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 equal to?

• A. $0$
• B. $1$
• C. $3$
• D. $2$

Answer 1

Answer: (A)

$0$

Let $\sum a^3\sin (B-C)=a^3\sin (B-C)+b^3\sin (C-A)+c^3\sin (A-B)$

Using sine rule, $k=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\therefore \sum a^3\sin (B-C)=\sum k^3{\sin }^{3}A\sin (B-C)$

Use $\sin A=\sin (180-(B+C\left)\right)=\sin (B+C)$

$\therefore \sum k^3{\sin }^{3}A\sin (B-C)=\sum k^3{\sin }^{2}A\sin (B+C)\sin (B-C)$

$=\sum k^3\frac{{\sin }^{2}A}{2}\left(\cos \right(2C)-\cos (2B\left)\right)$

$=\frac{k^3}{2}\left[{\sin }^{2}A\right(\cos \left(2C\right)-\cos \left(2B\right))+{\sin }^{2}B(\cos \left(2A\right)-\cos \left(2C\right))+{\sin }^{2}C(\cos \left(2B\right)-\cos \left(2A\right)\left)\right]$

$=\frac{k^3}{2}\left[{\sin }^{2}A\right(1-2{\sin }^{2}C-1+2{\sin }^{2}B)+{\sin }^{2}B(1-2{\sin }^{2}A-1+2{\sin }^{2}C)+{\sin }^{2}C(1-2{\sin }^{2}B-1-2{\sin }^{2}A)]$

(Using $\cos 2x=1-2{\sin }^{2}x$)

$=\frac{k^3}{2}\left[0\right]$

$=0$

This is the required solution.