The question involves finding the six trigonometric functions for an angle $\theta$, where the point on the terminal side of the angle is $(-4, -6)$.

Step 1: Find the radius $r$

The radius $r$ is the distance from the origin to the point $(-4, -6)$, calculated using the Pythagorean theorem:

$r = \sqrt{(-4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$

Step 2: Calculate the six trigonometric functions

The trigonometric functions are defined as follows:

• Sine: $\sin \theta = \frac{y}{r}$ $\sin \theta = \frac{-6}{2\sqrt{13}} = \frac{-6}{2\sqrt{13}} = \frac{-3}{\sqrt{13}} = \frac{-3\sqrt{13}}{13}$

• Cosine: $\cos \theta = \frac{x}{r}$ $\cos \theta = \frac{-4}{2\sqrt{13}} = \frac{-2}{\sqrt{13}} = \frac{-2\sqrt{13}}{13}$

• Tangent: $\tan \theta = \frac{y}{x}$ $\tan \theta = \frac{-6}{-4} = \frac{3}{2}$

• Cosecant: $\csc \theta = \frac{r}{y}$ $\csc \theta = \frac{2\sqrt{13}}{-6} = \frac{\sqrt{13}}{-3}$

• Secant: $\sec \theta = \frac{r}{x}$ $\sec \theta = \frac{2\sqrt{13}}{-4} = \frac{\sqrt{13}}{-2}$

• Cotangent: $\cot \theta = \frac{x}{y}$ $\cot \theta = \frac{-4}{-6} = \frac{2}{3}$

Final Answer:

• $\sin \theta = \frac{-3\sqrt{13}}{13}$
• $\cos \theta = \frac{-2\sqrt{13}}{13}$
• $\tan \theta = \frac{3}{2}$
• $\csc \theta = \frac{\sqrt{13}}{-3}$
• $\sec \theta = \frac{\sqrt{13}}{-2}$
• $\cot \theta = \frac{2}{3}$

In the options shown, it looks like the question asks for the value of $\sin \theta$. Based on this, the answer to that part is $\frac{-3\sqrt{13}}{13}$. However, your choice was marked as "The function is not defined," which does not apply here as $\sin \theta$ is clearly defined.

Would you like further details or clarification?

Here are 5 related questions you might explore:

1. How do the signs of trigonometric functions change based on the quadrant of the point?
2. What are the exact values of the trigonometric functions for special angles like $45^\circ$ or $60^\circ$?
3. How does the unit circle relate to the trigonometric functions?
4. What is the process to convert between radians and degrees for angle measures?
5. How can you apply the Pythagorean identity to verify trigonometric functions?

Tip:

Remember, the signs of the trigonometric functions depend on the quadrant in which the terminal side of the angle lies. For $(-4, -6)$, it lies in the third quadrant, where both sine and cosine are negative, but tangent is positive.