What real function is #(e^(ix)-e^(-ix))/(ie^(ix)+ie^(-ix))# equal to?

1 Answer
Oct 24, 2015

Answer:

#tan(x)#

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)#