- The radius gives the distance from the origin and the angle locates the direction. - Think about the horizontal and vertical components of a length \(r\) at angle \(\theta\).

1. Use \(x=r\cos\theta\) and \(y=r\sin\theta\). 2. \(x=6\cos\frac{\pi}{3}=3\) and \(y=6\sin\frac{\pi}{3}=3\sqrt{3}\).

\((3, 3\sqrt{3})\)

- Because the radius is positive, use the stated angle without reversing the direction. - Locate the quadrant containing an angle of \(\frac{5\pi}{6}\).

1. The radius is positive, so the point lies in the direction given directly by \(\theta=\frac{5\pi}{6}\). 2. The angle \(\frac{5\pi}{6}\) is \(150^\circ\), which lies in Quadrant II. 3. On the diagram, the Quadrant II point is labeled \(B\).

Point \(B\)

- The radius is the distance from the point to the origin. - Identify the standard angle for the negative y-axis in the required interval.

1. The point is \(6\) units from the origin, so \(r=6\). 2. It lies on the negative y-axis, whose angle in \([0,2\pi)\) is \(\theta=\frac{3\pi}{2}\).

\(\left(6, \frac{3\pi}{2}\right)\)

- Begin at the origin and face the direction indicated by the angle. - The radius tells you how far to move in that direction.

1. An angle of \(0^\circ\) points right along the positive x-axis. 2. A radius of \(4\) gives \(x=4\cos0^\circ=4\) and \(y=4\sin0^\circ=0\).

\((4,0)\); positive x-axis.

- Recall which way a negative angle turns. - The sign of the vertical coordinate tells you whether the point is above or below the x-axis.

1. A negative angle turns clockwise, so \(-90^\circ\) points straight down. 2. Then \(x=2\cos(-90^\circ)=0\) and \(y=2\sin(-90^\circ)=-2\).

\((0,-2)\); below the x-axis.

- The angle tells you which ray to start with. - Decide what the minus sign on the radius changes about the direction.

1. The angle \(0^\circ\) points right, but the negative radius reverses that direction. 2. The point lies three units to the left of the origin, at \((-3,0)\).

\((-3,0)\); a negative radius places the point opposite the ray at \(0^\circ\).

- How many degrees make a complete turn? - Think about your direction after turning all the way around.

1. Add \(360^\circ\) to the angle to get \(400^\circ\). 2. A full turn returns to the same ray, so the point stays in the same place.

\((3,400^\circ)\); the point does not move.

- Find the distance from the point to the origin first. - Use the signs of the rectangular coordinates to determine the quadrant before choosing the angle. - Check that the final angle lies in the required interval.

1. The radius is \(r=\sqrt{(-4)^2+(4\sqrt{3})^2}=8\). 2. The point is in Quadrant II. Its reference angle is \(\frac{\pi}{3}\), so \(\theta=\frac{2\pi}{3}\).

\(\left(8, \frac{2\pi}{3}\right)\)

- A negative radius places the point in the direction opposite the stated angle. - Ask which angle points exactly opposite \(\frac{\pi}{3}\). - Keep the final angle in the required interval.

1. Changing a negative radius to a positive radius reverses the direction of the ray, which changes the angle by \(\pi\). 2. \(\frac{\pi}{3}+\pi=\frac{4\pi}{3}\), so an equivalent pair is \(\left(4, \frac{4\pi}{3}\right)\).

\(\left(4, \frac{4\pi}{3}\right)\)

- With the same radius sign, angles differing by a whole revolution describe the same direction. - With the opposite radius sign, the angle must point in the opposite direction. - Check each option as a complete pair rather than comparing only the angles.

1. Adding or subtracting \(2\pi\) from an angle keeps the same point when the radius is unchanged, so A is equivalent to the given pair. 2. Changing the sign of the radius requires changing the direction by \(\pi\). For B, \(-\frac{\pi}{6}+\pi=\frac{5\pi}{6}\), so B is also equivalent. 3. C reverses the radius without changing the angle by \(\pi\), and D uses a different direction with the same positive radius, so neither represents the given point.

A and B

- First locate the direction of the stated angle without using the radius sign. - A negative radius places the point on the ray opposite that direction. - Use the quadrant of the reversed direction to choose among the labeled points.

1. The angle \(\frac{\pi}{6}\) points into Quadrant I, but the radius is negative, so the point lies in the opposite direction. 2. The opposite direction is \(\frac{\pi}{6}+\pi=\frac{7\pi}{6}\), which is in Quadrant III. 3. On the diagram, the Quadrant III point is \(A\).

Point \(A\). A negative radius reverses the direction by \(\pi\), so the point lies along \(\theta=\frac{7\pi}{6}\).

- Use the horizontal and vertical components of the radius. - Determine the signs from the quadrant before evaluating the reference-angle values.

1. Use \(x=r\cos\theta\) and \(y=r\sin\theta\). 2. \(x=5\cos\frac{4\pi}{3}=-\frac{5}{2}\). 3. \(y=5\sin\frac{4\pi}{3}=-\frac{5\sqrt{3}}{2}\).

\(\left(-\frac{5}{2}, -\frac{5\sqrt{3}}{2}\right)\)

- Find the distance from the origin before choosing the angle. - Use the signs of both coordinates to identify the quadrant. - Check that the angle representative satisfies the required interval.

1. \(r=\sqrt{(-5)^2+(-5)^2}=5\sqrt{2}\). 2. The point lies in Quadrant III and has reference angle \(\frac{\pi}{4}\). 3. Therefore, \(\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\).

\(\left(5\sqrt{2}, \frac{5\pi}{4}\right)\)

- Check whether the standard rectangular-polar identity gives the radius itself or its square. - After using the identity, remember what sign a distance from the origin must have.

