find the exact value of sin^-1 (1/2), -Π/2 ≤ x ≤ Π/2

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