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$$a \cdot \sin (B-C) +b \cdot \sin(C-A) +c \cdot \sin(A-B) =0$$ where $a, b, c$ are the sides of a triangle and $A, B, C$ are the angles of a triangle.

No idea how to solve this problem.

VividD
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Apoorv Jain
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2 Answers2

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

The sine rule says that $$ \frac{\sin A }{a} = \frac{\sin B }{b} = \frac{\sin C}{c} $$ So we have: $$ b\sin A = a\sin B \\ c\sin B = b\sin C\\ a\sin C = c \sin A $$ Substituting this into the final expression from above gives: $$ a \sin(B-C) + b \sin(C-A) + c \sin(A-B)\\ = a \sin B \cos C - b \sin A \cos C + b \sin C \cos A - c \sin B \cos A + c \sin A \cos B - a \sin C \cos B \\ = a \sin B \cos C - a \sin B \cos C + b\sin C \cos A - b \sin C \cos A + c\sin A \cos B - c \sin A \cos B\\ =0 $$

DavidButlerUofA
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1

Using Law of Sines, $\displaystyle a=2R\sin A$

Again, $\displaystyle\sin A=\sin[\pi-(B+C)]=\sin(B+C)$

$$\implies a\sin(B-C)=2R\sin A\sin(B-C)=2R\sin(B+C)\sin(B-C)$$

Now use Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$

Community
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lab bhattacharjee
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