Given that point $P(6, -5)$ is on the terminal side of angle $\theta$, we can calculate the six trigonometric functions for $\theta$ as follows:

Step 1: Find the radius $r$.

The radius $r$ is the distance from the origin to the point $P(x, y)$, and it is given by the formula: $r = \sqrt{x^2 + y^2}$ Substitute $x = 6$ and $y = -5$: $r = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61}$

Step 2: Use the coordinates to find the trigonometric functions.

Now, we can calculate the six trigonometric functions based on the values of $x$, $y$, and $r$.

1. Sine: $\sin(\theta) = \frac{y}{r} = \frac{-5}{\sqrt{61}}$
2. Cosine: $\cos(\theta) = \frac{x}{r} = \frac{6}{\sqrt{61}}$
3. Tangent: $\tan(\theta) = \frac{y}{x} = \frac{-5}{6}$
4. Cosecant (the reciprocal of sine): $\csc(\theta) = \frac{r}{y} = \frac{\sqrt{61}}{-5}$
5. Secant (the reciprocal of cosine): $\sec(\theta) = \frac{r}{x} = \frac{\sqrt{61}}{6}$
6. Cotangent (the reciprocal of tangent): $\cot(\theta) = \frac{x}{y} = \frac{6}{-5}$

Final Answers:

• $\sin(\theta) = \frac{-5}{\sqrt{61}}$
• $\cos(\theta) = \frac{6}{\sqrt{61}}$
• $\tan(\theta) = \frac{-5}{6}$
• $\csc(\theta) = \frac{\sqrt{61}}{-5}$
• $\sec(\theta) = \frac{\sqrt{61}}{6}$
• $\cot(\theta) = \frac{6}{-5}$

Would you like more detailed explanations of any of the steps, or do you have another question?

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Additional Questions:

1. How would the trigonometric functions change if the point was in a different quadrant?
2. Can you verify the values of $\sin(\theta)$ and $\cos(\theta)$ using a calculator?
3. How does the value of $r$ relate to the concept of the unit circle?
4. What are the signs of the trigonometric functions in each quadrant?
5. How do you convert between rectangular coordinates and polar coordinates?

Tip:

When solving trigonometric problems with points, always start by calculating the radius $r$, as it’s fundamental for all the functions.