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

1 Answer
Oct 21, 2015

Answer:

#sin x#

Explanation:

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)#