DeMoivre's Theorem expand's on Euler's formula:
e^(ix)=cosx+isinxeix=cosx+isinx
DeMoivre's Theorem says that:
- (e^(ix))^n=(cosx+isinx)^n(eix)n=(cosx+isinx)n
- (e^(ix))^n=e^(i nx)(eix)n=einx
- e^(i nx)=cos(nx)+isin(nx)einx=cos(nx)+isin(nx)
- cos(nx)+isin(nx)-=(cosx+isinx)^ncos(nx)+isin(nx)≡(cosx+isinx)n
Example:
cos(2x)+isin(2x)-=(cosx+isinx)^2cos(2x)+isin(2x)≡(cosx+isinx)2
(cosx+isinx)^2=cos^2x+2icosxsinx+i^2sin^2x(cosx+isinx)2=cos2x+2icosxsinx+i2sin2x
However, i^2=-1i2=−1
(cosx+isinx)^2=cos^2x+2icosxsinx-sin^2x(cosx+isinx)2=cos2x+2icosxsinx−sin2x
Resolving for real and imaginary parts of xx:
cos^2x-sin^2x+i(2cosxsinx)cos2x−sin2x+i(2cosxsinx)
Comparing to cos(2x)+isin(2x)cos(2x)+isin(2x)
cos(2x)=cos^2x-sin^2xcos(2x)=cos2x−sin2x
sin(2x)=2sinxcosxsin(2x)=2sinxcosx
These are the double angle formulae for coscos and sinsin
This allows us to expand cos(nx)cos(nx) or sin(nx)sin(nx) in terms of powers of sinxsinx and cosxcosx
DeMoivre's theorem can be taken further:
Given z=cosx+isinxz=cosx+isinx
z^n=cos(nx)+isin(nx)zn=cos(nx)+isin(nx)
z^(-n)=(cosx+isinx)^(-n)=1/(cos(nx)+isin(nx))z−n=(cosx+isinx)−n=1cos(nx)+isin(nx)
z^(-n)=1/(cos(nx)+isin(nx))xx(cos(nx)-isin(nx))/(cos(nx)-isin(nx))=(cos(nx)-isin(nx))/(cos^2(nx)+sin^2(nx))=cos(nx)-isin(nx)z−n=1cos(nx)+isin(nx)×cos(nx)−isin(nx)cos(nx)−isin(nx)=cos(nx)−isin(nx)cos2(nx)+sin2(nx)=cos(nx)−isin(nx)
z^n+z^(-n)=2cos(nx)zn+z−n=2cos(nx)
z^n-z^(-n)=2isin(nx)zn−z−n=2isin(nx)
So, if you wanted to express sin^nxsinnx in terms of multiple angles of sinxsinx and cosxcosx:
(2isinx)^n=(z-1/z)^n(2isinx)n=(z−1z)n
Expand and simply, then input values for z^n+z^(-n)zn+z−n and z^n-z^(-n)zn−z−n where necessary.
However, if it involved cos^nxcosnx, then you would do (2cosx)^n=(z+1/z)^n(2cosx)n=(z+1z)n and follow the similar steps.
