Precalculus Quiz: Special Triangles Unit Circle In Trigonometry

What this quiz covers

This quiz focuses on Special Triangles Unit Circle In Trigonometry, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\sqrt{3}/2$
2. $\sqrt{2}/2$
3. $1/2$ (correct answer)
4. $1$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 30-60-90 triangle has side ratios of 1 : √3 : 2 (short leg : long leg : hypotenuse), and when scaled to the unit circle (hypotenuse = 1), the short leg is 1/2 and the long leg is √3/2. In a 30-60-90 triangle with hypotenuse 1, the side opposite the 30° angle (π/6 radians) has length 1/2, and the side opposite the 60° angle (π/3 radians) has length √3/2. Therefore, sin(π/6) = 1/2, cos(π/6) = √3/2, while sin(π/3) = √3/2, cos(π/3) = 1/2. Choice C is correct because for θ=π/3 (60°), the adjacent side to the angle is the shorter leg 1/2, matching cos(π/3) from the scaled 30-60-90 ratios. Choice A reverses sine and cosine, giving the y-coordinate value (sine) when the question asks for the x-coordinate value (cosine). Remember that on the unit circle, sin(π/6) and cos(π/3) are equal (both = 1/2), and sin(π/3) and cos(π/6) are equal (both = √3/2), because these are complementary angles in a 30-60-90 triangle. To remember which value goes with which angle: the smaller angle (30° or π/6) has the smaller sine value (1/2), and the larger angle (60° or π/3) has the larger sine value (√3/2).

Question 2

1. $-\frac{\sqrt{2}}{2}$
2. $\frac{\sqrt{2}}{2}$ (correct answer)
3. $-1$
4. $1$

Explanation: Since $\cos(\beta) = \frac{\sqrt{2}}{2}$ and $\beta \in [0, \frac{\pi}{2}]$, we have $\beta = \frac{\pi}{4}$, so $\sin(\beta) = \frac{\sqrt{2}}{2}$ and $\tan(\beta) = 1$. Using unit circle identities: $\tan(\pi - \beta) = -\tan(\beta) = -1$ and $\sin(2\pi - \beta) = -\sin(\beta) = -\frac{\sqrt{2}}{2}$. Therefore, $\tan(\pi - \beta) \cdot \sin(2\pi - \beta) = (-1) \cdot (-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2}$. Choice A forgets that the product of two negatives is positive. Choice C incorrectly computes $\sin(2\pi - \beta)$ as $-1$. Choice D incorrectly computes the individual trigonometric values.

Question 3

1. $\frac{1 - \sqrt{3}}{2}$ (correct answer)
2. $\frac{\sqrt{3} - 1}{2}$
3. $\frac{1 + \sqrt{3}}{2}$
4. $\frac{-1 - \sqrt{3}}{2}$

Explanation: The triangle setup is meant to distract from the core calculation. We need $\cos(\pi + \frac{\pi}{6}) + \sin(\pi - \frac{\pi}{6})$. Using reference angles from special triangles: $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Then $\cos(\pi + \frac{\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$ and $\sin(\pi - \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$. Therefore, $-\frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 - \sqrt{3}}{2}$. Choice B has the terms reversed. Choice C incorrectly applies the unit circle identities with wrong signs. Choice D applies both identities with wrong signs.

Question 4

1. $\sqrt{3}/2$
2. $-\sqrt{3}/2$ (correct answer)
3. $-1/2$
4. $1/2$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 30-60-90 triangle has side ratios of 1 : √3 : 2 (short leg : long leg : hypotenuse), and when scaled to the unit circle (hypotenuse = 1), the short leg is 1/2 and the long leg is √3/2. For angle 5π/6, we use the 30-60-90 triangle with reference angle π/6; in quadrant II, cos is negative, so cos(5π/6) = -cos(π/6) = -√3/2. Choice B is correct because the reference triangle gives cos(π/6) = √3/2, and the quadrant II sign makes it negative. Choice C confuses sine and cosine, giving -1/2 which is actually sin(5π/6), not cos. Key to special angle problems: memorize the two special triangles (1:1:√2 for 45-45-90, and 1:√3:2 for 30-60-90), then remember that sin uses the opposite side and cos uses the adjacent side when the angle is at the origin. Remember that on the unit circle, sin(π/6) and cos(π/3) are equal (both = 1/2), and sin(π/3) and cos(π/6) are equal (both = √3/2), because these are complementary angles in a 30-60-90 triangle.

