This question: 2 point(s) possible
Sketch an angle $\theta$ in standard position such that $\theta$ has the least possible positive measure, and the point $(3, \sqrt{3})$ is on the terminal side of $\theta$. Then find the values of the six trigonometric functions for the angle. Rationalize denominators if applicable. Do not use a calculator

Step 1 ：Given the point (3, √3) lies in the first quadrant of the Cartesian plane. The angle θ in standard position with the least positive measure that has this point on its terminal side is the angle formed by the positive x-axis and the line segment connecting the origin and the point (3, √3).

Step 2 ：We can calculate the six trigonometric functions for this angle using the definitions of these functions in terms of the coordinates of the point on the terminal side of the angle.

Step 3 ：First, we calculate the distance from the origin to the point (x, y) using the Pythagorean theorem: r = √(x² + y²).

Step 4 ：Substituting x = 3 and y = √3 into the formula, we get r = √(3² + (√3)²) = √(9 + 3) = √12 = 2√3.

Step 5 ：Then, we can calculate the six trigonometric functions as follows:

Step 6 ：1. \(\sin(\theta) = \frac{y}{r} = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}\)

Step 7 ：2. \(\cos(\theta) = \frac{x}{r} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}\)

Step 8 ：3. \(\tan(\theta) = \frac{y}{x} = \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{3}\)

Step 9 ：4. \(\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{1}{2}} = 2\)

Step 10 ：5. \(\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}\)

Step 11 ：6. \(\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{\frac{\sqrt{3}}{3}} = \sqrt{3}\)

Step 12 ：\(\boxed{\text{Final Answer: The six trigonometric functions for the angle } \theta \text{ are as follows:}}\)

Step 13 ：1. \(\sin(\theta) = \frac{1}{2}\)

Step 14 ：2. \(\cos(\theta) = \frac{\sqrt{3}}{2}\)

Step 15 ：3. \(\tan(\theta) = \frac{\sqrt{3}}{3}\)

Step 16 ：4. \(\csc(\theta) = 2\)

Step 17 ：5. \(\sec(\theta) = \frac{2}{\sqrt{3}}\)

Step 18 ：6. \(\cot(\theta) = \sqrt{3}\)