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Reproduce the angle difference identities for sine and cosine by expressing $$\frac{e^{ix}}{e^{iy}}$$ in rectangular form.

Jul 16, 2021

#2
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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)

Jul 17, 2021