What does $(e^(ix)-e^(-ix))/(2i)$ equal?

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Answer 1 · 2015-10-21T18:40:51

$sin x$

Use the following identities:

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

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

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

So:

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

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

$= 2i sin(x)$

So:

$(e^(ix) - e^(-ix))/(2i) = sin(x)$

What does (e^(ix)-e^(-ix))/(2i)(e^(ix)-e^(-ix))/(2i) equal?