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

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Answer 1 · 2018-03-24T10:40:02

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

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)......to#Using #color(red)((1)#

#=12cosAcosBcosC#

#=RHS#