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Question Number 81010 by M±th+et£s last updated on 08/Feb/20

$$showthat$$$${(cos\theta -isin\theta )}^n=(cosn\theta -isin\theta )$$

Commented by MJS last updated on 08/Feb/20

$$\mathrm{wrong}$$

Commented by M±th+et£s last updated on 08/Feb/20

$$bringitfromanexamandithinkthatitswrong$$

Commented by MJS last updated on 08/Feb/20

$$\mathrm{obviously}\mathrm{it}\mathrm{should}\mathrm{be}$$$${(\cos \theta -\mathrm{i}\sin \theta )}^n=\cos n\theta -\mathrm{i}\sin n\theta$$$$\mathrm{we}\mathrm{know}\mathrm{that}$$$$-\sin \theta =\sin -\theta \wedge \cos \theta =\cos -\theta$$$$\Rightarrow$$$$z=\cos \theta -\mathrm{i}\sin \theta =\cos -\theta +\mathrm{i}\sin -\theta$$$$\mathrm{let}-\theta =\alpha$$$$z=\cos \alpha +\mathrm{i}\sin \alpha$$$$\mathrm{we}\mathrm{know}\mathrm{that}$$$$z=\cos \alpha +\mathrm{i}\sin \alpha ={\mathrm{e}}^{\mathrm{i}\alpha }$$$$\Rightarrow z^n={\mathrm{e}}^{\mathrm{i}n\alpha }=\cos n\alpha +\mathrm{i}\sin n\alpha =$$$$=\cos n\theta -\mathrm{i}\sin n\theta$$

Commented by abdomathmax last updated on 09/Feb/20

$$thecorrectQisprovethat$$$${(cos\theta -isin\theta )}^n=cos\left(n\theta \right)-isin\left(n\theta \right)$$$$byrecurrencen=1\left(true\right)$$$$letdupposeittrue\Rightarrow$$$${(cos\theta -isin\theta )}^{n+1}={(cos\theta -isin\theta )}^n(cos\theta -isin\theta )$$$$=e^{-in\theta }\times e^{-i\theta }=e^{-i(n+1)\theta }=cos(n+1)\theta -isin(n+1)\theta$$$$therelationistrueatterm(n+1)$$

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