We know that,
#color(red)((1)cos(2kpi+theta)=cos theta#
#color(red)((2)cos(pi/2+theta)=-sintheta#
#color(red)((3)sin^-1(-x)=-sin^-1x#
#color(red)((4)sin^-1(sin(x))=x, where,x in [-pi/2,pi/2]#
Here,
#(53pi)/5=10pi+(3pi)/5......1^(st)#.quadrant
#:.sin^-1(cos((53pi)/5))=sin^-1(cos(10pi+(3pi)/5))#
#=sin^-1(cos((3pi)/5)).....toApply(1)#
#=sin^-1(cos((6pi)/10))#
#=sin^-1(cos(pi/2+pi/10))#
#=sin^-1(-sin(pi/10))........toApply(2)#
#=-sin^-1(sin(pi/10)).......toApply(3)#
#=-pi/10..........toApply(4)#
#=>sin^-1(cos((53pi)/5))= -pi/10in[-pi/2,pi/2]#