Question

If $\cos A+\cos B+\cos C=0$ and $\cos 3A+\cos 3B+\cos 3C=\lambda \cos A\cos B\cos C$ then $\lambda$ is

• A. 8
• B. 10
• C. 12
• D. 16

Answer 1

Answer: (C)

12

Given $\cos A+\cos B+\cos C=0$-----equ($1$)

We know $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$.

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

$x^3+y^3+z^3-3xyz=0$

$x^3+y^3+z^3=3xyz$

${\cos }^3\left(A\right)+{\cos }^3\left(B\right)+{\cos }^3\left(C\right)=3\cos A\cos B\cos C$

Now, $\cos \left(3A\right)=\cos (2A+A)$

=$\cos \left(2A\right)\cos A-\sin \left(2A\right)\sin A$

= $\left({\cos }^2\right(A)-{\sin }^2(A\left)\right)\cos A-\left(2\sin A\cos A\right)\sin A$

= ${\cos }^3\left(A\right)-3\cos A{\sin }^2\left(A\right)$

=${\cos }^3\left(A\right)-3\cos A(1-{\cos }^2(A\left)\right)$

=$4{\cos }^3\left(A\right)-3\cos A$

Similarly $\cos \left(3B\right)=4{\cos }^3\left(B\right)-3\cos B$ and

$\cos \left(3C\right)=4{\cos }^3\left(C\right)-3\cos \left(C\right)$

$\therefore \cos \left(3A\right)+\cos \left(3B\right)+\cos \left(3C\right)$ =$4\left({\cos }^3\right(A)+{\cos }^3(B)+{\cos }^3(C\left)\right)-3(\cos A+\cos B+\cos C)$

=$4\left(3\cos A\cos B\cos C\right)-3\left(0\right)$

= $12\cos A\cos B\cos C$

$\therefore \lambda =12$