Question

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

• A. $4\cos A\cos 3B+\cos 3C$
• B. $12\cos A\cos B\cos C$
• C. $8\cos A\cos B\cos C$
• D. $10\cos A\cos B\cos C$

Answer 1

The correct option is B $12\cos A\cos B\cos C$

we have,

if, $A+B+C=0\Rightarrow A^3+B^3+C^3=3xyz$

$\therefore \cos A+\cos B+\cos C=0\Rightarrow {\cos }^{3}A+{\cos }^{3}B+{\cos }^{3}C=3\cos A\cos B\cos C$

Now,

$\cos 3A+\cos 3B+\cos 3C$

$4{\cos }^{3}A-3\cos A+4{\cos }^{3}B-3\cos B+4{\cos }^{3}B-3\cos B$

$4({\cos }^{3}A+{\cos }^{3}B+{\cos }^{3}C)-3(\cos A+\cos B+\cos C)$

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

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