What does #(e^(ix)-e^(-ix))/(2i)# equal?
1 Answer
Oct 21, 2015
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)#
