What real function is #(e^(ix)-e^(-ix))/(ie^(ix)+ie^(-ix))# to?

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Answer 1 · 2018-05-11T04:53:15

#tan x#

Using

#e^{ix} = cos x + i sin x#

and its conjugate

#e^{-ix} = cos x-i sin x#

we get

#e^{ix}+e^{-ix} = 2 cos x#
and
#e^{ix}-e^{-ix} = 2i sin x#

Thus

#(e^(ix)-e^(-ix))/(ie^(ix)+ie^(-ix)) = (2i sin x)/(i 2 cos x) = tan x#