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

1 Answer
Jan 18, 2018

sinA*sin(B-C)=sinC*sin(A-B)

=>2sin(pi-(B+C))*sin(B-C)=2sin(pi-(A+B))*sin(A-B)

=>2sin(B+C)*sin(B-C)=2sin(A+B)*sin(A-B)

=>cos(2C)-cos(2B)=cos(2B)-cos(2A)

=>1-2sin^2(C)-1+2sin^2(B)=1-2sin^2(B)-1+2sin^2(A)

=>sin^2(B)-sin^2(C)=sin^2(A)-sin^2(B)

=>4R^2sin^2(B)-4R^2sin^2(C)=4R^2sin^2(A)-4R^2sin^2(B)

=>b^2-c^2=a^2-b^2

This proves that

a^2,b^2,c^2 are in AP