If θ∈π2,3π2, then sin-1(sinθ) equals
θ
π-θ
2π-θ
-π+θ
We have,
θ∈π2,3π2⇒−θ∈−3π2,−π2⇒π−θ∈−π2,π2
Also, sin(π−θ)=sinθ
∴ sin−1(sinθ)=sin−1(sin(π−θ))=π−θ