Question 5

1. $\sqrt{3}$
2. $\frac{\sqrt{3}}{3}$ (correct answer)
3. $1$
4. $\frac{\sqrt{2}}{2}$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 30-60-90 triangle has side ratios of 1 : √3 : 2 (short leg : long leg : hypotenuse), and when scaled to the unit circle (hypotenuse = 1), the short leg is 1/2 and the long leg is √3/2. For angle π/6, we use the 30-60-90 triangle; tan(π/6) = sin(π/6)/cos(π/6) = (1/2)/(√3/2) = 1/√3 = √3/3. Choice B is correct because it computes tangent as opposite over adjacent from the 30-60-90 ratios, giving √3/3. Choice A incorrectly computes tangent as cos/sin instead of sin/cos, yielding √3. Key to special angle problems: memorize the two special triangles (1:1:√2 for 45-45-90, and 1:√3:2 for 30-60-90), then remember that sin uses the opposite side and cos uses the adjacent side when the angle is at the origin. To remember which value goes with which angle: the smaller angle (30° or π/6) has the smaller sine value (1/2), and the larger angle (60° or π/3) has the larger sine value (√3/2).

Question 6

1. $\frac{\sqrt{3}}{2}$
2. $\frac{\sqrt{2}}{2}$
3. $\frac{1}{2}$ (correct answer)
4. $\frac{\sqrt{3}}{3}$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 30-60-90 triangle has side ratios of 1 : √3 : 2 (short leg : long leg : hypotenuse), and when scaled to the unit circle (hypotenuse = 1), the short leg is 1/2 and the long leg is √3/2. In a 30-60-90 triangle with hypotenuse 1, the side opposite the 30° angle (π/6 radians) has length 1/2, and the side opposite the 60° angle (π/3 radians) has length √3/2. Therefore, sin(π/6) = 1/2, cos(π/6) = √3/2, while sin(π/3) = √3/2, cos(π/3) = 1/2. Choice C is correct because cos(π/3) equals the adjacent side to the 60° angle divided by the hypotenuse, which is the short leg (1/2) in the 30-60-90 triangle scaled to unit circle. Choice A confuses sine and cosine values, giving sin(π/3) = √3/2 when the question asks for cosine. Remember that on the unit circle, sin(π/6) and cos(π/3) are equal (both = 1/2), and sin(π/3) and cos(π/6) are equal (both = √3/2), because these are complementary angles in a 30-60-90 triangle.

Question 7

1. $\frac{\sqrt{6}}{2}$
2. $\sqrt{2}$
3. $1$
4. $\sqrt{3}$ (correct answer)

Explanation: When you encounter questions about points on the unit circle, you're working with coordinates where each point can be expressed as $(\cos \theta, \sin \theta)$ for angle $\theta$. To find the distance between two points, you'll use the standard distance formula. First, let's find the coordinates of both points. Point $P$ corresponds to angle $\frac{\pi}{6}$, so $P = \left(\cos\frac{\pi}{6}, \sin\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$. Point $Q$ corresponds to angle $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$, so $Q = \left(\cos\frac{5\pi}{6}, \sin\frac{5\pi}{6}\right) = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$. Using the distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ $d = \sqrt{\left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2} - \frac{1}{2}\right)^2} = \sqrt{(-\sqrt{3})^2 + 0^2} = \sqrt{3}$ This confirms answer D is correct. Let's examine why the other answers are wrong: A) $\frac{\sqrt{6}}{2}$ might result from incorrectly adding the coordinates rather than using the distance formula. B) $\sqrt{2}$ could come from mistakenly using $45°$ angles instead of $30°$ angles. C) $1$ might seem reasonable since it's the radius, but these points aren't diametrically opposite. Key strategy: Notice that angles $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ are reflections across the y-axis, giving points with the same y-coordinate but opposite x-coordinates. This makes the distance calculation much simpler—just twice the x-coordinate's absolute value.

