To solve $\sin\left(2 \tan^{-1}(x)\right)$, we can use a trigonometric identity involving double angles and inverse trigonometric functions.

Step-by-Step Solution

1. Recognize the Double Angle Identity for Sine:

   We know that: $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$

   So, $\sin\left(2 \tan^{-1}(x)\right) = 2 \sin\left(\tan^{-1}(x)\right) \cos\left(\tan^{-1}(x)\right)$

2. Find $\sin(\tan^{-1}(x))$ and $\cos(\tan^{-1}(x))$:

   Let $\theta = \tan^{-1}(x)$, so $\tan(\theta) = x$.

   We can represent this using a right triangle where:

   • Opposite side = $x$
   • Adjacent side = $1$

   Then, by the Pythagorean theorem, the hypotenuse is: $\sqrt{x^2 + 1}$

   Thus: $\sin(\theta) = \frac{x}{\sqrt{x^2 + 1}} \quad \text{and} \quad \cos(\theta) = \frac{1}{\sqrt{x^2 + 1}}$

3. Apply These Values in the Identity:

   Now, $\sin\left(2 \tan^{-1}(x)\right) = 2 \cdot \frac{x}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{x^2 + 1}}$

4. Simplify:

   $\sin\left(2 \tan^{-1}(x)\right) = \frac{2x}{x^2 + 1}$

Final Answer:

$\sin\left(2 \tan^{-1}(x)\right) = \frac{2x}{x^2 + 1}$

Would you like more details on any of these steps?

Additional Related Questions

1. How do we find the cosine of double angles involving inverse trigonometric functions?
2. What is the derivative of $\sin\left(2 \tan^{-1}(x)\right)$?
3. How do we solve problems involving $\sin(\tan^{-1}(x))$ without double angles?
4. Can we extend this approach to other inverse trigonometric functions, like $\cos^{-1}(x)$?
5. What are some applications of this result in calculus or physics?

Tip:

When simplifying expressions with inverse trigonometric functions, try to visualize them with triangles, as it often helps in finding sine, cosine, and tangent values directly.