<Question>
What is the range of the function $y=\log_2 (\sqrt{\sin x})$ for $0^\circ< x < 180^\circ$?
</Question>

<think>
<Plan_1>
Okay, so I need to find the range of the function y = log base 2 of the square root of sin x, where x is between 0 degrees and 180 degrees. Hmm, let me break this down step by step.

<prompt_1.1> Identify the components of the function y = log base 2 of the square root of sin x and determine the domain of the inner functions. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, let's recall that the range of a function is the set of all possible output values (y-values) it can produce. To find the range here, I need to consider the possible values that log_2(sqrt(sin x)) can take as x varies between 0 and 180 degrees.

Let's start by analyzing the inner functions. The function inside the logarithm is sqrt(sin x). So, first, sin x must be positive because the logarithm of a non-positive number is undefined. Also, the square root of sin x is defined only when sin x is non-negative. Therefore, sin x must be greater than 0. 

When is sin x positive? In the interval 0° < x < 180°, sin x is positive in the first and second quadrants, which corresponds to 0° < x < 180°, but excluding the points where sin x is zero. Since x is strictly between 0 and 180, sin x is positive throughout this interval. Wait, but at x = 0° and x = 180°, sin x is zero, but those endpoints are excluded. So in the open interval (0°, 180°), sin x is positive.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Therefore, sqrt(sin x) is defined and real for all x in (0°, 180°). However, sqrt(sin x) will output values between 0 and 1 because sin x ranges between 0 and 1 in this interval. But since sin x is between 0 and 1, sqrt(sin x) will also be between 0 and 1. But wait, when sin x is 1, sqrt(1) is 1, and when sin x approaches 0, sqrt(sin x) approaches 0. So sqrt(sin x) is in (0, 1].

Now, taking the logarithm base 2 of sqrt(sin x). Let's recall that log_2(z) is defined for z > 0. Since sqrt(sin x) is in (0, 1], log_2(sqrt(sin x)) is defined. But log_2 of a number between 0 and 1 is negative because log base 2 of 1 is 0, and as the number decreases from 1 to 0, the log becomes more and more negative. So log_2(sqrt(sin x)) will take on negative values.

But let's be precise here. Let's find the exact range.

Let me denote t = sqrt(sin x). Then t is in (0, 1], as established earlier. Then y = log_2(t). So we can rephrase the problem as finding the range of log_2(t) where t ∈ (0, 1].

<prompt_2.1> Find the range of log_2(t) for t in the interval (0, 1]. </prompt_2.1>
-----

But let me verify this again. Let's consider the behavior of sin x in the interval (0°, 180°). The maximum value of sin x is 1, which occurs at x = 90°. As x approaches 0° or 180°, sin x approaches 0.

<prompt_2.2> Verify the range by considering the behavior of sin x in the interval (0°, 180°). </prompt_2.2>

But let's check another angle. For example, take x = 45°.

<prompt_2.3> Check the range by taking specific examples such as x = 45°, x = 60°, and x approaching 0° or 180°. </prompt_2.3>
-----

But to make sure, let's analyze the composition of functions. Let's consider the domain of the original function.

<prompt_2.4> Analyze the composition of functions to ensure the range is correct. </prompt_2.4>

Alternatively, maybe we can do a substitution. Let’s set t = sin x. Then t ∈ (0, 1] because x ∈ (0°, 180°). Then sqrt(t) is in (0, 1], and log_2(sqrt(t)) is log_2(t^(1/2)) = (1/2) log_2(t). So y = (1/2) log_2(t), where t ∈ (0, 1].

<prompt_2.5> Use substitution to confirm the range of the function. </prompt_2.5>

Is there a possibility that sqrt(sin x) could be equal to 1? Yes, when sin x = 1, which occurs at x = 90°.

<prompt_2.6> Check if there is a possibility that sqrt(sin x) could be equal to 1. </prompt_2.6>

But let's check another angle. For example, take x = 30°.

<prompt_2.7> Verify the range by taking another specific example such as x = 30°. </prompt_2.7>

Wait, but is there a case where sqrt(sin x) could be greater than 1? No, because sin x is always between 0 and 1, so sqrt(sin x) is between 0 and 1.

<prompt_2.8> Confirm that sqrt(sin x) is always between 0 and 1. </prompt_2.8>
</Plan_2>

<Execution_2>
<execution_2.1>
The logarithm function log_2(t) is a continuous function decreasing from 0 to negative infinity as t decreases from 1 to 0. Therefore, since t is in (0, 1], the range of log_2(t) will be (-∞, 0].
</execution_2.1>

<execution_2.2>
So sqrt(sin x) approaches 0. Therefore, t ranges from just above 0 to 1. 

So log_2(t) when t is in (0, 1] will range from log_2(1) = 0 down to negative infinity. Wait, hold on. If t approaches 0, then log_2(t) approaches negative infinity. But if t is in (0, 1], then log_2(t) is in (-∞, 0]. 

