 QUESTION

# Find the value of $\lambda$, If $\cos A + \cos B + \cos C = 0$ and $\cos 3A + \cos 3B + \cos 3C$=$\lambda \cos A\cos B\cos C$$A.8 \\ B.10 \\ C.12 \\ D.16 \\$

Hint: In this we use basic identities ${x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - xz - yz)$ and basic trigonometry expansions ${\cos ^3}(A) + {\cos ^3}(B) + {\cos ^3}(C) = 3\cos A\cos B\cos C$. Use this to find the value of $\lambda$.

According to the question the equation that is given is $\cos A + \cos B + \cos C = 0........eq1$
We know that ${x^3} + {y^3} + {z^3} - 3xyz = (x + y + z)({x^2} + {y^2} + {z^2} - xy - xz - yz)$
Let $x = \cos A,y = \cos B$and $z = \cos C$
$\cos A + \cos B + \cos C = x + y + z = 0$,so this identity becomes
${x^3} + {y^3} + {z^3} - 3xyz = 0 \\ {x^3} + {y^3} + {z^3} = 3xyz \\$
${\cos ^3}(A) + {\cos ^3}(B) + {\cos ^3}(C) = 3\cos A\cos B\cos C$,
$\cos (3A) = \cos (2A + A) \\ = \cos (2A)\cos A - \sin (2A)\sin A \\ = ({\cos ^2}(A) - {\sin ^2}(A))\cos A - (2\sin A\cos A)\sin A \\ = {\cos ^3}(A) - 3\cos A{\sin ^2}(A) \\ = {\cos ^3}(A) - 3\cos A(1 - {\cos ^2}(A)) \\ = 4{\cos ^3}(A) - 3\cos A \\$
Similarly $\cos (3B) = 4{\cos ^3}(B) - 3\cos B$and
$\cos (3C) = 4{\cos ^3}(C) - 3\cos (C)$
$\therefore \cos (3A) + \cos (3B) + \cos (3C) = 4({\cos ^3}(A) + {\cos ^3}(B) + {\cos ^3}(C)) - 3(\cos A + \cos B + \cos C) \\ = 4(3\cos A\cos B\cos C) - 3(0) \\ = 12\cos A\cos B\cos C \\ \therefore \lambda = 12 \\$