You are looking to find the value of sin^-1(1/2) on the closed interval [-π/2, π/2].

Notice that this function denotes an inverse function, which can be confused with an exponential function. For instance, if this were an exponential function it would look like the following:

     (sin(x))^-1 = 1/sin(x)

That is to say that:     sin^-1(x) ≠ 1/sin(x)   ,   but     sin^-1(x) = arcsin(x)

When evaluating this inverse trig function, note that the following are equivalent:

          sin^-1(x) = θ     <==>     sin(θ) = x

So for the function in question,

     sin^-1(1/2) = θ     <==>     sin(θ) = 1/2

Looking at the unit circle, we see that sin(θ) is equal to 1/2 only when θ equal π/6 on the given interval. That is, 

          sin(π/6) = 1/2     <==>     sin^-1(1/2) = π/6

sin^-1 (1/2) = Π/6 <==Answer, since sin Π/6 = 1/2, and -Π/2 ≤ Π/6 ≤ Π/2