The coordinates are given by $$x = r\cos\theta$$ and $$y = r\sin\theta$$. We have $$r=5$$, $$x = -\frac{5\sqrt{2}}{2}$$, and $$y = \frac{5\sqrt{2}}{2}$$. So, $$\cos\theta = \frac{x}{r} = \frac{-5\sqrt{2}/2}{5} = -\frac{\sqrt{2}}{2}$$ and $$\sin\theta = \frac{y}{r} = \frac{5\sqrt{2}/2}{5} = \frac{\sqrt{2}}{2}$$. An angle with a negative cosine and a positive sine must be in Quadrant II. The angle in $$[0, 2\pi)$$ that satisfies these conditions is $$\theta = \frac{3\pi}{4}$$.

The coordinates are $$(r\cos\theta, r\sin\theta)$$. We are given $$r=8$$ and $$\sin\theta = \frac{3}{4}$$. The y-coordinate is $$y = r\sin\theta = 8(\frac{3}{4}) = 6$$. To find the x-coordinate, we use the identity $$\sin^2\theta + \cos^2\theta = 1$$. This gives $$(\frac{3}{4})^2 + \cos^2\theta = 1$$, so $$\cos^2\theta = 1 - \frac{9}{16} = \frac{7}{16}$$. Since $$\theta$$ is in Quadrant II, $$\cos\theta$$ is negative, so $$\cos\theta = -\frac{\sqrt{7}}{4}$$. The x-coordinate is $$x = r\cos\theta = 8(-\frac{\sqrt{7}}{4}) = -2\sqrt{7}$$. The coordinates of P are $$(-2\sqrt{7}, 6)$$.

For a point on the unit circle, the coordinates are given by $$(\cos\theta, \sin\theta)$$. The angle $$\theta = \frac{2\pi}{3}$$ is in Quadrant II, where the x-coordinate (cosine) is negative and the y-coordinate (sine) is positive. The reference angle is $$\frac{\pi}{3}$$. Thus, $$\cos(\frac{2\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2}$$ and $$\sin(\frac{2\pi}{3}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$$. The coordinates of $$P$$ are $$(-\frac{1}{2}, \frac{\sqrt{3}}{2})$$.

The coordinates are given by $$(r\cos\theta, r\sin\theta)$$. The angle $$\theta = -\frac{7\pi}{4}$$ is coterminal with $$\frac{\pi}{4}$$, which is in Quadrant I. Thus, $$\cos(-\frac{7\pi}{4}}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$ and $$\sin(-\frac{7\pi}{4}}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$$. The coordinates of $$P$$ are $$(2\sqrt{2} \cdot \frac{\sqrt{2}}{2}, 2\sqrt{2} \cdot \frac{\sqrt{2}}{2}) = (2 \cdot \frac{2}{2}, 2 \cdot \frac{2}{2}) = (2, 2)$$.

The coordinates of a point $$(x,y)$$ on a circle of radius $$r$$ are related to the angle $$\theta$$ by $$x = r\cos\theta$$ and $$y = r\sin\theta$$. We are given the point $$(-2, 5)$$ and radius $$r = \sqrt{29}$$. Therefore, $$\cos\theta = \frac{x}{r} = -\frac{2}{\sqrt{29}}$$ and $$\sin\theta = \frac{y}{r} = \frac{5}{\sqrt{29}}$$. We can verify that the given radius is correct: $$r = \sqrt{(-2)^2 + 5^2} = \sqrt{4+25} = \sqrt{29}$$.

The coordinates are $$(r\cos\theta, r\sin\theta)$$. With $$r=6$$ and $$\theta=\frac{5\pi}{4}$$, the point is in Quadrant III. The reference angle is $$\frac{\pi}{4}$$. Both cosine and sine are negative. $$\cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$$ and $$\sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$$. The coordinates are $$(6(-\frac{\sqrt{2}}{2}), 6(-\frac{\sqrt{2}}{2})) = (-3\sqrt{2}, -3\sqrt{2})$$.

This question tests AP Precalculus skills: understanding sine and cosine function values in context. Sine values represent the y-coordinate on the unit circle where the angle intersects. In this problem, the angle θ = π/2 (90°) is given for a sound wave model, requiring students to find its sine value. Choice B (1) is correct because at π/2 radians, the point on the unit circle is at (0, 1), making sin(π/2) = 1. Choice A (0) is incorrect as this is the cosine value at π/2, showing confusion between x and y coordinates at this quadrantal angle. Students should memorize the four quadrantal angles (0°, 90°, 180°, 270°) and their exact trigonometric values. Visualizing the unit circle helps students see that π/2 points straight up, giving maximum sine value.

This question tests AP Precalculus skills: understanding sine and cosine function values in context. Cosine values represent the x-coordinate on the unit circle where the angle intersects. In this problem, the angle θ = π/4 (45°) is given in an AC circuit context, requiring students to find its cosine value. Choice A (√2/2) is correct because at π/4 radians, the point on the unit circle has coordinates (√2/2, √2/2), making cos(π/4) = √2/2. Choice D (0.707) is incorrect as it's a decimal approximation when the problem specifically asks for exact values, showing students must distinguish between exact and approximate forms. Students should memorize that at 45°, both sine and cosine equal √2/2. Using the isosceles right triangle with hypotenuse 1 helps derive this value exactly.

This question tests AP Precalculus skills: understanding sine and cosine function values in context. Cosine values represent the x-coordinate on the unit circle where the angle intersects. In this problem, the angle θ = π (180°) is given for a sound wave model, requiring students to find its cosine value at this quadrantal angle. Choice C (-1) is correct because at π radians, the point on the unit circle is at (-1, 0), making cos(π) = -1. Choice B (0) is incorrect as this is the sine value at π, showing confusion between x and y coordinates at quadrantal angles. Students should memorize that π radians points directly left on the unit circle, giving the most negative cosine value. Understanding quadrantal angles as multiples of π/2 helps students quickly identify their exact trigonometric values.

This question tests AP Precalculus skills: understanding sine and cosine function values in context. The Pythagorean identity sin²θ + cos²θ = 1 connects sine and cosine values on the unit circle. In this problem, cos(θ) = -3/5 with θ in Quadrant II, requiring students to find sin(θ) using the identity. Choice A (4/5) is correct because sin²θ = 1 - cos²θ = 1 - (-3/5)² = 1 - 9/25 = 16/25, so sin(θ) = 4/5 (positive in Quadrant II). Choice B (-4/5) is incorrect as it has the wrong sign, forgetting that sine is positive in Quadrant II. Students must carefully apply both the Pythagorean identity and quadrant sign rules. The ASTC mnemonic reminds us that sine is positive in Quadrants I and II.