Use a right triangle: if sin^(-1)(x) = theta, then sin(theta) = x, so the opposite side is x and hypotenuse is 1, adjacent = sqrt(1-$x^2$). Therefore: cos(theta) = sqrt(1-$x^2$), tan(theta) = $\frac{x}{sqrt}$(1-$x^2$). This gives: sin^(-1)(x) = cos^(-1)(sqrt(1-$x^2$)) = tan^(-1)(x/sqrt(1-$x^2$)) for x >= 0.