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The given expression is \( \cos \left( \tan^{-1} \left( \frac{\frac{w}{\sqrt{36 - w}}}{2} \right) \right) \), and we need to rewrite it as an algebraic expression in \( w \).

### Step-by-step Solution:

1. **Let** \( \theta = \tan^{-1} \left( \frac{\frac{w}{\sqrt{36 - w}}}{2} \right) \).  
   This implies:
   \[
   \tan(\theta) = \frac{\frac{w}{\sqrt{36 - w}}}{2} = \frac{w}{2\sqrt{36 - w}}
   \]

2. **Using the right triangle approach**:  
   Since \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \), let the opposite side of the right triangle be \( w \), and the adjacent side be \( 2\sqrt{36 - w} \).

3. **Find the hypotenuse** using the Pythagorean theorem:
   \[
   \text{hypotenuse} = \sqrt{w^2 + \left( 2\sqrt{36 - w} \right)^2}
   = \sqrt{w^2 + 4(36 - w)}
   = \sqrt{w^2 + 144 - 4w}
   = \sqrt{w^2 - 4w + 144}
   \]

4. **Find \( \cos(\theta) \)**:  
   The cosine of \( \theta \) is given by:
   \[
   \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2\sqrt{36 - w}}{\sqrt{w^2 - 4w + 144}}
   \]

Thus, the algebraic expression for \( \cos \left( \tan^{-1} \left( \frac{\frac{w}{\sqrt{36 - w}}}{2} \right) \right) \) is:

\[
\boxed{\frac{2\sqrt{36 - w}}{\sqrt{w^2 - 4w + 144}}}
\]

Let me know if you would like more details or have any further questions!

Here are 5 questions related to this:

1. How do you compute the sine from a trigonometric inverse?
2. What is the relationship between trigonometric functions and right triangles?
3. How can you simplify complex expressions under a square root?
4. Why is the Pythagorean theorem essential in trigonometry?
5. What are common ways to express inverse trigonometric functions algebraically?

**Tip:** Inverse trigonometric functions can be transformed into algebraic forms using triangle relationships and the Pythagorean theorem.

Rewrite cos(tan^(-1)(w / sqrt(36 - w))) as an algebraic expression in w.

Learn how to rewrite the trigonometric expression cos(tan^(-1)(w / sqrt(36 - w))) as an algebraic expression in w. This solution covers key concepts in trigonometry, inverse functions, and the Pythagorean Theorem, suitable for high school students in Grades 10-12.

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