Henry is asked to find the exact value of \( \cos \frac{10 \pi}{3} \). His steps are shown below. 1. Subtract \( 2 \pi \) from \( \frac{10 \pi}{3} \) as many times as possible: \( \frac{10 \pi}{3}-2 \pi=\frac{4 \pi}{3} \) 2. Find the reference angle for \( \frac{4 \pi}{3}: \frac{3 \pi}{2}-\frac{4 \pi}{3}=\frac{\pi}{6} \). 3. The cosine value for \( \frac{\pi}{6} \) is \( \frac{\sqrt{3}}{2} \). 4. The cosine value is positive because \( \frac{\pi}{6} \) is in the first quadrant. Which of the following describes Henry's errors?

Henry made two main errors: - In step 2, the reference angle for an angle in the third quadrant is found by subtracting \( \pi \) from the angle. For \(\frac{4\pi}{3}\), the correct reference angle is: \[ \frac{4\pi}{3} - \pi = \frac{\pi}{3}. \] Henry incorrectly computed it as: \[ \frac{3\pi}{2} - \frac{4\pi}{3} = \frac{\pi}{6}. \] - In step 4, the cosine of an angle in the third quadrant should be negative. Even if the reference angle were correctly identified (as \(\frac{\pi}{3}\)), the cosine value is: \[ \cos \frac{\pi}{3} = \frac{1}{2}, \] so the actual value should be: \[ \cos \frac{4\pi}{3} = -\frac{1}{2}. \] Henry mistakenly used the cosine value for \(\frac{\pi}{6}\) and assumed it was positive. Thus, Henry’s errors are miscalculation of the reference angle and incorrect determination of the sign of the cosine in the third quadrant.