For $ t \in [0,\pi] $,
Why does this equality hold
$$
e^{iap\cdot e^{it}} = e^{iap(\cos(t) + i\sin(t))}=e^{-ap\sin(t)}
$$
Specifically, I don't know how to go from the second equality to the third equality.
simplifying $ e^{iape^{it}} $
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$\begingroup$
complex-analysis
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0This is not true – 2017-01-23
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0what about the variables? – 2017-01-23
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0$ a \in \Re \space and \space p>0 $ – 2017-01-23
1 Answers
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\begin{align*} \exp (iape^{it}) &= \exp [iap(\cos t+i\sin t)] \\ &= \exp [-ap\sin t+iap\cos t] \\ &= e^{-ap\sin t}e^{iap\cos t} \\ &= e^{-ap\sin t}[\cos (ap\cos t)+i\sin (ap\cos t)] \end{align*}
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0why does $e^{iap cos(t)} = 1 $ – 2017-01-23
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0No, $$e^{iap\cos t}= \cos (ap\cos t)+i\sin (ap\cos t)$$ – 2017-01-23
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0so the equality is wrong? – 2017-01-23
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0Very likely, double check your raw materials. – 2017-01-23