Question

prove that sin(b-c)+bsin(c-a)+csin(a-b)=0

Answer 1

Prove that $\sin (b - c) + \mathrm{bsin}(c - a) + \mathrm{csin}(a - b) = 0$

Solution.

This statement is not correct.

For example, $a = \frac{\pi}{2}, b = \frac{\pi}{3}, c = \frac{\pi}{6}$.

Then

$$b - c = \frac {\pi}{6},$$ $$c - a = - \frac {\pi}{3},$$ $$a - b = \frac {\pi}{6}.$$

We obtain

$$\sin \frac {\pi}{6} + \frac {\pi}{3} \sin \left(- \frac {\pi}{3}\right) + \frac {\pi}{6} \sin \frac {\pi}{6} = 0$$ $$\frac {1}{2} - \frac {\pi}{3} \frac {\sqrt {3}}{2} + \frac {\pi}{6} \frac {1}{2} = 0$$

Multiply both parts of equation by 12:

$$6 - 2 \sqrt {3} \pi + \pi = 0$$ $$6 + (1 - 2 \sqrt {3}) \pi = 0 \quad - \text{It is false}.$$

Answer.

The statement is not correct.