Okay, let's see. I need to find the range of the function y = log base 2 of sqrt(sin x) where x is between 0 degrees and 180 degrees. Hmm, ranges can sometimes be tricky, especially when dealing with logarithms and trigonometric functions. Let me break this down step by step.

First, let's recall that the logarithm function log_2(z) is defined only when z is positive. So, inside the logarithm, sqrt(sin x) must be positive. Since sqrt(sin x) is the square root of sin x, sin x must be positive for the square root to be real. Therefore, sin x > 0. But when is sin x positive? Well, sine is positive in the first and second quadrants, which corresponds to x between 0 degrees and 180 degrees. Wait, but the problem already states that x is between 0° and 180°, so that's already satisfied. So, sin x is positive here, so sqrt(sin x) is real. But we also need to make sure that sqrt(sin x) is not zero because log_2(0) is undefined. So sin x must be greater than 0, which is already true in the open interval (0°, 180°). So, the domain here is actually (0°, 180°), excluding the points where sin x = 0. But since x is strictly between 0 and 180°, sin x is positive, so sqrt(sin x) is positive. Therefore, the logarithm is defined.

Now, the function is y = log_2(sqrt(sin x)). Let me simplify this expression first. Remember that sqrt(sin x) is the same as (sin x)^(1/2). So, log_2((sin x)^(1/2)) can be rewritten using logarithm properties as (1/2) * log_2(sin x). So, y = (1/2) * log_2(sin x). That might make it easier to analyze.

So, to find the range of y, we need to find all possible values that y can take as x varies from 0° to 180°. Since y is half of log_2(sin x), the range of y depends on the range of log_2(sin x), scaled by 1/2.

Therefore, if we can find the range of log_2(sin x), then dividing by 2 will give the range of y.

So, let's focus on finding the range of log_2(sin x) for x ∈ (0°, 180°). 

First, let's consider the range of sin x over this interval. When x is between 0° and 180°, sin x starts at 0, increases to 1 at 90°, and then decreases back to 0. So, the range of sin x is (0, 1]. Wait, but at x = 0° and x = 180°, sin x is 0, but since x is strictly between 0° and 180°, sin x is in (0, 1]. So, sin x ∈ (0, 1].

Therefore, log_2(sin x) will take values where the input to log_2 is between 0 and 1. Let's recall the behavior of log base 2. For arguments between 0 and 1, log_2(z) is negative. Specifically, log_2(1) = 0, and as z approaches 0 from the right, log_2(z) approaches negative infinity. Also, log_2(z) is a monotonically increasing function. Wait, is that right? Wait, no. Wait, the logarithm function log_b(z) is increasing if b > 1. Since 2 > 1, log_2(z) is increasing. But when z is between 0 and 1, even though z is increasing towards 1, log_2(z) is increasing from negative infinity up to 0. So, as sin x increases from 0 to 1, log_2(sin x) increases from negative infinity up to 0. Therefore, the range of log_2(sin x) is (-infty, 0].

But wait, hold on. Wait, sin x is in (0, 1]. So, log_2(sin x) is in (-infty, 0]. Therefore, log_2(sin x) can take any real value less than or equal to 0. Then, multiplying by 1/2, the range of y = (1/2) log_2(sin x) is (-infty, 0].

But wait, is that correct? Let me check again. Let's take some specific values. For example, when sin x approaches 0, log_2(sin x) approaches negative infinity, so y approaches negative infinity. When sin x = 1, log_2(1) = 0, so y = 0. So, as x approaches 90°, sin x approaches 1, so y approaches 0. So, the maximum value of y is 0, and it can go down to negative infinity. Therefore, the range of y is (-infty, 0]. So, the range is all real numbers less than or equal to 0.

But wait, let me verify this with another approach. Let's consider substituting t = sin x. Then, t ∈ (0, 1]. Then, y = (1/2) log_2(t). So, as t ranges from 0 to 1, log_2(t) ranges from -infty to 0, so y ranges from -infty to 0. So, yes, that confirms it.

Therefore, the range of the original function y is (-infty, 0]. But let me make sure I didn't make any miscalculations here.

Wait, but let me check if there's any restriction on sin x beyond (0, 1]. For example, in the interval (0°, 180°), sin x is symmetric around 90°, so sin x increases from 0 to 1 as x goes from 0° to 90°, and then decreases back to 0 as x goes from 90° to 180°. So, the maximum value of sin x is 1 at 90°, and the minimum is approaching 0 as x approaches 0° or 180°.

Therefore, log_2(sin x) will take all values from negative infinity up to 0. Hence, multiplying by 1/2 doesn't change the interval's bounds, just scales it. So, y can be any real number less than or equal to 0.

Wait, but the problem is presented in degrees, but logarithmic functions are unit-agnostic here. So, regardless of the units, the range is determined by the argument of the logarithm, which is sin x. So, the answer should be (-infty, 0].

