Respuesta :

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[tex]\sin x+i\cos x=i(\cos x-i\sin x)=ie^{-ix}=e^{i\frac\pi2}e^{-ix}=e^{i\left(\frac\pi2-x\right)}[/tex]

By DeMoivre's theorem,

[tex](\cos y+i\sin y)^n=(e^{iy})^n=e^{iny}=\cos(ny)+i\sin(ny)[/tex]

which would here mean that

[tex](\sin x+i\cos x)^n=\left(e^{i\left(\frac\pi2-x\right)}\right)^n=e^{i\left(\frac{n\pi}2-nx\right)}=\cos\left(\dfrac{n\pi}2-nx\right)+i\sin\left(\dfrac{n\pi}2-nx\right)[/tex]

as required.

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