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.
