Math worksheet featuring problems on polar and rectangular coordinates, with instructions to plot points and convert between systems.

Let's solve each part of this problem step by step. Since I can't directly plot on polar coordinate paper, I will explain how to plot the given points and convert between polar and rectangular coordinates, as required.

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Part 1: Plotting Polar Coordinates (Exercises 1–12)

Polar coordinates are written as $(r, \theta)$, where:
- $r$ is the distance from the origin (can be positive or negative),
- $\theta$ is the angle in radians measured counterclockwise from the positive $x$-axis.

If $r < 0$, the point is plotted in the opposite direction of $\theta$.

We’ll go through each:

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#### 1. $\left(3, \frac{\pi}{6}\right)$
- $r = 3 > 0$, $\theta = \frac{\pi}{6} = 30^\circ$
- Move 3 units at $30^\circ$ from the positive $x$-axis.
- Plot in first quadrant.

#### 2. $(2, \pi)$
- $r = 2$, $\theta = \pi = 180^\circ$
- This is on the negative $x$-axis, 2 units left of origin.

#### 3. $\left(\frac{5}{2}, -\frac{2\pi}{5}\right)$
- $r = 2.5$, $\theta = -\frac{2\pi}{5} = -72^\circ$
- Negative angle means clockwise.
- So, $72^\circ$ below the positive $x$-axis → fourth quadrant.
- Plot 2.5 units at $-72^\circ$.

#### 4. $\left(5, -\frac{\pi}{3}\right)$
- $r = 5$, $\theta = -\frac{\pi}{3} = -60^\circ$
- Clockwise $60^\circ$ → fourth quadrant.
- Plot 5 units at $-60^\circ$.

#### 5. $\left(-2, \frac{7\pi}{6}\right)$
- $r = -2$, so go in the opposite direction of $\theta = \frac{7\pi}{6} = 210^\circ$
- Opposite direction: $210^\circ + 180^\circ = 390^\circ \equiv 30^\circ$
- So equivalent to $(2, 30^\circ)$ → first quadrant.
- Alternatively: $-2$ at $210^\circ$ means go 2 units in the direction opposite to $210^\circ$, which is $30^\circ$.
- So plot at $30^\circ$, but with radius 2.

#### 6. $\left(-5, \frac{\pi}{4}\right)$
- $r = -5$, $\theta = \frac{\pi}{4} = 45^\circ$
- Opposite direction: $45^\circ + 180^\circ = 225^\circ$
- So plot 5 units at $225^\circ$ → third quadrant.

#### 7. $\left(-3, -\frac{5\pi}{4}\right)$
- $r = -3$, $\theta = -\frac{5\pi}{4} = -225^\circ$
- Add $360^\circ$: $-225^\circ + 360^\circ = 135^\circ$
- But since $r < 0$, go in opposite direction: $135^\circ + 180^\circ = 315^\circ$
- So plot 3 units at $315^\circ$ → fourth quadrant.

Alternatively: $-3$ at $-225^\circ$ → same as $+3$ at $-225^\circ + 180^\circ = -45^\circ$ → $315^\circ$

#### 8. $\left(-4, -\frac{5\pi}{3}\right)$
- $r = -4$, $\theta = -\frac{5\pi}{3} = -300^\circ$
- Add $360^\circ$: $-300^\circ + 360^\circ = 60^\circ$
- $r < 0$, so go opposite: $60^\circ + 180^\circ = 240^\circ$
- Plot 4 units at $240^\circ$ → third quadrant.

Or: $-4$ at $-300^\circ$ → same as $+4$ at $-300^\circ + 180^\circ = -120^\circ$, which is $240^\circ$

#### 9. $\left(0.5, -\frac{8\pi}{3}\right)$
- $r = 0.5$, $\theta = -\frac{8\pi}{3} = -480^\circ$
- Reduce angle: $-480^\circ + 360^\circ = -120^\circ$, then $-120^\circ + 360^\circ = 240^\circ$
- So equivalent to $(0.5, 240^\circ)$
- Plot 0.5 units at $240^\circ$ → third quadrant.

#### 10. $(2.2, -6\pi)$
- $r = 2.2$, $\theta = -6\pi$ radians
- $-6\pi$ is a full rotation: $-6\pi + 6\pi = 0$, so equivalent to $0$ radians
- So it’s just $(2.2, 0)$ → on the positive $x$-axis

#### 11. $(2, 2)$
- This is not in polar form — wait! It says "plot the given polar coordinate points", but $(2,2)$ looks like rectangular.
- However, looking back: the instruction says "polar coordinate points" — so likely typo? Or perhaps meant to be interpreted as polar?
- But $(2,2)$ has no unit — if it's polar, $r=2$, $\theta=2$ radians.
- $2$ radians ≈ $114.6^\circ$ → second quadrant.
- So plot 2 units at $2$ radians.

But note: if this was intended as rectangular, it would not be polar. But since the exercise says "polar coordinate points", we assume $(2,2)$ means $r=2$, $\theta=2$ radians.

So: plot $r=2$, $\theta=2$ rad ≈ $114.6^\circ$

#### 12. $(-1, -1)$
- Again, likely polar: $r = -1$, $\theta = -1$ rad ≈ $-57.3^\circ$
- Since $r < 0$, go opposite direction: $-1 + \pi$ radians = $-1 + 3.1416 ≈ 2.1416$ rad ≈ $122.7^\circ$
- So plot 1 unit at $122.7^\circ$ → second quadrant.

Alternatively: $(-1, -1)$ → same as $(1, -1 + \pi) = (1, 2.1416)$

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✔ Summary for Part 1:

You would plot these on polar graph paper using the above reasoning. For negative $r$, reflect across origin; for negative angles, go clockwise.

