Questions: A point on the terminal side of an angle θ in standard position is (sqrt(11)/6, 5/6). Find the exact value of each of the six trigonometric functions of θ. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. sin θ= (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is not defined.

Solution Steps

To find the six trigonometric functions of the angle \(\theta\) given a point \(\left(\frac{\sqrt{11}}{6}, \frac{5}{6}\right)\) on its terminal side, we can follow these steps:

1. Identify the coordinates \(x = \frac{\sqrt{11}}{6}\) and \(y = \frac{5}{6}\).
2. Calculate the radius \(r\) using the Pythagorean theorem: \(r = \sqrt{x^2 + y^2}\).
3. Use the definitions of the trigonometric functions in terms of \(x\), \(y\), and \(r\):
   • \(\sin \theta = \frac{y}{r}\)
   • \(\cos \theta = \frac{x}{r}\)
   • \(\tan \theta = \frac{y}{x}\)
   • \(\csc \theta = \frac{r}{y}\)
   • \(\sec \theta = \frac{r}{x}\)
   • \(\cot \theta = \frac{x}{y}\)

Given the coordinates \( x = \frac{\sqrt{11}}{6} \) and \( y = \frac{5}{6} \), we first calculate the radius \( r \) using the Pythagorean theorem: \[ r = \sqrt{x^2 + y^2} = \sqrt{\left(\frac{\sqrt{11}}{6}\right)^2 + \left(\frac{5}{6}\right)^2} = 1 \]

Using the definitions of the trigonometric functions in terms of \( x \), \( y \), and \( r \):

\[ \sin \theta = \frac{y}{r} = \frac{\frac{5}{6}}{1} = \frac{5}{6} \approx 0.8333 \]

\[ \cos \theta = \frac{x}{r} = \frac{\frac{\sqrt{11}}{6}}{1} = \frac{\sqrt{11}}{6} \approx 0.5528 \]

\[ \tan \theta = \frac{y}{x} = \frac{\frac{5}{6}}{\frac{\sqrt{11}}{6}} = \frac{5}{\sqrt{11}} \approx 1.5076 \]

\[ \csc \theta = \frac{r}{y} = \frac{1}{\frac{5}{6}} = \frac{6}{5} = 1.2 \]

\[ \sec \theta = \frac{r}{x} = \frac{1}{\frac{\sqrt{11}}{6}} = \frac{6}{\sqrt{11}} \approx 1.8091 \]

\[ \cot \theta = \frac{x}{y} = \frac{\frac{\sqrt{11}}{6}}{\frac{5}{6}} = \frac{\sqrt{11}}{5} \approx 0.6633 \]

Final Answer