What real function is #(e^(ix)-e^(-ix))/(ie^(ix)+ie^(-ix))# equal to?
1 Answer
Oct 24, 2015
Explanation:
#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)#