rewrite cos(2tan^-1v) as an algebraic expression in v

To rewrite \(\cos(2 \tan^{-1}(v))\) as an algebraic expression in terms of \(v\), we follow these steps:

1. **Let** \(\theta = \tan^{-1}(v)\), so that \(\tan(\theta) = v\).

2. Now, we want to find \(\cos(2\theta)\). Using the double-angle identity for cosine:
   \[
   \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}
   \]

3. Since \(\tan(\theta) = v\), substitute \(v\) for \(\tan(\theta)\):
   \[
   \cos(2\theta) = \frac{1 - v^2}{1 + v^2}
   \]

Thus, \(\cos(2 \tan^{-1}(v)) = \frac{1 - v^2}{1 + v^2}\).

Would you like further details on the process, or do you have any questions?

Here are 5 related questions to expand on this:
1. How would you find \(\sin(2\tan^{-1}(v))\) in terms of \(v\)?
2. What is the geometric interpretation of \(\tan^{-1}(v)\)?
3. Can you express \(\sin(\tan^{-1}(v))\) as an algebraic expression in \(v\)?
4. How would you derive the double-angle identity for cosine?
5. What is the relationship between trigonometric identities and inverse trigonometric functions?

**Tip:** Remember that trigonometric identities are useful tools for converting complex expressions into simpler forms, especially when working with inverse trigonometric functions.

Rewrite cos(2tan^-1v) as an algebraic expression in v.

Learn how to rewrite cos(2tan^-1(v)) as an algebraic expression in v using trigonometric identities and inverse functions. This problem explores the double angle identity for cosine and algebraic manipulation techniques. Suitable for students in Grades 10-12.

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