Question

In a $\Delta ABC, a,b,c $ are the sides of the triangle opposite to the angles $A,B,C$ respectively. Then the value of $a^3\sin(B-C)+b^3\,\sin(C-A)+c^3\sin(A-B)$ is equqal to

• A. 0
• B. 1
• C. 3
• D. 2

Answer 1

Answer: (A)

$\sum a^{3} \sin (B-C)$
$=\sum k^{3} \sin ^{3} A \sin (B-C)$
$\left[\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\right]$
$=\sum k^{3}\left[\sin ^{2} A \sin (B+C) \sin (B-C)\right]$
$=k^{3}\left[\left\{\sin ^{2} A \times \frac{1}{2}(\cos 2 C-\cos 2 B)\right\}\right.+\sin ^{2} B \times \frac{1}{2}(\cos 2 A-\cos 2 C)\left.+\sin ^{2} C \times \frac{1}{2}(\cos 2 B-\cos 2 A)\right]$
$=\frac{k^{3}}{2}\left[\sin ^{2} A\left(1-2 \sin ^{2} C-1+2 \sin ^{2} B\right)\right.+\sin ^{2} B\left(1-2 \sin ^{2} A-1+2 \sin ^{2} C\right)\left.+\sin ^{2} C\left(1-2 \sin ^{2} B-1+2 \sin ^{2} A\right)\right]$
$=\frac{k^{3}}{2}\left[-2 \sin ^{2} A \sin ^{2} C+2 \sin ^{2} A \sin ^{2} B\right.-2 \sin ^{2} B \sin ^{2} A+2 \sin ^{2} B \sin ^{2} C\left.-2 \sin ^{2} C \sin ^{2} B+2 \sin ^{2} C \sin ^{2} A\right]$
$=\frac{k^{3}}{2}[0]=0$