Question

(sinx + icosx)^n = cos( nπ\2 - nx) + isin( nπ\2 - nx)

Answer 1

`\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(\frac{n\pi}2-nx\right)+i\sin\left(\frac{n\pi}2-nx\right)`

as required.