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

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Answer 1 · 2015-10-24T13:35:47

$tan(x)$

$e^(ix) = cos(x)+i sin(x)$

$cos(-x) = cos(x)$

$sin(-x) = -sin(x)$

So:

$e^(ix)-e^(-ix) = (cos(x)+i sin(x))-(cos(-x)+i sin(-x))$

$=(cos(x) + i sin(x))-(cos(x)-i sin(x)) = 2i sin(x)$

And:

$e^(ix)+e^(-ix) = (cos(x)+i sin(x))+(cos(-x)+i sin(-x))$

$=(cos(x) + i sin(x))+(cos(x)-i sin(x)) = 2 cos(x)$

So:

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

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