To find \( \sin(\tan^{-1}(4x)) \) where \( -\frac{1}{4} \leq x \leq \frac{1}{4} \), we can proceed as follows:

1. **Substitute** \( y = \tan^{-1}(4x) \). This means \( \tan(y) = 4x \).

2. **Use a right triangle** to interpret this. If \( \tan(y) = 4x \), then in a right triangle, the opposite side is \( 4x \) and the adjacent side is \( 1 \). Using the Pythagorean theorem, the hypotenuse \( h \) is:
   \[
   h = \sqrt{(4x)^2 + 1^2} = \sqrt{16x^2 + 1}
   \]

3. **Find \( \sin(y) \)**. In the triangle, \( \sin(y) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4x}{\sqrt{16x^2 + 1}} \).

Thus,
\[
\sin(\tan^{-1}(4x)) = \frac{4x}{\sqrt{16x^2 + 1}}
\]

So, the expression for \(\sin(\tan^{-1}(4x))\) is \(\frac{4x}{\sqrt{16x^2 + 1}}\), where \( -\frac{1}{4} \leq x \leq \frac{1}{4} \).

Would you like more details or have any questions?

Here are some questions you might ask next:
1. How do we derive the hypotenuse in a right triangle?
2. Can you explain why \(\tan^{-1}(4x)\) gives a specific angle in a right triangle?
3. How can we use similar techniques to solve for \(\cos(\tan^{-1}(4x))\)?
4. What is the geometric interpretation of the \(\tan^{-1}\) function?
5. Can we simplify \(\frac{4x}{\sqrt{16x^2 + 1}}\) further?

**Tip:** When working with inverse trigonometric functions, using right triangles and the Pythagorean theorem can simplify the problem greatly.

Learn how to compute sin(tan^-1(4x)) where -1/4 <= x <= 1/4. Understand the trigonometric relationships involved and apply the Pythagorean theorem in a right triangle. Suitable for advanced high school students.

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