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Hi I am looking for a way to calculate/simplify the following $$ e^{2i \phi}= \cos(2\phi)+i\sin(2\phi). $$ where $\phi=\arctan(y/x)$. I want to simplify so it is an expression with no trig functions. As done for the more simple case, $$ e^{i\phi}=\cos \phi+i\sin \phi $$ So I get $$ \cos \phi =\cos(\arctan(y/x))=\frac{x}{\sqrt{x^2+y^2}},\quad \sin \phi = \frac{y}{\sqrt{x^2+y^2}} $$ Thus I can simplify $e^{i\phi}$ easily, but How can I calculate $\cos (2\phi), \sin(2\phi)$ ? (without using wolfram alpha) $$ \cos(2\phi)=\cos(2\arctan(y/x))=?,\quad \sin(2\phi)=\sin(2\arctan(y/x))=? $$ I get stuck here. Thanks a lot!

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Hint

$$ \cos 2\phi = 2\cos^2 \phi - 1 = \frac{2x^2}{x^2 + y^2} - 1 = \frac{x^2 - y^2}{x^2+y^2} $$

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cos(2tan$^{-1}$($\frac{y}{x}$)).

We if Z = tan$^{-1}$($\frac{y}{x}$) then, tan(Z) = $\frac{y}{x}$. What you want is cos(2tan$^{-1}$($\frac{y}{x}$)) = cos(2Z) = $\frac{1-tan^{2}Z}{1+tan^{2}Z}$ and you already have tan(Z).

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