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

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

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