Starting at point \( (1,0) \), you move along a distance \( t \) along a unit circle to terminal point \( P(x, y) \). If \( \cos t=\frac{2}{7} \) and \( \sin t<0 \), find the other 5 trigonometric functions. Express your answer as a reduced fraction, NO DECIMALS! \( \sin t=\square \) \( \tan t=\square \) \( \csc t= \) \( \sec t= \)

To find the other five trigonometric functions given that \( \cos t = \frac{2}{7} \) and \( \sin t < 0 \), we can follow these steps: 1. **Identify the quadrant**: Since \( \cos t > 0 \) and \( \sin t < 0 \), we are in the fourth quadrant. 2. **Use the Pythagorean identity**: The Pythagorean identity states that: \[ \sin^2 t + \cos^2 t = 1 \] We can substitute \( \cos t \) into this equation to find \( \sin t \). 3. **Calculate \( \sin t \)**: \[ \sin^2 t + \left(\frac{2}{7}\right)^2 = 1 \] \[ \sin^2 t + \frac{4}{49} = 1 \] \[ \sin^2 t = 1 - \frac{4}{49} \] \[ \sin^2 t = \frac{49}{49} - \frac{4}{49} = \frac{45}{49} \] \[ \sin t = -\sqrt{\frac{45}{49}} = -\frac{\sqrt{45}}{7} = -\frac{3\sqrt{5}}{7} \] 4. **Calculate \( \tan t \)**: \[ \tan t = \frac{\sin t}{\cos t} = \frac{-\frac{3\sqrt{5}}{7}}{\frac{2}{7}} = -\frac{3\sqrt{5}}{2} \] 5. **Calculate \( \csc t \)**: \[ \csc t = \frac{1}{\sin t} = \frac{1}{-\frac{3\sqrt{5}}{7}} = -\frac{7}{3\sqrt{5}} = -\frac{7\sqrt{5}}{15} \quad \text{(rationalizing the denominator)} \] 6. **Calculate \( \sec t \)**: \[ \sec t = \frac{1}{\cos t} = \frac{1}{\frac{2}{7}} = \frac{7}{2} \] Now we can summarize the results: - \( \sin t = -\frac{3\sqrt{5}}{7} \) - \( \tan t = -\frac{3\sqrt{5}}{2} \) - \( \csc t = -\frac{7\sqrt{5}}{15} \) - \( \sec t = \frac{7}{2} \) Thus, the final answers are: \[ \sin t = -\frac{3\sqrt{5}}{7} \] \[ \tan t = -\frac{3\sqrt{5}}{2} \] \[ \csc t = -\frac{7\sqrt{5}}{15} \] \[ \sec t = \frac{7}{2} \]