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Part 2: Find Polar Coordinates from Rectangular (Exercises 13–16)

Given $(x, y)$, find $(r, \theta)$, where:
- $r = \sqrt{x^2 + y^2}$
- $\theta = \tan^{-1}(y/x)$, adjusted for quadrant

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#### 13. $(\sqrt{3}, 1)$
- $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$
- $\theta = \tan^{-1}(1/\sqrt{3}) = \tan^{-1}(\sqrt{3}/3) = \frac{\pi}{6}$ (since both $x,y > 0$, QI)
- Answer: $\boxed{(2, \frac{\pi}{6})}$

#### 14. $(-1, -1)$
- $r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$
- $\theta = \tan^{-1}((-1)/(-1)) = \tan^{-1}(1) = \frac{\pi}{4}$, but in third quadrant
- So $\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$
- Answer: $\boxed{(\sqrt{2}, \frac{5\pi}{4})}$

#### 15. $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$
- $r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1$
- $\theta = \tan^{-1}\left(\frac{-1/2}{-\sqrt{3}/2}\right) = \tan^{-1}(1/\sqrt{3}) = \frac{\pi}{6}$, but in third quadrant
- So $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$
- Answer: $\boxed{(1, \frac{7\pi}{6})}$

#### 16. $(-5, 4)$
- $r = \sqrt{(-5)^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}$
- $\theta = \tan^{-1}(4 / (-5)) = \tan^{-1}(-0.8)$
- Since $x < 0$, $y > 0$, it's in second quadrant
- $\tan^{-1}(-0.8) \approx -38.66^\circ$, so add $180^\circ$: $141.34^\circ$
- In radians: $141.34^\circ \times \frac{\pi}{180} \approx 2.466$ rad
- Use exact: $\theta = \pi - \tan^{-1}(4/5)$
- So $\theta = \pi - \tan^{-1}(0.8)$
- Answer: $\boxed{(\sqrt{41}, \pi - \tan^{-1}(0.8))}$ or numerically $\boxed{(\sqrt{41}, 2.466)}$

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Part 3: Find Rectangular Coordinates from Polar (Exercises 17–20)

Use:
- $x = r \cos \theta$
- $y = r \sin \theta$

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#### 17. $\left(8, \frac{4\pi}{3}\right)$
- $\theta = \frac{4\pi}{3} = 240^\circ$ → third quadrant
- $\cos(240^\circ) = \cos(180^\circ + 60^\circ) = -\cos(60^\circ) = -0.5$
- $\sin(240^\circ) = -\sin(60^\circ) = -\frac{\sqrt{3}}{2}$
- $x = 8 \cdot (-0.5) = -4$
- $y = 8 \cdot (-\frac{\sqrt{3}}{2}) = -4\sqrt{3}$
- Answer: $\boxed{(-4, -4\sqrt{3})}$

#### 18. $(-4, -\pi)$
- $r = -4$, $\theta = -\pi$
- $x = r \cos \theta = (-4)\cos(-\pi) = (-4)(-1) = 4$
- $y = r \sin \theta = (-4)\sin(-\pi) = (-4)(0) = 0$
- So point is $(4, 0)$
- But let’s verify: $(-4, -\pi)$ → same as $(4, -\pi + \pi) = (4, 0)$
- Yes: $\boxed{(4, 0)}$

#### 19. $(3.0, -0.40)$
- $r = 3.0$, $\theta = -0.40$ rad
- $x = 3.0 \cos(-0.40) = 3.0 \cos(0.40)$
- $\cos(0.40) \approx 0.9211$ → $x \approx 3.0 \times 0.9211 = 2.763$
- $y = 3.0 \sin(-0.40) = 3.0 \cdot (-\sin(0.40)) \approx 3.0 \cdot (-0.3894) = -1.168$
- Answer: $\boxed{(2.76, -1.17)}$ (rounded)

#### 20. $(-1.0, 1.0)$
- $r = -1.0$, $\theta = 1.0$ rad
- $x = r \cos \theta = (-1.0) \cos(1.0) \approx (-1.0)(0.5403) = -0.5403$
- $y = r \sin \theta = (-1.0) \sin(1.0) \approx (-1.0)(0.8415) = -0.8415$
- So $\boxed{(-0.54, -0.84)}$

Alternatively: $(-1.0, 1.0)$ is same as $(1.0, 1.0 + \pi)$ → $r=1$, $\theta \approx 4.1416$ rad
Then $x = \cos(4.1416) \approx \cos(237^\circ) \approx -0.54$, $y = \sin(4.1416) \approx -0.84$

Same result.

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✔ Final Answers:

#### Exercises 1–12: Plot on polar paper using the explanations above.

#### Exercises 13–16: Polar Coordinates

13. $(\sqrt{3}, 1)$ → $\boxed{(2, \frac{\pi}{6})}$
14. $(-1, -1)$ → $\boxed{(\sqrt{2}, \frac{5\pi}{4})}$
15. $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$ → $\boxed{(1, \frac{7\pi}{6})}$
16. $(-5, 4)$ → $\boxed{(\sqrt{41}, \pi - \tan^{-1}(0.8))}$ or $\boxed{(\sqrt{41}, 2.466)}$

#### Exercises 17–20: Rectangular Coordinates

17. $\left(8, \frac{4\pi}{3}\right)$ → $\boxed{(-4, -4\sqrt{3})}$
18. $(-4, -\pi)$ → $\boxed{(4, 0)}$
19. $(3.0, -0.40)$ → $\boxed{(2.76, -1.17)}$
20. $(-1.0, 1.0)$ → $\boxed{(-0.54, -0.84)}$

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Let me know if you'd like visual sketches or more detailed explanations for any specific point!