Question

If cosA+cosB+cosC = 0,
Prove:
cos3A+cos3B+cos3C = 12cosAcosBcosC

Answer 1

Answer:

For a Proof, please refer to the Explanation.

Explanation:

We know from Algebra, that,

x+y+z=0⇒x3+y3+z3=3xyz.

∴cosA+cosB+cosC=0⇒cos3A+cos3B+cos3C=3cosAcosBcosC..................................................(⋆).

Now, cos3A+cos3B+cos3C,

=4cos3A−3cosA+4cos3B−3cosB+4cos3C−3cosC,

=4(cos3A+cos3B+cos3C)−3(cosA+cosB+cosC),

=4(3cosAcosBcos

Answer 2

Answer:

12cosA cosB cosC

Step-by-step explanation:

We know that ,

a + b + c = 0

a³ + b³ + c³ = 3abc

•°• cosA + cosB + cosC = 0 ..........(1)

=> cos³A + cos³B + cos³C = 3cosA . cosB . cosC..........(2)

•°• LHS = cos3A + cos3B + cos3C

= ( 4 cos³A - 3cosA ) + ( 4cos³B - 3cosB ) + (4cos³C - 3cosC)

= 4( cos³A + cos³B + cos³C ) - 3( cosA + cosB + cosC )

= 4 × 3 cosA . cosB . cosC . - 3 × 0 ( From the first and second equation )

= 12 cosA . cosB . cosC