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The problem asks for the exact value of \( \cos \left( \frac{5\pi}{12} \right) \).

To solve this, let's break down \( \frac{5\pi}{12} \) into angles whose cosine values we know. We can use the sum or difference identities for cosine to find the exact value.

### Step-by-Step Solution
1. **Rewrite** \( \frac{5\pi}{12} \) as a sum:
   \[
   \frac{5\pi}{12} = \frac{3\pi}{12} + \frac{2\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}
   \]

2. **Use the Cosine Sum Identity**:
   \[
   \cos \left( \frac{\pi}{4} + \frac{\pi}{6} \right) = \cos \frac{\pi}{4} \cos \frac{\pi}{6} - \sin \frac{\pi}{4} \sin \frac{\pi}{6}
   \]

3. **Substitute Values**:
   - \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \)
   - \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
   - \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \)
   - \( \sin \frac{\pi}{6} = \frac{1}{2} \)

4. **Calculate**:
   \[
   \cos \left( \frac{\pi}{4} + \frac{\pi}{6} \right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}
   \]
   \[
   = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}
   \]
   \[
   = \frac{\sqrt{6} - \sqrt{2}}{4}
   \]

### Final Answer
The exact value is:
\[
\cos \left( \frac{5\pi}{12} \right) = \frac{\sqrt{6} - \sqrt{2}}{4}
\]

Would you like further details on how to apply trigonometric identities, or do you have other questions?

### Additional Practice Questions:
1. What is the exact value of \( \cos \left( \frac{7\pi}{12} \right) \)?
2. How can you use the sine sum identity to find \( \sin \left( \frac{5\pi}{12} \right) \)?
3. What is \( \tan \left( \frac{5\pi}{12} \right) \) using sum identities?
4. How do you simplify \( \cos \left( \frac{3\pi}{4} - \frac{\pi}{6} \right) \)?
5. Find the exact value of \( \cos \left( \frac{\pi}{3} + \frac{\pi}{4} \right) \).

**Tip:** Always look for ways to rewrite angles in terms of known values, as this often simplifies complex trigonometric calculations.

Find the exact value of cos(5π/12) by applying the cosine sum identity. This problem covers trigonometric concepts and identity applications, making it suitable for high school students in Grades 11-12.

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