terms are the same]

$=-(2\sin(\frac{x}{2}))^2$

$=-4\sin^2(\frac{x}{2})$

It is clear that both approaches give inconsistent answers. Please, can anyone check my work and let me know whether I made any mistake? Thank you.

Answer 1

$e^{ix/2}-e^{-ix/2}=cos(x/2)+isin(x/2)-(cos(-x/2)+isin(-x/2))=2isin(x/2)$ implies that $-(e{ix/2}-e^{-ix/2})^2=-(2i)^2sin^2(x/2)=4sin^2(x/2)$.

Answer 2

$$e^{-iy}=\cos(-y)+i\sin(-y)=?$$

$$e^{iy}-e^{-iy}=?$$

Answer 3

You should have $$-(e^{\frac{ix}{2}}-e^{-\frac{ix}{2}})^2 = -(i^2)(2\sin(\frac{x}{2}))^2=(2\sin(\frac{x}{2}))^2 =4\sin^2(\frac{x}{2})$$