Question

State true or false.

If $A+B+C=\pi$ then ${\sin }^{3}A\cos (B-C)+\sin {}^{3}B\cos (C-A)+\sin {}^{3}C\cos (A-B)=3$

• A. True
• B. False

Answer 1

The correct option is B False

We know that $A+B+C=\pi$

$\therefore A=\pi -(B+C)$

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

$={\sin }^{2}A\sin A\cos (B-C)+{\sin }^{2}B\sin B\cos (C-A)+{\sin }^{2}C\sin C\cos (A-B)$

$={\sin }^{2}A\sin (\pi -(B+C))\cos (B-C)+{\sin }^{2}B\sin (\pi -(C+A))\cos (C-A)+{\sin }^{2}C\sin (\pi -(A+B))\cos (A-B)$

$=\frac{1}{2}{\sin }^{2}A\times 2\sin (\pi -(B+C))\cos (B-C)+\frac{1}{2}{\sin }^{2}B\times 2\sin (\pi -(C+A))\cos (C-A)+\frac{1}{2}{\sin }^{2}C\times 2\sin (\pi -(A+B))\cos (A-B)$

$=\frac{1}{2}{\sin }^{2}A(\sin 2B+\sin 2C)+\frac{1}{2}{\sin }^{2}B(\sin 2C+\sin 2A)+\frac{1}{2}{\sin }^{2}C(\sin 2A+\sin 2B)$

$=\frac{1}{2}{\sin }^{2}A(2\sin B\cos B+2\sin C\cos C)+\frac{1}{2}{\sin }^{2}B(2\sin C\cos C+2\sin A\cos A)+\frac{1}{2}{\sin }^{2}C(2\sin A\cos A+2\sin B\cos B)$

$={\sin }^{2}A(\sin B\cos B+\sin C\cos C)+{\sin }^{2}B(\sin C\cos C+\sin A\cos A)+{\sin }^{2}C(\sin A\cos A+\sin B\cos B)$

$=\sin A\sin B\sin (A+B)+\sin B\sin C\sin (B+C)+\sin C\sin A\sin (C+A)$

$=3\sin A\sin B\sin C$