To find the six trigonometric functions of $\theta$, we need to determine the values of $\sin \theta$, $\cos \theta$, $\tan \theta$, $\csc \theta$, $\sec \theta$, and $\cot \theta$ for an angle whose terminal side passes through the point $(5, -3)$.

1. Calculate $r$:
   $r = \sqrt{x^2 + y^2} = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}$.

2. Trigonometric Ratios:

   • $\sin \theta = \frac{y}{r} = \frac{-3}{\sqrt{34}} = -\frac{3}{\sqrt{34}}$.
   • $\cos \theta = \frac{x}{r} = \frac{5}{\sqrt{34}} = \frac{5}{\sqrt{34}}$.
   • $\tan \theta = \frac{y}{x} = \frac{-3}{5} = -\frac{3}{5}$.

3. Reciprocal Trigonometric Ratios:

   • $\csc \theta = \frac{r}{y} = \frac{\sqrt{34}}{-3} = -\frac{\sqrt{34}}{3}$.
   • $\sec \theta = \frac{r}{x} = \frac{\sqrt{34}}{5}$.
   • $\cot \theta = \frac{x}{y} = \frac{5}{-3} = -\frac{5}{3}$.

Thus, the values of the six trigonometric functions of $\theta$ are:

• $\sin \theta = -\frac{3}{\sqrt{34}}$
• $\cos \theta = \frac{5}{\sqrt{34}}$
• $\tan \theta = -\frac{3}{5}$
• $\csc \theta = -\frac{\sqrt{34}}{3}$
• $\sec \theta = \frac{\sqrt{34}}{5}$
• $\cot \theta = -\frac{5}{3}$

Would you like further details on any of these steps, or do you have any questions?

1. How is $r$ calculated in trigonometric functions?
2. Why are reciprocal trigonometric functions used?
3. What are the signs of trigonometric functions in different quadrants?
4. How does standard position affect the trigonometric function values?
5. Why is it important to simplify the square root in trigonometry?

Tip: Remember, in standard position, the terminal side's location (quadrant) determines the sign of each trigonometric function.