How do i proof #tan2x=(2tanx)/(1-tan^2x)# with moivre?

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Feb 9, 2018

By de Moivre's theorem we can write

#cos2x+isin2x=(cosx+isinx)^2#

#=>cos2x+isin2x=cos^2x-sin^2x+i*2sinxcosx#

So equating real parts of both sides we get
#cos2x=cos^2x-sin^2x#

And equating imaginary parts of both sides we get
#sin2x=2sinxcosx#

Hence

#tan2x=(sin2x)/(cos2x)=(2sinxcosx)/(cos^2x-sin^2x)#

#=((2sinxcosx)/cos^2x)/(cos^2x/cos^2x-sin^2x/cos^2x)#

#=(2tanx)/(1-tan^2x)#

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