If cosA+cosB+cosC=0 then cos3A+cos3B+cos3C=λcosAcosBcosC then λ=
12
14
16
15
cosA+cosB+cosC=0
⇒cos3A+cos3B+cos3C
=3cosAcosBcosC and cos3A+cos3B+cos3C
=4(cos3A+cos3B+cos3C)−3(cosA+cosB+cosC)
=4(3cosAcosBcosC)−3(0)
=12cosacosBcosC