Question 8

1. $\frac{\sqrt{3}}{2} - \frac{1}{2}$
2. $-\frac{\sqrt{3}}{2} + \frac{1}{2}$
3. $-\frac{\sqrt{3}}{2} - \frac{1}{2}$ (correct answer)
4. $-\sqrt{3}$

Explanation: This problem combines properties of regular polygons inscribed in circles with trigonometric identities. When you see expressions like $\sin(\pi + \theta)$ or $\cos(2\pi - \theta)$, think about using angle addition and reference angle relationships. Let's evaluate each term separately. For $\sin(\pi + \frac{\pi}{3})$, use the identity $\sin(\pi + \theta) = -\sin(\theta)$. So $\sin(\pi + \frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$. For $\cos(2\pi - \frac{\pi}{3})$, use the identity $\cos(2\pi - \theta) = \cos(-\theta) = \cos(\theta)$. So $\cos(2\pi - \frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$. Therefore: $\sin(\pi + \frac{\pi}{3}) - \cos(2\pi - \frac{\pi}{3}) = -\frac{\sqrt{3}}{2} - \frac{1}{2}$ Choice A gives $\frac{\sqrt{3}}{2} - \frac{1}{2}$, which incorrectly applies $\sin(\pi + \theta) = \sin(\theta)$ instead of the negative identity. Choice B gives $-\frac{\sqrt{3}}{2} + \frac{1}{2}$, which correctly handles the sine term but treats the subtraction as addition, likely from misapplying $\cos(2\pi - \theta) = -\cos(\theta)$. Choice D gives $-\sqrt{3}$, which incorrectly combines the fractions or uses wrong trigonometric values. Choice C is correct: $-\frac{\sqrt{3}}{2} - \frac{1}{2}$. Key strategy: Master the fundamental trigonometric identities for angles in different quadrants, especially $\sin(\pi + \theta) = -\sin(\theta)$ and $\cos(2\pi - \theta) = \cos(\theta)$. These appear frequently in precalculus problems involving angle transformations.

Question 9

1. $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$
2. $\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$
3. $\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ (correct answer)
4. $\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. Special triangles allow us to find exact trigonometric values without a calculator: the 30-60-90 triangle gives values for angles of π/6 and π/3, while the 45-45-90 triangle gives values for angles of π/4. For angle 7π/6, we find the reference angle: 7π/6 - π = π/6, so we use a 30-60-90 triangle with reference angle π/6 in the third quadrant where both sine and cosine are negative. Choice C is correct because at 7π/6 in the third quadrant, cos(7π/6) = -cos(π/6) = -√3/2 (negative x-coordinate) and sin(7π/6) = -sin(π/6) = -1/2 (negative y-coordinate), using the 30-60-90 triangle values with negative signs for the third quadrant. Choice D reverses the coordinate values, incorrectly placing -1/2 as the x-coordinate and -√3/2 as the y-coordinate, when these should be swapped based on the 30-60-90 triangle ratios. Remember that on the unit circle, sin(π/6) and cos(π/3) are equal (both = 1/2), and sin(π/3) and cos(π/6) are equal (both = √3/2), because these are complementary angles in a 30-60-90 triangle.

Question 10

1. $\frac{2\sqrt{3}}{3}$
2. $-\frac{2\sqrt{3}}{3}$
3. $0$ (correct answer)
4. $-2\sqrt{3}$

Explanation: When you encounter trigonometric equations with special angles, you need to find all solutions in the given interval and then apply trigonometric identities carefully. First, solve $\cos(x) = \frac{1}{2}$. The cosine function equals $\frac{1}{2}$ at two angles in $[0, 2\pi)$: $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$. So $\alpha = \frac{\pi}{3}$ and $\beta = \frac{5\pi}{3}$. Now calculate each tangent value using the identity $\tan(\pi - \theta) = -\tan(\theta)$: For $\tan(\pi - \alpha) = \tan(\pi - \frac{\pi}{3}) = \tan(\frac{2\pi}{3}) = -\tan(\frac{\pi}{3}) = -\sqrt{3}$ For $\tan(\pi - \beta) = \tan(\pi - \frac{5\pi}{3}) = \tan(-\frac{2\pi}{3})$ Since tangent has period $\pi$, we have $\tan(-\frac{2\pi}{3}) = \tan(-\frac{2\pi}{3} + \pi) = \tan(\frac{\pi}{3}) = \sqrt{3}$ Therefore: $\tan(\pi - \alpha) + \tan(\pi - \beta) = -\sqrt{3} + \sqrt{3} = 0$ The answer is C. Choice A gives $\frac{2\sqrt{3}}{3}$, which you might get if you incorrectly used $\tan(30°)$ instead of $\tan(60°)$. Choice B gives $-\frac{2\sqrt{3}}{3}$, possibly from sign errors combined with the same angle mistake. Choice D gives $-2\sqrt{3}$, which could result from forgetting the cancellation and adding $-\sqrt{3} + (-\sqrt{3})$. Study tip: When working with expressions like $\tan(\pi - \theta)$, always use the identity $\tan(\pi - \theta) = -\tan(\theta)$ and watch for symmetric pairs that cancel out.

