The inverse sine function sin−1 is, to put it briefly, the inverse function of sine, so that y=sin−1x if x=sin y. Thus sin−1 ½ = π/6 because sin(π/6) = ½. However, sin(5π/6) = ½ also, so it might be thought that sin−1 ½ = π/6 or 5π/6. It is necessary to avoid such ambiguity so it is normally agreed that the value to be taken is the one lying in the interval [−π/2, π/2]. Similarly, y=tan−1x if x=tan y, and the value y is taken to lie in the interval (−π/2, π/2). Also, y=cos−1x if x=cos y and the value y is taken to lie in the interval [0,π].
A more advanced approach provides more explanation. The inverse function of a trigonometric function exists only if the original function is restricted to a suitable domain. This can be an interval I in which the function is strictly increasing or strictly decreasing (see inverse function). So, to obtain an inverse function for sin x, the function is restricted to a domain consisting of the interval [−π/2, π/2]; tan x is restricted to the interval (−π/2, π/2); and cos x is restricted to [0, π]. The domain of the inverse function is, in each case, the range of the restricted function. Hence the following inverse functions are obtained: sin−1: [−1, 1] → [−π/2, π/2]; tan−1: R → (−π/2, π/2); cos−1[−1, 1] → [0, π]. The notation arcsin, arctan and arccos, for sin−1, tan−1 and cos−1, is also used. The following derivatives can be obtained: