# If sinA*sin(B-c)=sinC*sin(A-B) then show that  a^2,b^2,c^2 are in AP.?

Jan 18, 2018

$\sin A \cdot \sin \left(B - C\right) = \sin C \cdot \sin \left(A - B\right)$

$\implies 2 \sin \left(\pi - \left(B + C\right)\right) \cdot \sin \left(B - C\right) = 2 \sin \left(\pi - \left(A + B\right)\right) \cdot \sin \left(A - B\right)$

$\implies 2 \sin \left(B + C\right) \cdot \sin \left(B - C\right) = 2 \sin \left(A + B\right) \cdot \sin \left(A - B\right)$

$\implies \cos \left(2 C\right) - \cos \left(2 B\right) = \cos \left(2 B\right) - \cos \left(2 A\right)$

$\implies 1 - 2 {\sin}^{2} \left(C\right) - 1 + 2 {\sin}^{2} \left(B\right) = 1 - 2 {\sin}^{2} \left(B\right) - 1 + 2 {\sin}^{2} \left(A\right)$

$\implies {\sin}^{2} \left(B\right) - {\sin}^{2} \left(C\right) = {\sin}^{2} \left(A\right) - {\sin}^{2} \left(B\right)$

$\implies 4 {R}^{2} {\sin}^{2} \left(B\right) - 4 {R}^{2} {\sin}^{2} \left(C\right) = 4 {R}^{2} {\sin}^{2} \left(A\right) - 4 {R}^{2} {\sin}^{2} \left(B\right)$

$\implies {b}^{2} - {c}^{2} = {a}^{2} - {b}^{2}$

This proves that

${a}^{2} , {b}^{2} , {c}^{2}$ are in AP