Answer to Question #100447 in Trigonometry for Gaby

Question

Verify: cos3x=cos³x-3sin²xcosx

Answer 1

cos3x=cos(2x+x)=cox2x*cosx-sin2x*sinx=(1-2sin²x)*cosx-2sinx*cosx*sinx=

=cosx-2sin²x*cosx-2sin²x*cosx=cosx*(1-4*(1-cos²x))=cosx*(1-4+4cos²x)=4cos³x-3cosx;

cos3x=cos³x-3sin²x*cosx

4cos³x-3cosx=cos³x-3sin²x*cosx;

4cos³x-3cosx-cos³x+3sin²x*cosx=0;

3cos³x+3sin²x*cosx-3cosx=0;

cosx(3cos²x+3sin²x-3)=0;

cosx*(3*(cos²x+sin²x)-3)=0;

cosx*(3*1-3)=0;

cosx*0=0, which is true, hence the initial formula was verified.