+68 Reproduce the angle difference identities for sine and cosine by expressing eixeiy in rectangular form.
+118703 To be honest, I am not really sure what this question means ..but
eixeiy=cos(x)+isin(x)cos(y)+isin(y)=cos(x)+isin(x)cos(y)+isin(y)∗cos(y)−isin(y)cos(y)−isin(y) =cos(x)cos(y)−icos(x)sin(y)+isin(x)cos(y)+sin(x)sin(y)cos2(y)+sin2(y) =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
eixeiy=e(ix−iy)=ei(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)
