How do you solve the inverse trig function sin(sin^-1 (1/3))sin(sin1(13))?

1 Answer
May 25, 2015

sin^-1(1/3)sin1(13) or arcsin(1/3)arcsin(13) is the angle thetaθ for which sin(theta) = 1/3sin(θ)=13

Therefore sin(sin^-1(1/3))sin(sin1(13))
= sin(theta)=sin(θ) for the value of thetaθ for which sin(theta) = 1/3sin(θ)=13

That is
sin(sin^-1(1/3)) = 1/3sin(sin1(13))=13