Question 11

1. $\sin\left(\pi/6\right)=\sin\left(\pi/3\right)$
2. $\sin\left(\pi/6\right)>\sin\left(\pi/3\right)$
3. $\sin\left(\pi/6\right)<\sin\left(\pi/3\right)$ (correct answer)
4. $\sin\left(\pi/6\right)=\cos\left(\pi/6\right)$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 30-60-90 triangle has side ratios of 1 : √3 : 2 (short leg : long leg : hypotenuse), and when scaled to the unit circle (hypotenuse = 1), the short leg is 1/2 and the long leg is √3/2. Comparing sin(π/6) and sin(π/3), we see that sin(π/6) = 1/2 (opposite the 30° angle) and sin(π/3) = √3/2 (opposite the 60° angle), and since 1/2 < √3/2, we have sin(π/6) < sin(π/3). Choice C is correct because sin(π/6) = 1/2 and sin(π/3) = √3/2, and since 1/2 ≈ 0.5 and √3/2 ≈ 0.866, we clearly have sin(π/6) < sin(π/3). Choice A incorrectly states these values are equal, confusing the fact that sin(π/6) = cos(π/3) (both equal 1/2) with a comparison of the two sine values. To remember which value goes with which angle: the smaller angle (30° or π/6) has the smaller sine value (1/2), and the larger angle (60° or π/3) has the larger sine value (√3/2).

Question 12

1. $(\frac{1}{2}, \frac{\sqrt{3}}{2})$
2. $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$ (correct answer)
3. $(-\frac{\sqrt{3}}{2}, -\frac{1}{2})$
4. $(\frac{\sqrt{3}}{2}, \frac{1}{2})$

Explanation: Since $\tan(\alpha) = \sqrt{3}$ and $\alpha$ is in the first quadrant, we have $\alpha = \frac{\pi}{3}$. The coordinates of point $P$ are $(\cos(\frac{\pi}{3}), \sin(\frac{\pi}{3})) = (\frac{1}{2}, \frac{\sqrt{3}}{2})$. For angle $\pi + \alpha$, we use the identity that this places us in the third quadrant where both coordinates are negative: $\cos(\pi + \alpha) = -\cos(\alpha) = -\frac{1}{2}$ and $\sin(\pi + \alpha) = -\sin(\alpha) = -\frac{\sqrt{3}}{2}$. Choice A gives the original point coordinates. Choice C swaps the coordinates incorrectly. Choice D forgets the sign changes for the third quadrant.

Question 13

1. $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$
2. $\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ (correct answer)
3. $\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$
4. $\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 45-45-90 triangle has side ratios of 1 : 1 : √2 (leg : leg : hypotenuse), and when scaled to fit the unit circle (hypotenuse = 1), each leg has length √2/2, giving sin(π/4) = cos(π/4) = √2/2. For angle 3π/4, we are in the second quadrant where the reference angle is π - 3π/4 = π/4, so we use a 45-45-90 triangle where cosine is negative (x-coordinate) and sine is positive (y-coordinate). Choice B is correct because at 3π/4 in the second quadrant, cos(3π/4) = -√2/2 (negative x-coordinate) and sin(3π/4) = √2/2 (positive y-coordinate), using the 45-45-90 triangle values with appropriate signs. Choice A fails to account for the quadrant, giving positive values for both coordinates when cosine should be negative in the second quadrant. For 45-45-90 triangles, the key insight is that the two legs are equal, so sin(π/4) = cos(π/4) = √2/2, and this same value appears at all 45° angles around the unit circle (with appropriate signs).

