Question

sinA sin2A+sin3A Sin6A=sin4A sin5A

Answer 1

sina sin2a +sin3a sin6a
multiplying and dividing by 2
1/2(2sina sin2a +2sin3a sin6a)
1/2(cos (a-2a)-cos(a+2a)+cos(3a-6a)-cos(3a+6a)
1/2(cos(-a)-cos3a+cos(-3a)-cos9a)
1/2(cosa-cos3a+cos3a-cos9a)
1/2(cosa-cos9a)
1/2(2sin(9a-a/2)sin(a+9a/2)
1/2(2sin4asin5a)
sin4asina5a
hence proved