+68 Reproduce the angle difference identities for sine and cosine by expressing \(\frac{e^{ix}}{e^{iy}}\) in rectangular form.
+118687 To be honest, I am not really sure what this question means ..but
\(\frac{e^{ix}}{e^{iy}}\\ =\frac{cos(x)+isin(x)}{cos(y)+isin(y)}\\ =\frac{cos(x)+isin(x)}{cos(y)+isin(y)}*\frac{cos(y)-isin(y)}{cos(y)-isin(y)}\\~\\ =\frac{cos(x)cos(y)-icos(x)sin(y)+isin(x)cos(y) +sin(x)sin(y) }{cos^2(y)+sin^2(y)}\\~\\ =\frac{cos(x)cos(y) +sin(x)sin(y) +i[-cos(x)sin(y)+sin(x)cos(y) ] }{1}\\~\\ =cos(x-y)+isin(x-y)\)
ALSO
\(\frac{e^{ix}}{e^{iy}}\\ =e^{ (ix-iy) }\\ =e^{ i(x-y) }\\ =cos(x-y)+isin(x-y)\)
I must be almost finished but i don't get what the question actually wants. 
LaTex
\frac{e^{ix}}{e^{iy}}\\
=\frac{cos(x)+isin(x)}{cos(y)+isin(y)}\\
=\frac{cos(x)+isin(x)}{cos(y)+isin(y)}*\frac{cos(y)-isin(y)}{cos(y)-isin(y)}\\~\\
=\frac{cos(x)cos(y)-icos(x)sin(y)+isin(x)cos(y) +sin(x)sin(y) }{cos^2(y)+sin^2(y)}\\~\\
=\frac{cos(x)cos(y) +sin(x)sin(y) +i[-cos(x)sin(y)+sin(x)cos(y) ] }{1}\\~\\
=cos(x-y)+sin(x-y)
\frac{e^{ix}}{e^{iy}}\\
=e^{ (ix-iy) }\\
=e^{ i(x-y) }\\
=cos(x-y)+isin(x-y)