But let me check if the original function is defined for all x in (0°, 180°). Because sqrt(sin x) is positive, so log_2 is defined. Yes. So, the domain is (0°, 180°), and the range is (-infty, 0].

But wait, let me think again. Let's take x approaching 0°, sin x approaches 0, so sqrt(sin x) approaches 0, so log_2(0) is negative infinity. So, y approaches negative infinity. Similarly, as x approaches 180°, sin x approaches 0, so the same thing. At x = 90°, sin x = 1, so sqrt(1) = 1, so log_2(1) = 0. So, y = 0. So, the maximum value of y is 0, and it decreases without bound as x approaches the endpoints. Therefore, the range is indeed (-infty, 0].

But wait, the problem says "for 0° < x < 180°". So, the function is defined on an open interval. Therefore, sin x never actually reaches 0 or 1 in the open interval. Wait, no. Wait, at x = 90°, sin x = 1, which is included in the interval (0°, 180°), as 90° is between 0° and 180°. So, sin x reaches 1 at x = 90°, which is inside the domain. Therefore, log_2(1) = 0 is attainable. Therefore, the maximum value of y is 0. However, as x approaches 0° or 180°, sin x approaches 0, so log_2(sin x) approaches negative infinity, so y approaches negative infinity. Therefore, the range is all real numbers less than or equal to 0. But since x is strictly between 0° and 180°, sin x is strictly between 0 and 1, so log_2(sin x) is strictly less than 0. Wait, but at x = 90°, sin x = 1, so log_2(1) = 0. So, even though x is in the open interval, the function y can attain 0. So, the range is (-infty, 0]. Therefore, the function's range is all real numbers y such that y ≤ 0.

But let me confirm once again. If x is in (0°, 180°), then sin x ∈ (0, 1]. So, sqrt(sin x) ∈ (0, 1]. Then, log_2(sqrt(sin x)) is log_2(z), where z ∈ (0, 1]. Since log_2(z) is defined for z > 0, and log_2(z) ≤ 0 for z ∈ (0, 1]. Therefore, the range is (-infty, 0]. So, yes, that's correct.

Therefore, the range of the function is all real numbers y such that y ≤ 0, which can be written as (-∞, 0].

But let me check with another angle. Suppose we let t = sin x. Then, since x ∈ (0°, 180°), t ∈ (0, 1]. Then, log_2(sqrt(t)) = (1/2) log_2(t). So, as t approaches 0, log_2(t) approaches -infty, so y approaches -infty. When t = 1, log_2(1) = 0, so y = 0. Since t can take any value in (0, 1], y can take any value in (-infty, 0]. Therefore, the range is (-infty, 0].

Therefore, the answer should be (-infty, 0], which in boxed form is \boxed{(-\infty, 0]}.

But wait, hold on. Wait a second. Let me check if there's a mistake here. Wait, in the problem statement, the function is log base 2 of sqrt(sin x). So, sqrt(sin x) is the square root, which is equivalent to (sin x)^(1/2). So, log_2((sin x)^(1/2)) = (1/2) log_2(sin x). So, that's correct.

But let me confirm with specific values. For example, when x = 45°, sin x = sqrt(2)/2 ≈ 0.7071. Then, sqrt(sin x) ≈ sqrt(0.7071) ≈ 0.841, and log_2(0.841) ≈ log_2(1/√2) ≈ log_2(2^(-1/2)) = -1/2. So, y = (1/2)*(-1/2) = -1/4. So, y is -0.25. That's within (-infty, 0]. Another value: x = 30°, sin x = 0.5. sqrt(0.5) ≈ 0.7071, log_2(0.7071) ≈ -0.5, so y = -0.5/2 = -0.25. Wait, that's the same as before. Wait, maybe my example is not good. Let's take x = 60°, sin x = sqrt(3)/2 ≈ 0.866. sqrt(0.866) ≈ 0.9306. log_2(0.9306) ≈ log_2(2^(-0.087)) ≈ -0.087. So, y ≈ -0.0435, which is still negative. So, as x approaches 90°, sin x approaches 1, so sqrt(sin x) approaches 1, log_2(1) = 0, so y approaches 0. So, that's correct.

Another test: x = 15°, sin x ≈ 0.2588. sqrt(0.2588) ≈ 0.5085. log_2(0.5085) ≈ log_2(1/2) = -1 (since 2^(-1) = 0.5). Wait, 0.5085 is slightly larger than 0.5, so log_2(0.5085) ≈ -0.97. So, y ≈ -0.97/2 ≈ -0.485. So, still negative. So, all these test cases confirm that y is negative or zero, and can be as low as negative infinity.

Therefore, I think my conclusion is correct. The range is all real numbers y such that y ≤ 0, which is (-∞, 0].

**Final Answer**
The range of the function is \boxed{(-\infty, 0]}.
Therefore, the range of the function $y=\log_2 (\sqrt{\sin x})$ for $0^\circ< x < 180^\circ$ is $\boxed{(-\infty, 0]}$.
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