Is (1-Exp[-i x])^2 equal to Sin^2(x) in particle physics?

[y35dp]

start point (1-Exp[-i x])^2, (i^2 = -1)

finish point Sin^2(x)

 

[Mentallic]

umm... I don't quite understand what your question is or what to make of what you wrote.

 

[HallsofIvy]

I would interpret this as "given $f(x)= (1- e^{-ix})^2$ show that $f(x)= sin^2(x)$.

Except for the slight problem that they are NOT equal! For example, when $x= \pi/2$, $1- e^{-i\pi/2}= 1+ i$ while $sin^2(\pi/2)= 1$.

y35dp, can you please tell us what the problem really is?

 

[Mentallic]

That was my initial thought, that it was asking to show $$(1-e^{-ix})^2\equiv sin^2x$$ but it isn't true so I was at a complete loss.

Whatever happened to the starter thread layout with the problem, equations and attempt titles?

 

[y35dp]

ok this confirms my thoughts that the two aren't equal this is a particle physics problem but i though the issue was my algebra but the issue must be with my physics!