Question

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

Answer 1

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

`=12cosAcosBcosC`

`=RHS`