Let arcsin(3/5)=alpha, &, arctan2=beta; alpha, beta in (-pi/2,pi/2)arcsin(35)=α,&,arctan2=β;α,β∈(−π2,π2).
Hence, by defn., sinalpha=3/5>0 so, alpha in (0,pi/2)sinα=35>0so,α∈(0,π2), and,
tanbeta=2>0, so, beta in (0,pi/2)tanβ=2>0,so,β∈(0,π2)
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.
