Question

Let $A,B,C$ are three angles such that
$\sin A+\sin B+\sin C=0,$ then the value of $\frac{\sin A.\sin B.\sin C}{\sin 3A+\sin 3B+\sin 3C}$ (wherever defined) is

• A. $12$
• B. $-12$
• C. $-\frac{1}{12}$
• D. $\frac{1}{12}$

Answer 1

The correct option is C $-\frac{1}{12}$

$\frac{\sin A.\sin B.\sin C}{\sin 3A+\sin 3B+\sin 3C}=\frac{\sin A.\sin B.\sin C}{3(\sin A+\sin B+\sin C)-4({\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C)}=\frac{\sin A.\sin B.\sin C}{-4(3\sin A.\sin B.\sin C)}=-\frac{1}{12}$

$[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\Rightarrow a^3+b^3+c^3=3abc\text{if}a+b+c=0]$