The right triangle method converts between inverse trig functions. If sin^(-1)(x) = theta, draw a right triangle with opposite = x, hypotenuse = 1, adjacent = sqrt(1-$x^2$). Then: cos(theta) = sqrt(1-$x^2$), tan(theta) = $\frac{x}{sqrt}$(1-$x^2$), etc. So sin^(-1)(x) = tan^(-1)(x/sqrt(1-$x^2$)) = cos^(-1)(sqrt(1-$x^2$)) for x in [0, 1]. For negative x, use parity: sin^(-1)(-x) = -sin^(-1)(x) then convert the positive part. Similarly, if tan^(-1)(x) = theta: sin(theta) = $\frac{x}{sqrt}$(1+$x^2$), cos(theta) = 1/sqrt(1+$x^2$). These conversions are essential when different inverse trig functions appear in the same equation.