1. The correct identity is \(r^2=x^2+y^2\), not \(r=x^2+y^2\). 2. For \((-8, 6)\), \(r^2=(-8)^2+6^2=100\). 3. A polar radius used as a distance is nonnegative, so \(r=10\).

The error is omitting the square on \(r\). Since \(r^2=x^2+y^2=100\), \(r=10\).

- Work out which ray each angle points along. - For \(V\), which side of the origin does its signed distance place it on? - Decide which resulting Cartesian locations match plotted labels.

1. Pair \(U\) places a point at \((3,0)\). 2. Pair \(V\) has negative radius along the left-pointing ray, so it also gives \((3,0)\). Both match plotted point \(A\). 3. Pair \(W\) gives \((0,3)\), which is point \(C\).

\(U\) and \(V\) are the same point; \(U\) is point \(A\), and \(W\) is point \(C\).

- Read the horizontal and vertical coordinates before finding the radius. - Use the right triangle from the origin to determine the length and angle. - Check the point's quadrant when choosing the angle.

1. Read the plotted point as \((4,3)\). 2. The radius is \(r=\sqrt{4^2+3^2}=5\). 3. The point is in Quadrant I, so \(\theta=\arctan(\frac{3}{4})\).

\((r,\theta)=(5,\arctan(\frac{3}{4}))\)

- Mark both distances along a ray from the origin. Which portion of that ray separates the points? - Imagine keeping both distances but turning one point to a different ray. What would change about the segment joining them?

1. Both points lie on the same ray from the origin because their angles agree. 2. Their distances from the origin are \(2\) and \(6\), so the distance between them is \(6-2=4\). 3. If the angles differed, the points would generally lie on different rays. The straight segment between them would not be along one ray, so simply subtracting radii would not give the distance.

\(4\). Subtraction works because the points lie on the same ray; for different rays, the difference of radii is generally not their straight-line distance.

- Separate the effects of a negative angle and a negative radius. - Find an equivalent angle for a positive radius before using cosine and sine.

1. Negating the radius moves the point through a half-turn, so \((-4,-30^\circ)\) is equivalent to \((4,150^\circ)\). 2. The rectangular coordinates are \(x=4\cos150^\circ=-2\sqrt{3}\) and \(y=4\sin150^\circ=2\).

\((-2\sqrt{3},2)\); equivalent polar pair \((4,150^\circ)\)

- Read the rectangular coordinates of \(P\) from the grid before converting. - The positive radius is the distance from the origin to the point. - A negative-radius representation uses the direction opposite the positive-radius representation.

1. The graph places \(P\) at \((-2, 2)\). Its distance from the origin is \(r=\sqrt{(-2)^2+2^2}=2\sqrt{2}\). 2. With positive radius, the point is in Quadrant II on the line \(y=-x\), so \(\theta=\frac{3\pi}{4}\). 3. For a negative radius, use the opposite direction: \(\frac{3\pi}{4}+\pi=\frac{7\pi}{4}\) after reducing modulo \(2\pi\). Thus \(\left(-2\sqrt{2}, \frac{7\pi}{4}\right)\) is equivalent.

a) \(\left(2\sqrt{2}, \frac{3\pi}{4}\right)\) b) \(\left(-2\sqrt{2}, \frac{7\pi}{4}\right)\)

- The two radii and the angle between them form a triangle with the desired distance as its third side. - Find the angular separation before choosing a distance method. - Keep the calculation exact rather than converting the angles to decimals.

1. Both points are \(4\) units from the origin, and the angle between their position rays is \(\frac{5\pi}{6}-\frac{\pi}{6}=\frac{2\pi}{3}\). 2. By the Law of Cosines, \(AB^2=4^2+4^2-2\cdot4\cdot4\cos\frac{2\pi}{3}=48\). 3. Therefore, \(AB=\sqrt{48}=4\sqrt{3}\).

\(4\sqrt{3}\)

- Relate the directions used by positive and negative radii for the same point. - Use the positive y-coordinate to restrict the possible positive-radius angle. - Combine the angle relationship with the stated sum only after choosing the correct branch in \([0,2\pi)\).

1. Because \(r>0\) and the point has positive y-coordinate, \(0<\theta<\pi\). 2. Changing the sign of the radius requires reversing the direction by \(\pi\). In the required interval, \(\phi=\theta+\pi\). 3. Use the given sum: \(\theta+(\theta+\pi)=2\pi\). Thus \(2\theta=\pi\), so \(\theta=\frac{\pi}{2}\) and \(\phi=\frac{3\pi}{2}\).

\(\theta=\frac{\pi}{2}\), \(\phi=\frac{3\pi}{2}\)

- Substitute \(r=0\) into both rectangular conversion formulas. - Ask whether changing only the angle can change either rectangular coordinate when the radius is zero. - Use the distance identity to connect the pole with \(r=0\).

1. When \(r=0\), \(x=0\cdot\cos\theta=0\) and \(y=0\cdot\sin\theta=0\) for every \(\theta\). Thus every pair \((0, \theta)\) represents the origin. 2. For example, \((0, 0)\), \(\left(0, \frac{\pi}{2}\right)\), and \((0, 7\pi)\) are three distinct polar pairs for the origin. 3. The identity \(r^2=x^2+y^2\) gives \(r=0\) at the origin, while both coordinate formulas become zero regardless of \(\theta\). Therefore, no unique angle is determined there.

a) \((0, 0)\) b) For example, \((0, 0)\), \(\left(0, \frac{\pi}{2}\right)\), and \((0, 7\pi)\) c) When \(r=0\), both \(x=r\cos\theta\) and \(y=r\sin\theta\) equal \(0\) for every \(\theta\), so the pole has no unique polar angle.