To find the terminal point on the unit circle corresponding to \( t = -\frac{41\pi}{4} \), we can follow these steps:

### Step 1: Reduce the angle
The angle \( t = -\frac{41\pi}{4} \) is greater than \( 2\pi \) in magnitude, so we first reduce it to an equivalent angle between \( 0 \) and \( 2\pi \). To do this, we add multiples of \( 2\pi \) to get a positive equivalent angle.

\[
2\pi = \frac{8\pi}{4}
\]

So, add \( 2\pi \) multiple times until the angle falls within the range \( [0, 2\pi] \). We do this by calculating:

\[
t + 10\pi = -\frac{41\pi}{4} + \frac{40\pi}{4} = -\frac{\pi}{4}
\]

Thus, \( t = -\frac{\pi}{4} \) is the equivalent angle within \( [0, 2\pi] \).

### Step 2: Find the terminal point
The angle \( -\frac{\pi}{4} \) corresponds to the same terminal point as \( \frac{7\pi}{4} \) in the standard position (since it's a counterclockwise rotation from \( 2\pi \)).

On the unit circle, the terminal point corresponding to \( \frac{7\pi}{4} \) is:

\[
\left( \cos\left(\frac{7\pi}{4}\right), \sin\left(\frac{7\pi}{4}\right) \right)
\]

We know that:

\[
\cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2} \quad \text{and} \quad \sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2}
\]

### Final Answer:
The terminal point is \( \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \).

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Would you like more details on this process or further clarification? 

Here are some related questions for deeper understanding:

1. How do you reduce negative angles to positive ones in trigonometry?
2. What are the coordinates of common angles on the unit circle?
3. How does the unit circle help in understanding trigonometric functions?
4. What is the significance of the terminal point on the unit circle?
5. Can the same process be used for positive angles larger than \( 2\pi \)?

**Tip**: When working with angles on the unit circle, reducing them modulo \( 2\pi \) simplifies your calculations.

Learn how to find the terminal point on the unit circle corresponding to t = -41π/4. This problem involves angle reduction to an equivalent angle between 0 and 2π, using key trigonometric concepts. Suitable for students in Grades 10-12.

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