: Given \( \sin \theta=\cos \frac{3 \pi}{11} \) and \( \theta \in\left(\frac{\pi}{2}, \pi\right) \), find the exact value of \( \theta \)

To find the value of \( \theta \) given that \( \sin \theta = \cos \frac{3 \pi}{11} \) and \( \theta \in \left( \frac{\pi}{2}, \pi \right) \), we can use the co-function identity that relates sine and cosine: \[ \sin \theta = \cos \left( \frac{\pi}{2} - \theta \right). \] Setting the two expressions equal, we have: \[ \cos \left( \frac{\pi}{2} - \theta \right) = \cos \frac{3 \pi}{11}. \] The cosine function is periodic and even, which means the general solutions for \( a = b \) are: \[ \frac{\pi}{2} - \theta = \frac{3\pi}{11} + 2k\pi \quad \text{or} \quad \frac{\pi}{2} - \theta = -\frac{3\pi}{11} + 2k\pi, \] for any integer \( k \). Since we restrict \( \theta \) to \( \left( \frac{\pi}{2}, \pi \right) \), we'll focus on the cases when \( k=0 \). Starting with the first equation: \[ \frac{\pi}{2} - \theta = \frac{3\pi}{11}. \] Rearranging gives: \[ \theta = \frac{\pi}{2} - \frac{3\pi}{11}. \] To compute \( \frac{\pi}{2} - \frac{3\pi}{11} \), we first need a common denominator. The least common multiple of 2 and 11 is 22: \[ \frac{\pi}{2} = \frac{11\pi}{22}, \quad \text{and } \frac{3\pi}{11} = \frac{6\pi}{22}. \] Thus, we have: \[ \theta = \frac{11\pi}{22} - \frac{6\pi}{22} = \frac{5\pi}{22}. \] Now, consider the second equation: \[ \frac{\pi}{2} - \theta = -\frac{3\pi}{11}, \] which gives: \[ \theta = \frac{\pi}{2} + \frac{3\pi}{11}. \] Using the same common denominator of 22: \[ \theta = \frac{11\pi}{22} + \frac{6\pi}{22} = \frac{17\pi}{22}. \] Now, we check if \( \frac{5\pi}{22} \) is in the interval \( \left( \frac{\pi}{2}, \pi \right) \): - Calculate \( \frac{\pi}{2} = \frac{11\pi}{22} \). - Since \( \frac{5\pi}{22} < \frac{11\pi}{22} \), this value is not valid. Now, check \( \frac{17\pi}{22} \): - \( \frac{17\pi}{22} > \frac{11\pi}{22} \) and \( \frac{17\pi}{22} < \pi \) confirms that it is indeed in the interval. Thus, the exact value of \( \theta \) is: \[ \theta = \frac{17\pi}{22}. \]