Let #arcsin(3/5)=alpha, &, arctan2=beta; alpha, beta in (-pi/2,pi/2)#.
Hence, by defn., #sinalpha=3/5>0 so, alpha in (0,pi/2)#, and,
#tanbeta=2>0, so, beta in (0,pi/2)#
Now, reqd. value
#=sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta.......(1)#
Therefore, to find this, we will need #cosalpha, sinbeta, cosbeta#
#sinalpha=3/5 rArr cosalpha=+sqrt(1-sin^2alpha)=+4/5, as, alpha in (0,pi/2)#
#cosbeta=1/secbeta=1/(+sqrt(1+tan^2beta))=1/(+sqrt5)#, and,
#sinbeta=tanbetacosbeta=2(1/(+sqrt5))=+2/sqrt5, as beta in (0,pi/2)#
Using all these in #(1)#, we get,
the reqd. value #=3/5*1/sqrt5+4/5*2/sqrt5=11/(5sqrt5)=(11sqrt5)/25#.
