If x = cos theta + isin theta and y = cos phi | vedclass

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(C) Given $x = \cos 	heta + i\sin 	heta = e^{i	heta}$ and $y = \cos \phi + i\sin \phi = e^{i\phi}$.Applying De Moivre's theorem,we have ${x^m} = e^

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$2\cos (m	heta + n\phi)$

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(C) Given $x = \cos 	heta + i\sin 	heta = e^{i	heta}$ and $y = \cos \phi + i\sin \phi = e^{i\phi}$.Applying De Moivre's theorem,we have ${x^m} = e^{im	heta}$ and ${y^n} = e^{in\phi}$.Therefore,${x^m}{y^n} = e^{i(m	heta + n\phi)}$ and ${x^{-m}}{y^{-n}} = e^{-i(m	heta + n\phi)}$.Using Euler's formula $e^{i\alpha} + e^{-i\alpha} = 2\cos \alpha$,we get:${x^m}{y^n} + {x^{-m}}{y^{-n}} = e^{i(m	heta + n\phi)} + e^{-i(m	heta + n\phi)} = 2\cos (m	heta + n\phi)$.

If $x = \cos \theta + i\sin \theta$ and $y = \cos \phi + i\sin \phi$,then ${x^m}{y^n} + {x^{-m}}{y^{-n}}$ is equal to

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