If \( \sin \theta=\frac{2}{5}, \theta \) in quadrant II, find the exact value of \( \begin{array}{llll}\text { (a) } \cos \theta & \text { (b) } \sin \left(\theta+\frac{\pi}{6}\right) & \text { (c) } \cos \left(\theta-\frac{\pi}{3}\right) & \text { (d) } \tan \left(\theta+\frac{\pi}{4}\right)\end{array} \)

Given that \( \sin \theta = \frac{2}{5} \) and \( \theta \) is in quadrant II, we can find the exact values of the other trigonometric functions as follows: (a) To find \( \cos \theta \), we use the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \). \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{2}{5}\right)^2 = 1 - \frac{4}{25} = \frac{21}{25} \] Since \( \theta \) is in quadrant II, \( \cos \theta \) is negative. Therefore, \[ \cos \theta = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5} \] (b) To find \( \sin \left(\theta + \frac{\pi}{6}\right) \), we use the sum formula for sine: \[ \sin \left(\theta + \frac{\pi}{6}\right) = \sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6} \] We know that \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \) and \( \sin \frac{\pi}{6} = \frac{1}{2} \). Substituting the values we have: \[ \sin \left(\theta + \frac{\pi}{6}\right) = \frac{2}{5} \cdot \frac{\sqrt{3}}{2} + \left(-\frac{\sqrt{21}}{5}\right) \cdot \frac{1}{2} = \frac{\sqrt{3}}{5} - \frac{\sqrt{21}}{10} \] (c) To find \( \cos \left(\theta - \frac{\pi}{3}\right) \), we use the difference formula for cosine: \[ \cos \left(\theta - \frac{\pi}{3}\right) = \cos \theta \cos \frac{\pi}{3} + \sin \theta \sin \frac{\pi}{3} \] We know that \( \cos \frac{\pi}{3} = \frac{1}{2} \) and \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \). Substituting the values we have: \[ \cos \left(\theta - \frac{\pi}{3}\right) = \left(-\frac{\sqrt{21}}{5}\right) \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{21}}{10} + \frac{\sqrt{3}}{5} \] (d) To find \( \tan \left(\theta + \frac{\pi}{4}\right) \), we use the sum formula for tangent: \[ \tan \left(\theta + \frac{\pi}{4}\right) = \frac{\tan \theta + \tan \frac{\pi}{4}}{1 - \tan \theta \tan \frac{\pi}{4}} \] Since \( \tan \frac{\pi}{4} = 1 \), we have: \[ \tan \left(\theta + \frac{\pi}{4}\right) = \frac{\tan \theta + 1}{1 - \tan \theta} \] We know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2}{5}}{-\frac{\sqrt{21}}{5}} = -\frac{2}{\sqrt{21}} \). Substituting the values we have: \[ \tan \left(\theta + \frac{\pi}{4}\right) = \frac{-\frac{2}{\sqrt{21}} + 1}{1 + \frac{2}{\sqrt{21}}} = \frac{-2 + \sqrt{21}}{\sqrt{21} + 2} \] To rationalize the denominator, we multiply the numerator and denominator by \( \sqrt{21} - 2 \): \[ \tan \left(\theta + \frac{\pi}{4}\right) = \frac{(-2 + \sqrt{21})(\sqrt{21} - 2)}{(\sqrt{21} + 2)(\sqrt{21} - 2)} = \frac{-2\sqrt{21} + 4 + 21 - 4}{21 - 4} = \frac{17 - 2\