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"Express $\sin(z)$ and $\cos(z)$ in rectangular form."

For $z \in \mathbb{C}$ (complex numbers), we have defined \begin{equation} \sin (z)=\frac{e^{iz}-e^{-iz}}{2i} \end{equation} and \begin{equation} \cos(z)=\frac{e^{iz}+e^{-iz}}{2} \end{equation} I believe this is the polar form. Wikipedia helps out, by stating that \begin{align} \sin(x + iy) = \sin (x) \cosh (y) + i \cos (x) \sinh (y) \\ \cos(x + iy) = \cos (x) \cosh (y) - i \sin (x) \sinh (y) \end{align} However, how would one derive this? Thanks for your help!

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  • $\begingroup$ What happens when you take the right-hand-side, convert to exponentials, and simplify? $\endgroup$ – GEdgar Jan 21 '15 at 16:16
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Hint: Let $z = x+iy$ where $z$ is now expressed in Cartesian coordinates. Then use the expression given to you for sine, $$\sin(x+iy) = \frac{e^{x+iy}-e^{-(x+iy)}}{2i}$$ and use a bit of algebra to get your result. It may also help to expand $$\sin(x)\cosh(y)+i\cos(x)\sinh(y)$$ and see how you can make it become the LHS.

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