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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\cos A+\cos B+\cos C=0, then prove that cos3A+cos3B+cos3C=12cosAcosBcosC\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C.

Explanation

Solution

Hint:Use the trigonometric formula cos3x=4cos3x3cosx\cos 3x=4{{\cos }^{3}}x-3\cos x to simplify the given expression. Then use the algebraic identity which says that if x+y+z=0x+y+z=0, then we have x3+y3+z3=3xyz{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz to prove the given trigonometric equation.

Complete step-by-step answer:

We know that cosA+cosB+cosC=0\cos A+\cos B+\cos C=0. We have to prove the trigonometric equation cos3A+cos3B+cos3C=12cosAcosBcosC\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C.
We know the trigonometric identity cos3x=4cos3x3cosx\cos 3x=4{{\cos }^{3}}x-3\cos x.
Thus, we have cos3A+cos3B+cos3C=(4cos3A3cosA)+(4cos3B3cosB)+(4cos3C3cosC)\cos 3A+\cos 3B+\cos 3C=\left( 4{{\cos }^{3}}A-3\cos A \right)+\left( 4{{\cos }^{3}}B-3\cos B \right)+\left( 4{{\cos }^{3}}C-3\cos C \right).
Rearranging the terms of the above equation, we have cos3A+cos3B+cos3C=4cos3A+4cos3B+4cos3C3cosA3cosB3cosC\cos 3A+\cos 3B+\cos 3C=4{{\cos }^{3}}A+4{{\cos }^{3}}B+4{{\cos }^{3}}C-3\cos A-3\cos B-3\cos C.
So, we have cos3A+cos3B+cos3C=4(cos3A+cos3B+cos3C)3(cosA+cosB+cosC)\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)-3\left( \cos A+\cos B+\cos C \right).
We know that cosA+cosB+cosC=0\cos A+\cos B+\cos C=0.
Thus, we have cos3A+cos3B+cos3C=4(cos3A+cos3B+cos3C).....(1)\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right).....\left( 1 \right).
We know the algebraic identity which says that if x+y+z=0x+y+z=0, then we have x3+y3+z3=3xyz{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz .
Substituting x=cosA,y=cosB,z=cosCx=\cos A,y=\cos B,z=\cos C in the above equation, we have cos3A+cos3B+cos3C=3cosAcosBcosC.....(2){{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C=3\cos A\cos B\cos C.....\left( 2 \right).
Substituting equation (2) in equation (1), we have cos3A+cos3B+cos3C=4(cos3A+cos3B+cos3C)=4(3cosAcosBcosC)\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)=4\left( 3\cos A\cos B\cos C \right).
Thus, we have cos3A+cos3B+cos3C=4(cos3A+cos3B+cos3C)=4(3cosAcosBcosC)=12cosAcosBcosC\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)=4\left( 3\cos A\cos B\cos C \right)=12\cos A\cos B\cos C.
Hence, we have proved that if cosA+cosB+cosC=0\cos A+\cos B+\cos C=0, then we have cos3A+cos3B+cos3C=12cosAcosBcosC\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C.

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.