Question

Let $\sin A+\sin B+\sin C=0$ then the value of $\sin 3A+\sin 3B+\sin 3C$ is

• A. $-12\sin A.\sin B.\sin C$
• B. $12\sin A.\sin B.\sin C$
• C. $0$
• D. none of these

Answer 1

The correct option is A $-12\sin A.\sin B.\sin C$

Given

$\sin A+\sin A+\sin C=0$

or, $\sin A+\sin B=-\sin C$......(1)

Now cubing both sides we get,

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

or, ${\sin }^{3}A+{\sin }^{3}B-3\sin A.\sin B.\sin C=-{\sin }^{3}C$[ Using (1)]

or, ${\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C=3\sin A.\sin B.\sin C$......(2).

Now $\sin 3A+\sin 3B+\sin 3C$

$=3(\sin A+\sin B+\sin C)-4({\sin }^3+{\sin }^{3}B+{\sin }^{3}C)$ [ Since $\sin 3A=3\sin A-4{\sin }^{3}A$]

$=3.0-4.3\sin A.\sin B.\sin C$ [ Using (2)]

$=-12\sin A.\sin B.\sin C$.