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

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Answer 1 · 2015-10-25T12:40:15

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

but I think you meant to ask...

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

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

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

So with the question as asked:

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

I think you may have been wanting one of the following results:

$(e^(ix)+e^(-ix))/2$

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

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

$= cos(x)$

$color(white)()$

$(e^(ix)-e^(-ix))/(2i)$

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

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

$= sin(x)$

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