The real part of {e^{e^{itheta }}} is | vedclass

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Question

(A) We know that ${e^{i	heta } = \cos 	heta + i\sin 	heta }$.Therefore,${e^{e^{i	heta }} = e^{\cos 	heta + i\sin 	heta } = e^{\cos 	heta } \cdo

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${e^{\cos 	heta }}\cos (\sin 	heta )$

Vedclass · Answer

(A) We know that ${e^{i	heta } = \cos 	heta + i\sin 	heta }$.Therefore,${e^{e^{i	heta }} = e^{\cos 	heta + i\sin 	heta } = e^{\cos 	heta } \cdot e^{i\sin 	heta }}$.Using Euler's formula,${e^{i\sin 	heta } = \cos(\sin 	heta ) + i\sin(\sin 	heta )}$.Thus,${e^{e^{i	heta }} = e^{\cos 	heta } [\cos(\sin 	heta ) + i\sin(\sin 	heta )] = e^{\cos 	heta } \cos(\sin 	heta ) + i e^{\cos 	heta } \sin(\sin 	heta )}$.The real part is ${e^{\cos 	heta } \cos(\sin 	heta )}$.

The real part of ${e^{e^{i\theta }}}$ is

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