Wait, but that seems conflicting with my initial thought. Let me check again. 

If t is 1, log_2(1) = 0. If t is 1/2, log_2(1/2) = -1. If t is 1/4, log_2(1/4) = -2, and so on. So as t approaches 0, log_2(t) approaches negative infinity. Therefore, the range of log_2(t) where t ∈ (0, 1] is (-∞, 0]. 

But wait, but sqrt(sin x) can take the value 1 when sin x = 1, which is at x = 90°. So log_2(1) = 0 is included in the range. Then, as sin x varies from just above 0 up to 1, sqrt(sin x) varies from just above 0 up to 1, so log_2(sqrt(sin x)) varies from negative infinity up to 0. Therefore, the range of y is (-∞, 0].
</execution_2.2>

<execution_2.3>
Then sin 45° is sqrt(2)/2 ≈ 0.707. So sqrt(sin x) ≈ 0.841, and log base 2 of that is log2(0.841) ≈ -0.25. So that's a negative value. If x is 60°, sin 60° is sqrt(3)/2 ≈ 0.866. sqrt(0.866) ≈ 0.9306, log2(0.9306) ≈ -0.105. So even closer to zero. 

If x approaches 90°, sin x approaches 1, so sqrt(sin x) approaches 1, and log2(1) is 0. So as x approaches 90°, y approaches 0. 

As x approaches 0° or 180°, sin x approaches 0, so sqrt(sin x) approaches 0, and log2 of a number approaching 0 from the positive side goes to negative infinity. Therefore, the function y can take on values from negative infinity up to 0.
</execution_2.3>

<execution_2.4>
The function is log_2(sqrt(sin x)). For this to be defined, sin x must be positive, which it is in (0°, 180°). Then sqrt(sin x) is between 0 and 1, as discussed. Then log_2 of that is between negative infinity and 0. Wait, but when t approaches 0, log_2(t) approaches negative infinity. But in our case, t is in (0, 1], so log_2(t) is in (-∞, 0]. But wait, when t = 1, log_2(1) = 0. When t approaches 0, log_2(t) approaches -∞. So the range of y is (-∞, 0].
</execution_2.4>

<execution_2.5>
Since t ∈ (0, 1], log_2(t) ∈ (-∞, 0]. Therefore, multiplying by 1/2, we still have (1/2) log_2(t) ∈ (-∞, 0]. Therefore, the range of y is (-∞, 0].
</execution_2.5>

<execution_2.6>
So sqrt(1) = 1, so log_2(1) = 0, so y = 0. So when x = 90°, y = 0. 

Then, as t approaches 0, log_2(t) approaches negative infinity, so (1/2) log_2(t) also approaches negative infinity. Therefore, the range of y is (-∞, 0].
</execution_2.6>

<execution_2.7>
Then sin x = 0.5, so sqrt(0.5) = √(1/2) = 1/√2 ≈ 0.707. Then log_2(0.707) is log_2(1/√2) = log_2(2^(-1/2)) = -1/2. So y = -0.5 in this case.
</execution_2.7>

<execution_2.8>
Therefore, log_2(sqrt(sin x)) is log_2 of a number between 0 and 1, hence negative, and since the maximum value of sqrt(sin x) is 1, the maximum value of log_2(sqrt(sin x)) is 0, achieved when sqrt(sin x) = 1, i.e., sin x = 1, which is at x = 90°. As sin x approaches 0, log_2(sqrt(sin x)) approaches negative infinity.
</execution_2.8>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But let me double-check.

Alternatively, perhaps we can write this as a composition of functions.

<prompt_3.1> Verify the range by considering the composition of functions. </prompt_3.1>
-----

Therefore, the answer should be that the range is all real numbers less than or equal to 0, which can be written as (-∞, 0].

<prompt_3.2> Confirm the final range of the function y = log base 2 of the square root of sin x. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let’s denote u = sin x, so u ∈ (0, 1]. Then v = sqrt(u), so v ∈ (0, 1]. Then y = log_2(v). 

So the range of y is the set { log_2(v) | v ∈ (0, 1] }. Since log_2 is a strictly increasing function, the maximum value occurs at the maximum v, which is 1. log_2(1) = 0. The minimum value is approached as v approaches 0, which is -∞. Therefore, the range is (-∞, 0].
</execution_3.1>

<execution_3.2>
But let me verify once more.

Alternatively, maybe we can think in terms of substitution. Let’s set t = sqrt(sin x). Then t ∈ (0, 1], as sin x ∈ (0, 1]. Then y = log_2(t). The logarithm function log_2(t) is a continuous and strictly increasing function when t > 0. However, since t ∈ (0, 1], log_2(t) will take values from -∞ up to 0. Therefore, the range is (-∞, 0].
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
**Final Answer**
The range of the function is \boxed{(-\infty, 0]}.
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the range of the function $y=\log_2 (\sqrt{\sin x})$ for $0^\circ< x < 180^\circ$ is $\boxed{(-\infty,0]}$.
</Final_answer>
</think>