This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. The sign of each trigonometric function depends on which quadrant the terminal side lies in: Quadrant I (all positive), Quadrant II (sin positive, cos negative), Quadrant III (tan positive, sin and cos negative), Quadrant IV (cos positive, sin negative). Since the angle terminates in Quadrant III with reference angle π/4, we find the magnitude from tan(π/4) = 1, and since tangent is positive in Quadrant III (both sine and cosine negative, so their ratio positive), tan(θ) = 1. Choice C is correct because it connects to the quadrant properties where tangent is positive in Quadrant III, matching the reference angle's value. Choice A gives a negative value, forgetting that in Quadrant III, tan is positive due to both sin and cos being negative. The reference angle is always positive and acute, found by measuring to the nearest x-axis: for Quadrant II use π - θ, for Quadrant III use θ - π, for Quadrant IV use 2π - θ. Key to unit circle problems: first determine which quadrant the angle terminates in, then use the reference angle to find magnitudes, and finally apply the correct signs based on the quadrant (memorize: All Students Take Calculus for which functions are positive in each quadrant).

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. The sign of each trigonometric function depends on which quadrant the terminal side lies in: Quadrant I (all positive), Quadrant II (sin positive, cos negative), Quadrant III (tan positive, sin and cos negative), Quadrant IV (cos positive, sin negative). Since the angle terminates in Quadrant III with reference angle π/4, we know the terminal point has coordinates (-√2/2, -√2/2) because both x and y are negative in Quadrant III, and the reference angle π/4 gives us the 45-45-90 special triangle values. In Quadrant III, the x-coordinate is negative and the y-coordinate is negative, which means cos(θ) is negative and sin(θ) is negative, making tan(θ) = sin/cos positive. Choice C is correct because tan(θ) = (-√2/2)/(-√2/2) = 1, as the negative signs cancel when dividing. Choice A gives -1, which would be correct if one coordinate were positive and one negative, but in Quadrant III both are negative. Remember that on the unit circle, cos(θ) is always the x-coordinate and sin(θ) is always the y-coordinate of the terminal point—this definition works for any angle, positive or negative, and regardless of how many full rotations are involved.

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. Unlike right triangle definitions which only work for acute angles, the unit circle definition allows us to evaluate trigonometric functions for negative angles, angles greater than 90° (or π/2), and even angles representing multiple complete rotations. The angle 9π/4 can be simplified by subtracting 2π (one full rotation): 9π/4 - 8π/4 = π/4, which places it in Quadrant I where both sine and cosine are positive. For angle θ = π/4, the terminal side lies in Quadrant I, where angles are between 0 and π/2. On the unit circle, this corresponds to point (√2/2, √2/2), so cos(π/4) = √2/2 and sin(π/4) = √2/2. Choice A is correct because after removing the full rotation, the angle π/4 in Quadrant I has sin(π/4) = √2/2. Choice B gives the negative value, which would be correct for an angle in Quadrant III or IV, but this angle terminates in Quadrant I. For angles outside [0, 2π], find the co-terminal angle by adding or subtracting 2π until you get an angle in the standard range, then evaluate using the unit circle.

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. The unit circle in the coordinate plane enables us to define sine and cosine for any angle: if angle θ in standard position has terminal side passing through point (x, y) on the unit circle, then cos(θ) = x and sin(θ) = y. For angle θ = 5π/6, the terminal side lies in Quadrant II, where angles are between π/2 and π. Since 5π/6 is in Quadrant II, we find the reference angle is π - 5π/6 = π/6, which gives us the 30-60-90 special triangle relationship. Applying the correct signs for this quadrant (cosine is negative, sine is positive), we get cos(5π/6) = -√3/2. Choice C is correct because it gives the negative x-coordinate for an angle in Quadrant II with reference angle π/6. Choice A gives the magnitude correct but uses the wrong sign, forgetting that in Quadrant II, cosine is negative. Remember that on the unit circle, cos(θ) is always the x-coordinate and sin(θ) is always the y-coordinate of the terminal point—this definition works for any angle, positive or negative, and regardless of how many full rotations are involved.

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. The unit circle in the coordinate plane enables us to define sine and cosine for any angle: if angle θ in standard position has terminal side passing through point (x, y) on the unit circle, then cos(θ) = x and sin(θ) = y. For angle θ = 7π/6, the terminal side lies in Quadrant III, where the reference angle is π/6, which gives us sin(π/6) = 1/2, but since sine is negative in Quadrant III, sin(7π/6) = -1/2. Choice D is correct because it connects to the unit circle coordinates with the y-coordinate being -1/2 for this angle in Quadrant III. Choice A gives the magnitude correct but uses the wrong sign, forgetting that in Quadrant III, sine is negative. Key to unit circle problems: first determine which quadrant the angle terminates in, then use the reference angle to find magnitudes, and finally apply the correct signs based on the quadrant (memorize: All Students Take Calculus for which functions are positive in each quadrant). The reference angle is always positive and acute, found by measuring to the nearest x-axis: for Quadrant III use θ - π.

