How can I solve the mentioned problem?Please,help.

If,cosA+cosB+cosC=0 ;then prove that ,cos3A+cos3B+cos3C=12cosAcosBcosC

1 Answer
Dec 20, 2017

Given

cosA+cosB+cosC=0

=>cosA+cosB=-cosC

=>(cosA+cosB)^3=-cos^3C

=>cos^3A+cos^3B+3cosAcosB(cosA+cosB)=-cos^3C

=>cos^3A+cos^3B+3cosAcosB(-cosC)=-cos^3C

=>cos^3A+cos^3B+cos^3C=3cosAcosBcosC

=>4cos^3A+4cos^3B+4cos^3C=12cosAcosBcosC
Now
cos3A+cos3B+cos3B

=4cos^3A-3cosA+4cos^3B-3cosB+4cos^3C-3cosC

=4cos^3A+4cos^3B+4cos^3C-3(cosA+cosB+cosC)

=12cosAcosBcosC+3*0

=12cosAcosBcosC