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How would you get $$e^{\cos\theta+i\sin\theta}=e^{\cos\theta}(\cos(\sin\theta)+i(\sin(\sin\theta)))$$

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You're reading a pair of parentheses in WolframAlpha's answer that aren't there. WA yields $e^{\cos\theta}\cos(\sin\theta) + i\sin(\sin\theta)e^{\cos\theta} = e^{\cos\theta}\cos(\sin\theta) + e^{\cos\theta}i\sin(\sin\theta) = e^{\cos\theta}(\cos(\sin\theta) + i\sin(\sin\theta))$, which is what your notes gave you. – jwodder Dec 2 '11 at 2:38
up vote 5 down vote accepted

Just write $e^{\cos \theta + i \sin \theta} = e^{\cos \theta} e^{i \sin \theta}$ and use the formula for $e^{ix}$ with $x = \sin \theta$.

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Thanks for your help! May I ask, is there a reason why sometimes it's written in that form? – Ray Dec 2 '11 at 0:27
The real question is why anybody would want to consider $e^{e^{i\theta}}$ (other than just to give you practice in using Euler's formula). The answer might be that they want to see what the conformal map $z \to e^z$ does to the unit circle. To plot the image as a parametric curve, you might want to take the real and imaginary parts, and those are $e^{\cos \theta} \cos(\sin \theta)$ and $e^{\cos \theta} \sin(\sin \theta)$. – Robert Israel Dec 2 '11 at 0:41
It could be an interesting exercise to see if it can help solve certain integrals involving iterated trig functions. – Aleks Vlasev Dec 2 '11 at 6:18
You mean like $J_0(x) = \frac{1}{2\pi} \int_0^{2 \pi} \cos(x \cos(t))\ dt$ (where $J_0$ is the Bessel function of the first kind and order $0$) and its relatives? – Robert Israel Dec 2 '11 at 19:14
@Aleks, like this? – J. M. Dec 3 '11 at 1:32

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