Evaluating Inverse Trig Functions

A regular trig function takes an angle and gives a ratio. An inverse trig function runs that backward: you hand it a ratio and it hands you an angle. But sine, cosine, and tangent each repeat their values forever as you go around the circle, so infinitely many angles share the same sine. For the inverse to be an actual function, with one output per input, each one gets a restricted range where it only gives back one angle. arcsin lives in [-π/2, π/2], arccos lives in [0, π], and arctan lives in (-π/2, π/2).

To evaluate one, ask: what angle, inside that specific window, has this ratio? For arctan(1), scan (-π/2, π/2) for an angle with tangent 1 - that's π/4. For arccos(0), scan [0, π] - that's π/2. For arcsin(-1/2), scan [-π/2, π/2] - that's -π/6, not the other angle with sine -1/2 that lives outside the window. This is why arcsin(sin(5pi/6)) is not 5pi/6: 5pi/6 sits outside arcsin's range, so the expression first collapses to sin(5pi/6) = 1/2, then arcsin(1/2) returns π/6, the equivalent angle that does live in [-π/2, π/2].

This topic is the hinge between reading the unit circle forward and reading it backward. Everything about reference angles and quadrants from earlier in the unit still applies - you're just now asking which angle in the restricted range matches a given ratio, instead of which ratio matches a given angle.