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How do I prove that $\sin(π/2+iy)=1/2(e^{y}+e^{−y})=\cosh y$?

Can you help please?

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    Thanks for your comments helped me a lot2012-05-08

1 Answers 1

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The definition of $\cosh(y)$ is $\cosh(y)=\frac{e^y+e^{-y}}{2}.$ The definition of $\sin(z)$ (or a property, if you use some other definition) is $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}.$ Thus $\sin(\tfrac{\pi}{2}+iy)=\frac{e^{i\left(\tfrac{\pi}{2}+iy\right)}-e^{-i\left(\tfrac{\pi}{2}+iy\right)}}{2i}=\frac{e^{\pi i/2}e^{-y}-e^{-\pi i/2}e^y}{2i}.$ Now consider what $e^{\pi i/2}$ is, and you will be done.

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    This is equal to i and with this result is obtained2012-05-08