Free math worksheets

- 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 \(y=6\sin\frac{\pi}{3}=3\sqrt{3}\).

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

- 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 \(\theta=\frac{7\pi}{6}\).

- 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, \(AB^2=4^2+4^2-2\cdot4\cdot4\cos\frac{2\pi}{3}=32+16=48\). 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 \(\phi=\frac{3\pi}{2}\).

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