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Prove that:


What I have tried.

$$\cosh\left(\frac{\pi}{2}\right)=\cos\left(i\frac{\pi}{2}\right)$$ $$=Re\{e^{i.i\frac{\pi}{2}}\}$$ $$=Re\{e^{-\frac{\pi}{2}}\}$$

Why is $e^{-\frac{\pi}{2}}$ not answer any why is $$\frac{e^{-\frac{\pi}{2}}+e^{\frac{\pi}{2}}}{2}$$

a correct solution. Did I miss something somewhere?

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The relation $\cos(\theta)=\mathcal{R}\left(e^{i\theta}\right)$ only holds when $\theta$ is real. – Steven Stadnicki Feb 6 '13 at 21:49
How do you define $\cosh$ and $\cos$? $\cosh$ is usually defined as $\cosh \theta = (e^\theta + e^{-\theta}) / 2$. Do you have another definition? – Ayman Hourieh Feb 6 '13 at 21:54
@AymanHourieh, no it's solved. I was stupid. – user45099 Feb 6 '13 at 21:55
up vote 4 down vote accepted

$\cosh(x)$ is usually defined defined as $\frac{e^{x} + e^{-x}}{2}$. If you haven't some different definition, then it is quite straightforward: $$\cosh\left(\frac{\pi}{2}\right)=\frac{e^{\frac{\pi}{2}} + e^{-\frac{\pi}{2}}}{2} = \frac{1}{2}e^{-\frac{\pi}{2}}(1 + e^x)$$

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$\left(1+\frac1{1^2}\right)\left(1+\frac1{3^2}\right)\left(1+\frac1{5^2}\right)\left(1+\frac1{7^2}\right)\dotsm= \cosh \frac\pi 2$. SMS Product Series item no. 2 gives this product but omits the first factor.

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Welcome to math.SE: I have tried to improve the readability of your question by introducing Tex. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. – A.P. Apr 21 '13 at 11:08

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