https://file.aiquickdraw.com/mathbot/file/1726992486025pcdhxsi7.jpg

Let's examine the two polar equations and analyze how to sketch them.

### (ii) \( r = 6 \sin \theta \)
This is the polar equation of a **limacon** with a sine term. Here's how it works:

- When \( \theta = 0 \), \( r = 6 \sin(0) = 0 \), so the curve passes through the origin.
- When \( \theta = \frac{\pi}{2} \), \( r = 6 \sin\left(\frac{\pi}{2}\right) = 6 \), so the point \( (r, \theta) = (6, \frac{\pi}{2}) \) is plotted on the positive y-axis.
- When \( \theta = \pi \), \( r = 6 \sin(\pi) = 0 \), so it returns to the origin.
- When \( \theta = \frac{3\pi}{2} \), \( r = 6 \sin\left(\frac{3\pi}{2}\right) = -6 \), which indicates a point in the negative y-direction.
  
This gives a **limacon shape without an inner loop**, symmetric with respect to the vertical axis.

### (iii) \( r = -1 - \cos \theta \)
This is also a polar equation, and it generates a **limacon** with a negative \( r \) and cosine term. Here's the interpretation:

- When \( \theta = 0 \), \( r = -1 - \cos(0) = -2 \), so the point is 2 units to the left of the origin.
- When \( \theta = \frac{\pi}{2} \), \( r = -1 - \cos\left(\frac{\pi}{2}\right) = -1 \), the point is 1 unit below the origin.
- When \( \theta = \pi \), \( r = -1 - \cos(\pi) = 0 \), meaning the curve passes through the origin.
- When \( \theta = \frac{3\pi}{2} \), \( r = -1 - \cos\left(\frac{3\pi}{2}\right) = -1 \), a point below the origin.
  
This results in a **limacon with an inner loop**, symmetric with respect to the horizontal axis.

Would you like me to generate an exact sketch of these curves for you? 

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Here are 5 related questions:
1. How does the graph of \( r = a \sin \theta \) differ from \( r = a \cos \theta \)?
2. What happens to the curve when you adjust the coefficient in \( r = k \sin \theta \)?
3. How do we determine the symmetry of a polar curve based on the equation?
4. What effect does a negative coefficient have in polar equations like \( r = -a - b \cos \theta \)?
5. Can you explain how to convert a polar equation to Cartesian coordinates?

**Tip**: When plotting polar curves, always start by determining key points for special angles like \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \) to understand the general shape of the graph.

Sketch the curves in polar coordinates: (ii) r = 6sinθ (iii) r = -1 - cosθ.

Learn how to sketch polar curves, including r = 6sinθ and r = -1 - cosθ. This problem explores limacon shapes, symmetry, and key trigonometric concepts in polar coordinates, suitable for high school students (Grades 10-12).

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