
If $\cos A+\cos B+\cos C=0$, then prove that $\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C$.
Answer
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Hint:Use the trigonometric formula $\cos 3x=4{{\cos }^{3}}x-3\cos x$ to simplify the given expression. Then use the algebraic identity which says that if $x+y+z=0$, then we have ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$ to prove the given trigonometric equation.
Complete step-by-step answer:
We know that $\cos A+\cos B+\cos C=0$. We have to prove the trigonometric equation $\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C$.
We know the trigonometric identity $\cos 3x=4{{\cos }^{3}}x-3\cos x$.
Thus, we have $\cos 3A+\cos 3B+\cos 3C=\left( 4{{\cos }^{3}}A-3\cos A \right)+\left( 4{{\cos }^{3}}B-3\cos B \right)+\left( 4{{\cos }^{3}}C-3\cos C \right)$.
Rearranging the terms of the above equation, we have $\cos 3A+\cos 3B+\cos 3C=4{{\cos }^{3}}A+4{{\cos }^{3}}B+4{{\cos }^{3}}C-3\cos A-3\cos B-3\cos C$.
So, we have $\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)-3\left( \cos A+\cos B+\cos C \right)$.
We know that $\cos A+\cos B+\cos C=0$.
Thus, we have $\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right).....\left( 1 \right)$.
We know the algebraic identity which says that if $x+y+z=0$, then we have ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$ .
Substituting $x=\cos A,y=\cos B,z=\cos C$ in the above equation, we have ${{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C=3\cos A\cos B\cos C.....\left( 2 \right)$.
Substituting equation (2) in equation (1), we have $\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)=4\left( 3\cos A\cos B\cos C \right)$.
Thus, we have $\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)=4\left( 3\cos A\cos B\cos C \right)=12\cos A\cos B\cos C$.
Hence, we have proved that if $\cos A+\cos B+\cos C=0$, then we have $\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C$.
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.
Complete step-by-step answer:
We know that $\cos A+\cos B+\cos C=0$. We have to prove the trigonometric equation $\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C$.
We know the trigonometric identity $\cos 3x=4{{\cos }^{3}}x-3\cos x$.
Thus, we have $\cos 3A+\cos 3B+\cos 3C=\left( 4{{\cos }^{3}}A-3\cos A \right)+\left( 4{{\cos }^{3}}B-3\cos B \right)+\left( 4{{\cos }^{3}}C-3\cos C \right)$.
Rearranging the terms of the above equation, we have $\cos 3A+\cos 3B+\cos 3C=4{{\cos }^{3}}A+4{{\cos }^{3}}B+4{{\cos }^{3}}C-3\cos A-3\cos B-3\cos C$.
So, we have $\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)-3\left( \cos A+\cos B+\cos C \right)$.
We know that $\cos A+\cos B+\cos C=0$.
Thus, we have $\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right).....\left( 1 \right)$.
We know the algebraic identity which says that if $x+y+z=0$, then we have ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$ .
Substituting $x=\cos A,y=\cos B,z=\cos C$ in the above equation, we have ${{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C=3\cos A\cos B\cos C.....\left( 2 \right)$.
Substituting equation (2) in equation (1), we have $\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)=4\left( 3\cos A\cos B\cos C \right)$.
Thus, we have $\cos 3A+\cos 3B+\cos 3C=4\left( {{\cos }^{3}}A+{{\cos }^{3}}B+{{\cos }^{3}}C \right)=4\left( 3\cos A\cos B\cos C \right)=12\cos A\cos B\cos C$.
Hence, we have proved that if $\cos A+\cos B+\cos C=0$, then we have $\cos 3A+\cos 3B+\cos 3C=12\cos A\cos B\cos C$.
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.
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