Question

asin(B-C)+bsin(C-A)+csin(A-B)

Answer 1

Answer on question 36105 – Math – Trigonometry

In triangle ABC, prove that $a \sin(B - C) + b \sin(C - A) + c \sin(A - B) = 0$

Solution

Using the following trigonometric identity

$$\sin (A - B) = \sin A \cos B - \sin B \cos A$$

We get

$$\begin{array}{l} a \sin (B - C) + b \sin (C - A) + c \sin (A - B) = a \sin B \cos C - a \sin C \cos B + \\ + b \sin C \cos A - b \sin A \cos C + c \sin A \cos B - c \sin B \cos A = \cos A (b \sin C - c \sin B) + \\ + \cos B (c \sin A - a \sin C) + \cos C (a \sin B - b \sin A) \end{array}$$

According to the sine theorem we have

$${\frac {a}{\sin A}} = {\frac {b}{\sin B}} = {\frac {c}{\sin C}}$$

Therefrom

$$b \sin C - c \sin B = c \sin A - a \sin C = a \sin B - b \sin A = 0$$

And we get

$$a \sin (B - C) + b \sin (C - A) + c \sin (A - B) = 0.$$

QED