In any triangle $ ABC $ $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $.

Question

In any triangle $ ABC $ prove that $\frac {sin(B-C)}{sin (B+C)}=\frac {b^2-c^2}{a^2} $.

Please help. Thanks in advance.

Answer 1

Note that \begin{align*} \frac{\sin(B-C)}{\sin (B+C)}&=\frac{\sin B \cos C-\cos B \sin C}{\sin B \cos C+\cos B \sin C}\\ &=\frac{\frac{\sin B}{\sin C} \cos C-\cos B}{\frac{\sin B}{\sin C} \cos C+\cos B}\\ &=\frac{\frac{b}{c} \frac{c^2-a^2-b^2}{2ab}-\frac{b^2-a^2-c^2}{2ac}}{\frac{b}{c} \frac{c^2-a^2-b^2}{2ab}+\frac{b^2-a^2-c^2}{2ac}}\\ &=\frac{b^2-c^2}{a^2}. \end{align*}

Answer 2

In the above configuration, $R \sin(B-C) = OJ_A = M_A H_A$ and $R\sin(B+C)=R\sin(A)= BM_A$. Moreover, by the Pythagorean theorem we have $H_A B^2-H_A C^2=AB^2-AC^2=c^2-b^2$, hence $ b^2-c^2 = (H_A B+H_A C)(H_A B-H_A C) $ and $ H_A B - H_A C = \frac{c^2-b^2}{a}.$ It follows that

$$ BH_A = \frac{1}{2}\left(\frac{c^2-b^2}{a}+a\right) = \frac{a^2+c^2-b^2}{2a} $$ and $$ M_A H_A = \frac{a}{2}-BH_A = \frac{b^2-c^2}{2a} $$ so $$ \boxed{\frac{\sin(B-C)}{\sin(B+C)}=\frac{OJ_A}{BM_A} = \frac{2 M_A H_A}{a} = \color{red}{\frac{b^2-c^2}{a^2}}.}$$

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An alternative approach through the sine addition formula and the cosine theorem:

$$\begin{eqnarray*}\frac{\sin(B-C)}{\sin(B+C)}=\frac{2R \sin(B-C)}{2R\sin(A)}&=&\frac{2R\sin(B)\cos(C)-2R\sin(C)\cos(B)}{a}\\&=&\frac{2ab\cos(C)-2ac\cos(B)}{2a^2}\\&=&\frac{(a^2+b^2-c^2)-(a^2+c^2-b^2)}{2a^2}\\&=&\color{red}{\frac{b^2-c^2}{a^2}}.\end{eqnarray*}$$