Consider real numbers x and y such that x,y∈[−1,1] and x2+y2≤1.
(i) Prove that
arcsinx+arcsiny=arcsin(x1−y2+y1−x2). Clearly mention any domain restrictions required for the identity to hold in the principal value branch.
(ii) Show that when x=135 and y=1312, the right‐hand side of the identity in part (i) simplifies to 2π.
(iii) Using the identity, evaluate arcsin(135)+arcsin(1312) without a calculator.