Question

How to make a formule for sin(3x) and cos(3x) (with the help of De Moivre formulas) in function of sin(x) and cos(x)? Thank you!

Answer 1

The answers are `cos3x=4cos^3x-3cosx` and `sin3x=3sinx-4sin^3x`

Explanation:

We need

`cos^2x+sin^2x=1`

According to Demoivre's theorem

`(cosx+isinx)^n=cosnx+isinnx`

Here

`n=3`

So,

`cos3x+isin3x=(cosx+isinx)^3`

`=cos^3x+3icos^2xsinx+3i^2cosxsin^2x+i^3sin^3x`

`cos^3x+3icos^2xsinx-3cosxsin^2x-isin^3x`

`=(cos^3x-3cosxsin^2x)+i(3cos^2xsinx-sin^3x)`

Comparing the real parts and the imaginay parts

`cos3x=cos^3x-3cosxsin^2x=cos^3x-3cosx(1-cos^2x)`

`=cos^3x-3cosx+3cos^3x`

`=4cos^3x-3cosx`

`sin3x=3cos^2xsinx-sin^3x=3(1-sin^2x)sinx-sin^3x`

`=3sinx-3sin^3x-sin^3x`

`=3sinx-4sin^3x`