\(\sin x+i\cos x=i(\cos x-i\sin x)=ie^{-ix}=e^{i\frac\pi2}e^{-ix}=e^{i\left(\frac\pi2-x\right)}\)
By DeMoivre's theorem,
\((\cos y+i\sin y)^n=(e^{iy})^n=e^{iny}=\cos(ny)+i\sin(ny)\)
which would here mean that
\((\sin x+i\cos x)^n=\left(e^{i\left(\frac\pi2-x\right)}\right)^n=e^{i\left(\frac{n\pi}2-nx\right)}=\cos\left(\dfrac{n\pi}2-nx\right)+i\sin\left(\dfrac{n\pi}2-nx\right)\)
as required.
