Question

Question #2249a

Answer 1

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)`