<Question>
Compute $\arcsin \left( -\frac{1}{2} \right).$  Express your answer in radians.
</Question>

<think>
<Plan_1>
Okay, so I need to compute arcsin of -1/2. Hmm, let me start by recalling what the arcsin function does. Arcsin is the inverse of the sine function, right? But since sine isn't one-to-one over all real numbers, the domain of arcsin is restricted to [-π/2, π/2] to make it invertible. That means the output of arcsin will always be between -π/2 and π/2.

First, let me remember the unit circle and the values where sine is -1/2. Sine corresponds to the y-coordinate on the unit circle. So, where is the y-coordinate -1/2?

<prompt_1.1> Recall the unit circle and determine where the y-coordinate is -1/2. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's think about the standard angles. In the unit circle, the reference angles where sine is 1/2 are π/6 (30 degrees) and 5π/6 (150 degrees) in the first and second quadrants, respectively. But since we need negative values here, -1/2, that would be in the third and fourth quadrants where sine is negative.

So, the angles where sine is -1/2 would be π + π/6 = 7π/6 and 2π - π/6 = 11π/6. But wait, arcsin's range is [-π/2, π/2], so those angles are 7π/6 and 11π/6. However, 7π/6 is 210 degrees, which is in the third quadrant, and 11π/6 is 330 degrees, in the fourth quadrant. But since arcsin can only output between -π/2 and π/2, which is -90 degrees to 90 degrees, the corresponding angle must be in that range.

Wait, but 11π/6 is 330 degrees, which is equivalent to -π/6 if we consider angles coterminal with it. Because 11π/6 - 2π = -π/6.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So, the reference angle here is π/6, but since it's negative, it's in the fourth quadrant when considering the range of arcsin.

Therefore, arcsin(-1/2) should be -π/6.

Let me check.
Sine of -π/6 is -1/2. Yes, because sin(-θ) = -sinθ. So sin(-π/6) = -sin(π/6) = -1/2.

But let me make sure I'm not making a mistake here. The arcsin function returns the angle in [-π/2, π/2] whose sine is the given number.
Since -π/6 is within that interval and sin(-π/6) = -1/2, that should be correct.

Let me verify with another approach. If I consider the definition of inverse functions: if y = arcsin(x), then sin(y) = x and y is in [-π/2, π/2].

<prompt_2.1> Verify the solution using the definition of inverse functions and consider the quadrant of the original angle. </prompt_2.1>

Wait, but let me double-check. If I think about the unit circle, -π/6 is the same as 11π/6, but arcsin can't return 11π/6 because that's greater than π/2.
So the correct answer must be -π/6.

Another way to think about it: the principal value for arcsin is in the range [-π/2, π/2], so even though sin(7π/6) = -1/2 and sin(11π/6) = -1/2, those angles are outside the range of arcsin. Therefore, the only valid one in the range is -π/6, which corresponds to 11π/6 when subtracting 2π.

So, after going through all these steps and verifying, I'm confident that the answer is -π/6.

But just to make sure, let me recall the common values.

<prompt_2.2> Recall common values and confirm the solution. </prompt_2.2>

Alternatively, if I use a calculator to check: typing in arcsin(-1/2) should give -π/6. Let me convert -π/6 to degrees to check intuition.
-π/6 radians is -30 degrees. Sine of -30 degrees is -1/2, which matches.

So, yes, everything lines up. The answer is -π/6.
</Plan_2>

<Execution_2>
<execution_2.1>
So, if x = -1/2, then y must be an angle in that interval where sin(y) = -1/2. The possible angles in that interval are -π/6 and -5π/6. Wait, but -5π/6 is -150 degrees, which is less than -π/2 (-90 degrees). So -5π/6 is outside the range of arcsin. Therefore, the only angle in [-π/2, π/2] with sine -1/2 is -π/6.
</execution_2.1>

<execution_2.2>
For π/6, sine is 1/2, so negative π/6 should have sine -1/2. That's correct. So, since arcsin is the inverse function, and considering the range, the answer must be -π/6.
</execution_2.2>
</Execution_2>

<Final_answer>
\boxed{-\frac{\pi}{6}}.

\]
</Final_answer>
</think>