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!

1 Answer
Jan 14, 2018

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