How do you find the terminal point on the unit circle determined by t = 5π/12?

To find the terminal point on the unit circle determined by \( t = \frac{5\pi}{12} \), follow these steps:

1. **Understand the unit circle and the parameter \( t \):**
- The unit circle is a circle of radius 1 centered at the origin \((0, 0)\) in the coordinate plane.
- The parameter \( t \) represents the radian measure of the angle swept counterclockwise from the positive x-axis.

2. **Express the angle \( t \) in terms of sine and cosine:**
- On the unit circle, the terminal point corresponding to an angle \( t \) is given by:
\[
(\cos t, \sin t)
\]
- This means we need to calculate \( \cos\left(\frac{5\pi}{12}\right) \) and \( \sin\left(\frac{5\pi}{12}\right) \).

3. **Simplify the angle \( t = \frac{5\pi}{12} \):**
- Break \( \frac{5\pi}{12} \) into known reference angles. Rewrite it as:
\[
\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}
\]
- This allows us to use the angle sum identities for sine and cosine:
\[
\cos(a + b) = \cos a \cos b - \sin a \sin b
\]
\[
\sin(a + b) = \sin a \cos b + \cos a \sin b
\]
- Here, \( a = \frac{\pi}{4} \) and \( b = \frac{\pi}{6} \).

4. **Compute \( \cos\left(\frac{5\pi}{12}\right) \):**
- Using \( \cos(a + b) \):
\[
\cos\left(\frac{5\pi}{12}\right) = \cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right) - \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{6}\right)
\]
- Substituting known values:
\[
\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \, \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \, \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \, \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
\]
- Perform the calculation:
\[
\cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}
\]
\[
\cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}
\]
\[
\cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{6} - \sqrt{2}}{4}
\]

5. **Compute \( \sin\left(\frac{5\pi}{12}\right) \):**
- Using \( \sin(a + b) \):
\[
\sin\left(\frac{5\pi}{12}\right) = \sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{6}\right)
\]
- Substituting known values:
\[
\sin\left(\frac{5\pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}
\]
- Perform the calculation:
\[
\sin\left(\frac{5\pi}{12}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}
\]
\[
\sin\left(\frac{5\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}
\]

6. **Write the terminal point:**
- The terminal point on the unit circle is:
\[
\left(\cos\left(\frac{5\pi}{12}\right), \sin\left(\frac{5\pi}{12}\right)\right)
\]
- Substituting the results:
\[
\left(\frac{\sqrt{6} - \sqrt{2}}{4}, \frac{\sqrt{6} + \sqrt{2}}{4}\right)
\]

Final answer:
The terminal point on the unit circle for \( t = \frac{5\pi}{12} \) is \( \left(\frac{\sqrt{6} - \sqrt{2}}{4}, \frac{\sqrt{6} + \sqrt{2}}{4}\right) \).