Question 14

1. $(-1/2,-\sqrt{3}/2)$ (correct answer)
2. $(1/2,-\sqrt{3}/2)$
3. $(-\sqrt{3}/2,-1/2)$
4. $(1/2,\sqrt{3}/2)$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 30-60-90 triangle has side ratios of 1 : √3 : 2 (short leg : long leg : hypotenuse), and when scaled to the unit circle (hypotenuse = 1), the short leg is 1/2 and the long leg is √3/2. In a 30-60-90 triangle with hypotenuse 1, the side opposite the 30° angle (π/6 radians) has length 1/2, and the side opposite the 60° angle (π/3 radians) has length √3/2. Therefore, sin(π/6) = 1/2, cos(π/6) = √3/2, while sin(π/3) = √3/2, cos(π/3) = 1/2. Choice A is correct because for θ=4π/3 in quadrant III, the reference π/3 gives cos=1/2 (negative) and sin=√3/2 (negative), yielding (-1/2, -√3/2). Choice C reverses sine and cosine, giving the x-coordinate as -√3/2 and y as -1/2 when actually cos(4π/3) = -1/2 and sin(4π/3) = -√3/2. Remember that on the unit circle, sin(π/6) and cos(π/3) are equal (both = 1/2), and sin(π/3) and cos(π/6) are equal (both = √3/2), because these are complementary angles in a 30-60-90 triangle. To remember which value goes with which angle: the smaller angle (30° or π/6) has the smaller sine value (1/2), and the larger angle (60° or π/3) has the larger sine value (√3/2).

Question 15

1. $1/2$
2. $\sqrt{3}/2$ (correct answer)
3. $\sqrt{2}/2$
4. $\sqrt{3}/3$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 30-60-90 triangle has side ratios of 1 : √3 : 2 (short leg : long leg : hypotenuse), and when scaled to the unit circle (hypotenuse = 1), the short leg is 1/2 and the long leg is √3/2. In a 30-60-90 triangle with hypotenuse 1, the side opposite the 30° angle (π/6 radians) has length 1/2, and the side opposite the 60° angle (π/3 radians) has length √3/2. Therefore, sin(π/6) = 1/2, cos(π/6) = √3/2, while sin(π/3) = √3/2, cos(π/3) = 1/2. Choice B is correct because for θ=π/3, the opposite side in the 30-60-90 triangle is the longer leg √3/2, matching sin(π/3). Choice A swaps the values for 30° and 60°, using sin(π/3) = 1/2 when actually sin(π/6) = 1/2 and sin(π/3) = √3/2. Remember that on the unit circle, sin(π/6) and cos(π/3) are equal (both = 1/2), and sin(π/3) and cos(π/6) are equal (both = √3/2), because these are complementary angles in a 30-60-90 triangle. Always express special angle values exactly using radicals (1/2, √2/2, √3/2) rather than decimal approximations—this is both more precise and expected in mathematics.

Question 16

1. $\frac{1}{2}$
2. $\frac{\sqrt{3}}{2}$
3. $\frac{\sqrt{2}}{2}$ (correct answer)
4. $\sqrt{3}$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 45-45-90 triangle has side ratios of 1 : 1 : √2 (leg : leg : hypotenuse), and when scaled to fit the unit circle (hypotenuse = 1), each leg has length √2/2, giving sin(π/4) = cos(π/4) = √2/2. In a 45-45-90 triangle on the unit circle, both legs are equal and have length √2/2 (since leg² + leg² = 1²), which means at angle π/4, the coordinates are (√2/2, √2/2), so both sine and cosine equal √2/2. Choice C is correct because for a 45° angle in a 45-45-90 triangle, the opposite side (which gives sine) has length √2/2 when the hypotenuse is scaled to 1. Choice A uses the value from a 30-60-90 triangle (√3/2), confusing the special triangles and their associated angles. For 45-45-90 triangles, the key insight is that the two legs are equal, so sin(π/4) = cos(π/4) = √2/2, and this same value appears at all 45° angles around the unit circle (with appropriate signs).

Question 17

1. $(\sqrt{2}/2,\sqrt{2}/2)$
2. $(1/2,\sqrt{3}/2)$
3. $(\sqrt{3}/2,1/2)$ (correct answer)
4. $(\sqrt{3}/2,\sqrt{3}/2)$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 30-60-90 triangle has side ratios of 1 : √3 : 2 (short leg : long leg : hypotenuse), and when scaled to the unit circle (hypotenuse = 1), the short leg is 1/2 and the long leg is √3/2. In a 30-60-90 triangle with hypotenuse 1, the side opposite the 30° angle (π/6 radians) has length 1/2, and the side opposite the 60° angle (π/3 radians) has length √3/2. Therefore, sin(π/6) = 1/2, cos(π/6) = √3/2, while sin(π/3) = √3/2, cos(π/3) = 1/2. Choice C is correct because for θ=π/6, the triangle places the x-coordinate (cos) as the longer adjacent side √3/2 and the y-coordinate (sin) as the shorter opposite side 1/2. Choice B confuses the values for π/6 and π/3, swapping sine and cosine by using sin(π/6) = √3/2 when actually sin(π/6) = 1/2. Key to special angle problems: memorize the two special triangles (1:1:√2 for 45-45-90, and 1:√3:2 for 30-60-90), then remember that sin uses the opposite side and cos uses the adjacent side when the angle is at the origin. To remember which value goes with which angle: the smaller angle (30° or π/6) has the smaller sine value (1/2), and the larger angle (60° or π/3) has the larger sine value (√3/2).

