Question

In a triangle ABC,
$a^3\cos (B-C)+b^3\cos (C-A)+c^3\cos (A-B)=$

• A. abc
• B. 3 abc
• C. a+b+c
• D. None of these

Answer 1

The correct option is B

3 abc

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