https://file.aiquickdraw.com/mathbot/file/17296157473040q2ia6el.jpg

Let's solve the polar to Cartesian coordinate conversions based on the given points in the image. The formula to convert from polar coordinates \((r, \theta)\) to Cartesian coordinates \((x, y)\) is:

\[
x = r \cos(\theta), \quad y = r \sin(\theta)
\]

### (a) \( \left( 6, \frac{4\pi}{3} \right) \)
Here, \(r = 6\) and \( \theta = \frac{4\pi}{3} \).

1. \( x = 6 \cos\left(\frac{4\pi}{3}\right) \)
   - \( \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \)
   - \( x = 6 \times \left(-\frac{1}{2}\right) = -3 \)

2. \( y = 6 \sin\left(\frac{4\pi}{3}\right) \)
   - \( \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} \)
   - \( y = 6 \times \left(-\frac{\sqrt{3}}{2}\right) = -3\sqrt{3} \)

So, \( (x, y) = (-3, -3\sqrt{3}) \).

---

### (b) \( \left( -4, \frac{3\pi}{4} \right) \)
Here, \(r = -4\) and \( \theta = \frac{3\pi}{4} \).

1. \( x = -4 \cos\left(\frac{3\pi}{4}\right) \)
   - \( \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \)
   - \( x = -4 \times \left(-\frac{\sqrt{2}}{2}\right) = 2\sqrt{2} \)

2. \( y = -4 \sin\left(\frac{3\pi}{4}\right) \)
   - \( \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
   - \( y = -4 \times \frac{\sqrt{2}}{2} = -2\sqrt{2} \)

So, \( (x, y) = (2\sqrt{2}, -2\sqrt{2}) \).

---

### (c) \( \left( -5, -\frac{\pi}{3} \right) \)
Here, \(r = -5\) and \( \theta = -\frac{\pi}{3} \).

1. \( x = -5 \cos\left(-\frac{\pi}{3}\right) \)
   - \( \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} \)
   - \( x = -5 \times \frac{1}{2} = -\frac{5}{2} \)

2. \( y = -5 \sin\left(-\frac{\pi}{3}\right) \)
   - \( \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \)
   - \( y = -5 \times \left(-\frac{\sqrt{3}}{2}\right) = \frac{5\sqrt{3}}{2} \)

So, \( (x, y) = \left(-\frac{5}{2}, \frac{5\sqrt{3}}{2}\right) \).

---

### Final answers:

(a) \( (x, y) = (-3, -3\sqrt{3}) \)

(b) \( (x, y) = (2\sqrt{2}, -2\sqrt{2}) \)

(c) \( (x, y) = \left(-\frac{5}{2}, \frac{5\sqrt{3}}{2}\right) \)

Would you like further details or have any questions?

Here are some related questions to consider:
1. How can you convert between Cartesian and polar coordinates?
2. What is the significance of a negative radius in polar coordinates?
3. How does the unit circle help in determining sine and cosine values?
4. What are the typical applications of polar coordinates in real life?
5. How does changing the angle \(\theta\) affect the point's position in polar coordinates?

**Tip:** To easily remember trigonometric values, it helps to visualize the unit circle, where cosine corresponds to the x-coordinate and sine to the y-coordinate of a point on the circle.

Plot the point whose polar coordinates are given and find the Cartesian coordinates of the point for the following: (a) (6, 4π/3), (b) (-4, 3π/4), (c) (-5, -π/3).

This problem demonstrates how to convert the given polar coordinates (6, 4π/3), (-4, 3π/4), and (-5, -π/3) into Cartesian coordinates using trigonometric identities and the unit circle. Suitable for students in grades 9-12.

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