Let $ \sin \alpha = a, \cos \alpha = b$, so $ a^2 + b^2 = 1$. WLOG $ a < b$. Then the four terms are $ a, b, a/b, b/a$, with possible orders 1. $ a, b, a/b, b/a$ 2. $ a, a/b, b, b/a$ In the second case, we have $ a/b + b/a = 2b$. But that implies $ 2b \geq 2$ and thus $ b = 1$, which is not in the domain. In the first case, $ a+a/b = 2b$. So $ b + 1 = 2bb/a$ and $ b = 2/a - 1 - 2a$. Then $ b^2 = 4/a^2 - 4/a - 7 + 4a + 4a^2$, or $ 0 = 5a^4 + 4a^3 - 8a^2 - 4a + 4$. We also have $ 2a/b = b + b/a$. Then $ 2a = 1/a - a + 1 - aa$ and $ 0 = a^3 + 3a^2 - a - 1$. But these two polynomials are relatively prime, so no $ a$ exist.

Since $ \sin (\alpha) = \cos (\frac{\pi}{2}-\alpha)$ and $ \tan (\alpha) = \cot (\frac{\pi}{2}-\alpha)$, WLOG, let $ \alpha \in (0, \frac{\pi}{4}]$. It is easy to see that $ \alpha = \frac{\pi}{4}$ does not yield an arithmetic progression, so $ \alpha \in (0, \frac{\pi}{4})$. Also, it is easy to see that the sequence cannot be constant, so WLOG, let the sequence be increasing. For all $ \alpha \in (0, \frac{\pi}{4})$, we have: $ \sin \alpha < \cos \alpha$ $ \sin \alpha < \tan \alpha$ $ \tan \alpha < \cot \alpha$ $ \cos \alpha < \cot \alpha$ So, the sequence in order is either: $ \sin \alpha$, $ \cos \alpha$, $ \tan \alpha$, $ \cot \alpha$ OR $ \sin \alpha$, $ \tan \alpha$, $ \cos \alpha$, $ \cot \alpha$ Let $ A = \sin \alpha$, and $ B = \cos \alpha$. Then, the terms in order are either $ A, B, \frac{A}{B}, \frac{B}{A}$ OR $ A, \frac{A}{B}, B, \frac{B}{A}$. In the 1st sequence $ B-A$ and $ \frac{B}{A} - \frac{A}{B}$ are both equal to the common difference. In the 2nd sequence $ B-A$ and $ \frac{B}{A} - \frac{A}{B}$ are both equal to twice common difference. Therefore: $ B - A = \frac{B}{A} - \frac{A}{B}$ $ AB^2 - A^2B = B^2 - A^2$ $ (B-A)(AB) = (B-A)(B+A)$ $ B = A$ OR $ AB = A+B$ $ B = A$: $ \cos \alpha = \sin \alpha$, so $ \alpha = \frac{\pi}{4}$, which, as previously stated, does not yield an arithmetic progression. $ AB = A+B$: Since $ 0< AB <1$: $ AB < \sqrt{AB}$ Since $ A+B > 0$: $ \frac{A+B}{2} < A+B$ And, by AM-GM: $ \sqrt{AB} \le \frac{A+B}{2}$ So, $ AB < \sqrt{AB} \le \frac{A+B}{2} < A+B$. Thus $ AB \neq A+B$. Therefore, there are no such angles $ \alpha$ which yields an arithmetic progression.

$ AB = A + B$ is to reduce it to $ A(1-B) + B = 0$ and note that two positive numbers can't add to $ 0$.