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

If,cosA+cosB+cosC=0cosA+cosB+cosC=0 ;then prove that ,cos3A+cos3B+cos3C=12cosAcosBcosCcos3A+cos3B+cos3C=12cosAcosBcosC

1 Answer
P dilip_k
Dec 20, 2017

Given

cosA+cosB+cosC=0cosA+cosB+cosC=0

=>cosA+cosB=-cosC⇒cosA+cosB=−cosC

=>(cosA+cosB)^3=-cos^3C⇒(cosA+cosB)3=−cos3C

=>cos^3A+cos^3B+3cosAcosB(cosA+cosB)=-cos^3C⇒cos3A+cos3B+3cosAcosB(cosA+cosB)=−cos3C

=>cos^3A+cos^3B+3cosAcosB(-cosC)=-cos^3C⇒cos3A+cos3B+3cosAcosB(−cosC)=−cos3C

=>cos^3A+cos^3B+cos^3C=3cosAcosBcosC⇒cos3A+cos3B+cos3C=3cosAcosBcosC

=>4cos^3A+4cos^3B+4cos^3C=12cosAcosBcosC⇒4cos3A+4cos3B+4cos3C=12cosAcosBcosC
Now
cos3A+cos3B+cos3Bcos3A+cos3B+cos3B

=4cos^3A-3cosA+4cos^3B-3cosB+4cos^3C-3cosC=4cos3A−3cosA+4cos3B−3cosB+4cos3C−3cosC

=4cos^3A+4cos^3B+4cos^3C-3(cosA+cosB+cosC)=4cos3A+4cos3B+4cos3C−3(cosA+cosB+cosC)

=12cosAcosBcosC+3*0=12cosAcosBcosC+3⋅0

=12cosAcosBcosC=12cosAcosBcosC

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