If #sinAsin(B-c)=sinCsin(A-B)# then show that # a^2,b^2,c^2# are in AP.?

Question

3108 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-18T15:53:45

#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

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