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

Question

16452 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-08T09:53:09

For a Proof, please refer to the Explanation.

We know from Algebra, that,

#x+y+z=0 rArr x^3+y^3+z^3=3xyz.#

#:. cosA+cosB+cosC=0 rArr cos^3A+cos^3B+cos^3C=3cosAcosBcosC..................................................(star).#

Now, #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),..............[because (star)," & given],"#

#=12cosAcosBcosC.#

Hence, the Proof.