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. The unit circle in the coordinate plane enables us to define sine and cosine for any angle: if angle θ in standard position has terminal side passing through point (x, y) on the unit circle, then cos(θ) = x and sin(θ) = y. For angle θ = 3π/4, the terminal side lies in Quadrant II, where the reference angle is π/4, which gives cos(π/4) = √2/2, but since cosine is negative in Quadrant II, cos(3π/4) = -√2/2. Choice B is correct because it connects to the unit circle coordinates in Quadrant II, where the x-coordinate is negative while matching the magnitude from the 45-45-90 triangle ratios. Choice A gives the positive value, using the wrong sign and forgetting that in Quadrant II, cosine is negative. Key to unit circle problems: first determine which quadrant the angle terminates in, then use the reference angle to find magnitudes, and finally apply the correct signs based on the quadrant (memorize: All Students Take Calculus for which functions are positive in each quadrant). The reference angle is always positive and acute, found by measuring to the nearest x-axis: for Quadrant II use π - θ, for Quadrant III use θ - π, for Quadrant IV use 2π - θ.

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. Unlike right triangle definitions which only work for acute angles, the unit circle definition allows us to evaluate trigonometric functions for negative angles, angles greater than 90° (or π/2), and even angles representing multiple complete rotations. The angle 13π/6 can be simplified by subtracting 2π (or 12π/6), resulting in π/6, which is co-terminal and places it in Quadrant I. Choice C is correct because subtracting 2π from 13π/6 yields π/6, sharing the same terminal side on the unit circle. Choice A adds extra π instead of subtracting full rotations, placing it in a different quadrant. For angles outside [0, 2π], find the co-terminal angle by adding or subtracting 2π until you get an angle in the standard range, then evaluate using the unit circle. To find the quadrant: reduce angles greater than 2π by subtracting 2π, convert negative angles by adding 2π, then identify which quarter of the circle (0 to π/2, π/2 to π, π to 3π/2, or 3π/2 to 2π) contains the terminal side.

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. The sign of each trigonometric function depends on which quadrant the terminal side lies in: Quadrant I (all positive), Quadrant II (sin positive, cos negative), Quadrant III (tan positive, sin and cos negative), Quadrant IV (cos positive, sin negative). The angle 5π/3 is between 3π/2 and 2π, which places it in Quadrant IV where sine is negative. Choice B is correct because in Quadrant IV, the y-coordinate is negative, which means sin(θ) is negative. Choice A might tempt if confusing with Quadrant I, but 5π/3 actually terminates in Quadrant IV. To find the quadrant: reduce angles greater than 2π by subtracting 2π, convert negative angles by adding 2π, then identify which quarter of the circle (0 to π/2, π/2 to π, π to 3π/2, or 3π/2 to 2π) contains the terminal side. Key to unit circle problems: first determine which quadrant the angle terminates in, then use the reference angle to find magnitudes, and finally apply the correct signs based on the quadrant (memorize: All Students Take Calculus for which functions are positive in each quadrant).

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. The unit circle in the coordinate plane enables us to define sine and cosine for any angle: if angle θ in standard position has terminal side passing through point (x, y) on the unit circle, then cos(θ) = x and sin(θ) = y. The point P(-1/2, √3/2) lies in Quadrant II on the unit circle, so sin(θ) = y = √3/2. Choice B is correct because it connects to the unit circle coordinates, where the y-coordinate directly gives the sine value, positive in Quadrant II. Choice A reverses the sine and cosine values, using the x-coordinate for sine when sine equals the y-coordinate on the unit circle. Remember that on the unit circle, cos(θ) is always the x-coordinate and sin(θ) is always the y-coordinate of the terminal point—this definition works for any angle, positive or negative, and regardless of how many full rotations are involved. Key to unit circle problems: first determine which quadrant the angle terminates in, then use the reference angle to find magnitudes, and finally apply the correct signs based on the quadrant (memorize: All Students Take Calculus for which functions are positive in each quadrant).

This question tests understanding of how the unit circle extends trigonometric functions to all real numbers. The unit circle in the coordinate plane enables us to define sine and cosine for any angle: if angle θ in standard position has terminal side passing through point (x, y) on the unit circle, then cos(θ) = x and sin(θ) = y. For angle θ = 3π/2, the terminal side lies on the negative y-axis at point (0, -1), so tan(θ) = sin(θ)/cos(θ) = -1/0, which is undefined. Choice D is correct because it recognizes that tangent is undefined when cos(θ) = 0, as at 3π/2 on the unit circle. Choice C gives a numeric value for tangent when x = 0, but tan(3π/2) is undefined because we cannot divide by zero. Remember that on the unit circle, cos(θ) is always the x-coordinate and sin(θ) is always the y-coordinate of the terminal point—this definition works for any angle, positive or negative, and regardless of how many full rotations are involved. Key to unit circle problems: first determine which quadrant the angle terminates in, then use the reference angle to find magnitudes, and finally apply the correct signs based on the quadrant (memorize: All Students Take Calculus for which functions are positive in each quadrant).