To solve \( \log_2 \left( \sin^{-1} \left( \frac{4\sqrt{3}}{7} \right) \right) \), we need to carefully work through the expression step-by-step.

### Step 1: Evaluate the inverse sine function

We are given \( \sin^{-1} \left( \frac{4\sqrt{3}}{7} \right) \), which asks us to find an angle \( \theta \) such that:
\[
\sin(\theta) = \frac{4\sqrt{3}}{7}
\]
However, for \( \sin(\theta) \), the values range between \(-1\) and \(1\), and we need to check whether \( \frac{4\sqrt{3}}{7} \) is within this range. We first calculate the numerical value of \( \frac{4\sqrt{3}}{7} \):
\[
\frac{4\sqrt{3}}{7} \approx \frac{4 \times 1.732}{7} \approx \frac{6.928}{7} \approx 0.9897
\]

Since \( 0.9897 \) is within the valid range for sine (i.e., between \(-1\) and \(1\)), we can proceed.

### Step 2: Find the angle \( \theta \)

To determine the angle \( \theta = \sin^{-1}(0.9897) \), we use a calculator:
\[
\theta \approx \sin^{-1}(0.9897) \approx 1.424 \, \text{radians}
\]

### Step 3: Apply the logarithmic function

Now we apply the base-2 logarithm to the result:
\[
\log_2(1.424)
\]
Using a calculator for the logarithm base-2:
\[
\log_2(1.424) \approx 0.511
\]

Thus, the value of \( \log_2 \left( \sin^{-1} \left( \frac{4\sqrt{3}}{7} \right) \right) \) is approximately \( 0.511 \).

Would you like further details or explanations?

Here are 5 related questions:
1. How do you compute inverse trigonometric functions manually?
2. What are the properties of logarithms that help in solving complex logarithmic equations?
3. How is the sine function derived for non-standard angles?
4. Why does the sine function have restrictions for its inverse?
5. What are the typical ranges for \( \sin^{-1}(x) \) in degrees and radians?

**Tip**: Always ensure your input values fall within the domain of the trigonometric function you're dealing with!

Explore the solution to log₂(sin⁻¹(4√3/7)), involving inverse trigonometric functions and base-2 logarithms. Learn how to compute sin⁻¹(4√3/7), verify the value's range, and apply logarithmic rules. This detailed explanation is ideal for students in Grades 11-12.

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