Question
Question: If \(\cos A+\cos B+\cos C=0\), then prove that \(\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C\)....
If cosA+cosB+cosC=0, then prove that cos3A+cos3B+cos3C=12cosAcosBcosC.
Solution
Hint:Use the trigonometric formula cos3x=4cos3x−3cosx to simplify the given expression. Then use the algebraic identity which says that if x+y+z=0, then we have x3+y3+z3=3xyz to prove the given trigonometric equation.
Complete step-by-step answer:
We know that cosA+cosB+cosC=0. We have to prove the trigonometric equation cos3A+cos3B+cos3C=12cosAcosBcosC.
We know the trigonometric identity cos3x=4cos3x−3cosx.
Thus, we have cos3A+cos3B+cos3C=(4cos3A−3cosA)+(4cos3B−3cosB)+(4cos3C−3cosC).
Rearranging the terms of the above equation, we have cos3A+cos3B+cos3C=4cos3A+4cos3B+4cos3C−3cosA−3cosB−3cosC.
So, we have cos3A+cos3B+cos3C=4(cos3A+cos3B+cos3C)−3(cosA+cosB+cosC).
We know that cosA+cosB+cosC=0.
Thus, we have cos3A+cos3B+cos3C=4(cos3A+cos3B+cos3C).....(1).
We know the algebraic identity which says that if x+y+z=0, then we have x3+y3+z3=3xyz .
Substituting x=cosA,y=cosB,z=cosC in the above equation, we have cos3A+cos3B+cos3C=3cosAcosBcosC.....(2).
Substituting equation (2) in equation (1), we have cos3A+cos3B+cos3C=4(cos3A+cos3B+cos3C)=4(3cosAcosBcosC).
Thus, we have cos3A+cos3B+cos3C=4(cos3A+cos3B+cos3C)=4(3cosAcosBcosC)=12cosAcosBcosC.
Hence, we have proved that if cosA+cosB+cosC=0, then we have cos3A+cos3B+cos3C=12cosAcosBcosC.
Note: We can’t solve this question without using the algebraic and trigonometric identity. An algebraic identity is an equality that holds for all possible values of its variables. They are used for the factorization of the polynomials.