DeMoivre's Theorem expand's on Euler's formula:
e^(ix)=cosx+isinx
DeMoivre's Theorem says that:
- (e^(ix))^n=(cosx+isinx)^n
- (e^(ix))^n=e^(i nx)
- e^(i nx)=cos(nx)+isin(nx)
- cos(nx)+isin(nx)-=(cosx+isinx)^n
Example:
cos(2x)+isin(2x)-=(cosx+isinx)^2
(cosx+isinx)^2=cos^2x+2icosxsinx+i^2sin^2x
However, i^2=-1
(cosx+isinx)^2=cos^2x+2icosxsinx-sin^2x
Resolving for real and imaginary parts of x:
cos^2x-sin^2x+i(2cosxsinx)
Comparing to cos(2x)+isin(2x)
cos(2x)=cos^2x-sin^2x
sin(2x)=2sinxcosx
These are the double angle formulae for cos and sin
This allows us to expand cos(nx) or sin(nx) in terms of powers of sinx and cosx
DeMoivre's theorem can be taken further:
Given z=cosx+isinx
z^n=cos(nx)+isin(nx)
z^(-n)=(cosx+isinx)^(-n)=1/(cos(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+z^(-n)=2cos(nx)
z^n-z^(-n)=2isin(nx)
So, if you wanted to express sin^nx in terms of multiple angles of sinx and cosx:
(2isinx)^n=(z-1/z)^n
Expand and simply, then input values for z^n+z^(-n) and z^n-z^(-n) where necessary.
However, if it involved cos^nx, then you would do (2cosx)^n=(z+1/z)^n and follow the similar steps.
