If cosA+cosB+cosC=0 Then prove that cos3A+cos3B+cos3C=12cosAcosBcosC ?

1 Answer
Mar 24, 2018

Hint:
cos3theta=4cos^3theta-3costheta

Explanation:

We know that

If color(blue)(a+b+c=0,then ,a^3+b^3+c^3=3abc

Let, a=cosA,b=cosB,c=cosC.

color(red)( :.cosA+cosB+cosC=0

color(red)(=>cos^3A+cos^3B+cos^3C=3cosAcosBcosC...to(1)

LHS=cos3A+cos3B+cos3C

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

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

=4(3cosAcosBcosC)-3(0)......toUsing color(red)((1)

=12cosAcosBcosC

=RHS