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#