Question

if A+B+C=180, prove that, asin(B-C)+bsin(C-A)+csin(A-B)=0.

Answer 1

Answer on Question #49505 – Math – Trigonometry

Question:

If $A + B + C = 180{}^{\circ}$, prove that, $a \cdot \sin(B - C) + b \cdot \sin(C - A) + c \cdot \sin(A - B) = 0$

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

Solution:

Let us prove given trigonometric identity in too steps.

1) Use the sine of difference identity for the angles $\alpha$ and $\beta$

$$\sin(\alpha - \beta) = \sin\alpha \cdot \cos\beta - \cos\alpha \cdot \sin\beta.$$

Rewriting (1) by taking into account (2) we get

$$a \cdot \left(\sin B \cos C - \underline{\cos B} \sin C\right) + b \cdot \left(\sin C \underline{\cos A} - \underline{\cos C} \sin A\right) + c \cdot \left(\sin A \underline{\cos B} - \underline{\cos A} \sin B\right) = \cos C (\sin B - \cos A) + \cos B (\cos A - \cos C) + \cos A (\cos C - \cos A).$$

2) Since $A + B + C = 180{}^{\circ}$, then we can use the law of sines (an equation relating the lengths of the sides of any shaped triangle to the sines of its angles):

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

Hence, we obtain the following relations

$$\sin B = b \sin a, \sin A = \sin C, \quad b \sin C = \sin B.$$

Substituting (4) into (*) we see that

$$\sin B - \sin A = \sin A - \sin C = \sin C - \sin A = 0.$$

Therefore, we get

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

QED.

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

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