Question 18

1. $-\frac{\sqrt{2}}{2}$ (correct answer)
2. $\frac{\sqrt{2}}{2}$
3. $-\frac{\sqrt{3}}{2}$
4. $\frac{1}{2}$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 45-45-90 triangle has side ratios of 1 : 1 : √2 (leg : leg : hypotenuse), and when scaled to fit the unit circle (hypotenuse = 1), each leg has length √2/2, giving sin(π/4) = cos(π/4) = √2/2. For angle 5π/4, we are in the third quadrant where the reference angle is 5π/4 - π = π/4, so we use a 45-45-90 triangle where both cosine (x-coordinate) and sine (y-coordinate) are negative. Choice A is correct because cos(5π/4) = -cos(π/4) = -√2/2, since 5π/4 is in the third quadrant where cosine values are negative and the reference angle π/4 has cosine value √2/2. Choice B gives the positive value √2/2, failing to account for the negative sign required in the third quadrant where 5π/4 is located. For 45-45-90 triangles, the key insight is that the two legs are equal, so sin(π/4) = cos(π/4) = √2/2, and this same value appears at all 45° angles around the unit circle (with appropriate signs).

Question 19

1. $\tfrac{1}{2}$
2. $\tfrac{\sqrt{3}}{2}$
3. $\tfrac{\sqrt{2}}{2}$ (correct answer)
4. $\sqrt{3}$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 45-45-90 triangle has side ratios of 1 : 1 : √2 (leg : leg : hypotenuse), and when scaled to fit the unit circle (hypotenuse = 1), each leg has length √2/2, giving sin(π/4) = cos(π/4) = √2/2. In a 45-45-90 triangle on the unit circle, both legs are equal and have length √2/2 (since leg² + leg² = 1²), which means at angle π/4, the coordinates are (√2/2, √2/2), so both sine and cosine equal √2/2. Choice C is correct because it uses the 45-45-90 ratios scaled to hypotenuse 1, where the opposite side to π/4 is √2/2, directly giving sin(π/4). Choice B confuses the 30-60-90 triangle with the 45-45-90 triangle, using √3/2 when the correct ratio is √2/2. For 45-45-90 triangles, the key insight is that the two legs are equal, so sin(π/4) = cos(π/4) = √2/2, and this same value appears at all 45° angles around the unit circle (with appropriate signs). Always express special angle values exactly using radicals (1/2, √2/2, √3/2) rather than decimal approximations—this is both more precise and expected in mathematics.

Question 20

1. $(\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{2}}{2})$ (correct answer)
2. $(\tfrac{\sqrt{3}}{2},\tfrac{1}{2})$
3. $(\tfrac{1}{2},\tfrac{\sqrt{3}}{2})$
4. $(\tfrac{\sqrt{2}}{2},\tfrac{1}{2})$

Explanation: This question tests understanding of special right triangles and how their ratios determine exact trigonometric values on the unit circle. A 45-45-90 triangle has side ratios of 1 : 1 : √2 (leg : leg : hypotenuse), and when scaled to fit the unit circle (hypotenuse = 1), each leg has length √2/2, giving sin(π/4) = cos(π/4) = √2/2. In a 45-45-90 triangle on the unit circle, both legs are equal and have length √2/2 (since the legs² + legs² = 1²), which means at angle π/4, the coordinates are (√2/2, √2/2), so both sine and cosine equal √2/2. Choice A is correct because the 45-45-90 triangle gives equal x and y coordinates of √2/2 on the unit circle. Choice B confuses the 30-60-90 triangle with the 45-45-90 triangle, using (√3/2, 1/2) when the correct is (√2/2, √2/2). For 45-45-90 triangles, the key insight is that the two legs are equal, so sin(π/4) = cos(π/4) = √2/2, and this same value appears at all 45° angles around the unit circle (with appropriate signs). Always express special angle values exactly using radicals (1/2, √2/2, √3/2) rather than decimal approximations—this is both more precise and expected in mathematics.