_{[ 0]}^{[ 2 pie]} sin ix sin jx dx=1/2 _{[ 0]}^{[ 2 pie]} [cos(i-j)-cos(i+j)]dx

                                   =1/2[ { sin2pie(i-j)}/(i-j) - {sin2pie(i+j)}/(i+j)]

now wen i , j are any 2 integers s.t. i is nt equal to j

den

i-j vl b an integer n thus _{[ 0]}^{[ 2 pie]} sin ix sin jx dx will b equal to 0

 wen i=j then _{[ 0]}^{[ 2 pie]} sin ix sin jx =_{[ 0]}^{[ 2 pie]} sin^2 ix dx

                                                   = _[0]^{[ 2 pie]} (1-cos2ix)/2 dx

                                                   =pie - _[0]^{[ 2 pie]}  cos 2ix dx

                                                   = pie -[ sin4pie i ]/2i

                                                   = pie which is nt equal to 0

therfore R is symmetric n transitive for i not equal to j

bt R is not reflexive

 

PLZ DO TEL ME WHETHER EM CORRECT OR NOT.................