Precalculus Quiz: Special Triangles Unit Circle In Trigonometry
20 questions · exam conditions
0:00
Special Triangles Unit Circle In TrigonometryQuestion 1 of 20

On the unit circle, an angle of π/3\pi/3 is drawn in standard position, creating a 30-60-90 triangle with radius 11. Based on the special triangle ratios and cos⁡θ\cos\theta being the xx-coordinate, what is the exact value of cos⁡(π/3)\cos(\pi/3)?

3/2\sqrt{3}/2
2/2\sqrt{2}/2
1/21/2
11
← Back to quizzes

Precalculus Quiz

Precalculus Quiz: Special Triangles Unit Circle In Trigonometry

Practice Special Triangles Unit Circle In Trigonometry in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

On the unit circle, an angle of π/3\pi/3 is drawn in standard position, creating a 30-60-90 triangle with radius 11. Based on the special triangle ratios and cos⁡θ\cos\theta being the xx-coordinate, what is the exact value of cos⁡(π/3)\cos(\pi/3)?

  1. 3/2\sqrt{3}/2
  2. 2/2\sqrt{2}/2
  3. 1/21/2 (correct answer)
  4. 11
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

If cos⁡(β)=22\cos(\beta) = \frac{\sqrt{2}}{2} where β\beta is a special angle in [0,π2][0, \frac{\pi}{2}], what is the exact value of tan⁡(π−β)⋅sin⁡(2π−β)\tan(\pi - \beta) \cdot \sin(2\pi - \beta)?

  1. −22-\frac{\sqrt{2}}{2}
  2. 22\frac{\sqrt{2}}{2} (correct answer)
  3. −1-1
  4. 11
Explanation: Since cos⁡(β)=22\cos(\beta) = \frac{\sqrt{2}}{2} and β∈[0,π2]\beta \in [0, \frac{\pi}{2}], we have β=π4\beta = \frac{\pi}{4}, so sin⁡(β)=22\sin(\beta) = \frac{\sqrt{2}}{2} and tan⁡(β)=1\tan(\beta) = 1. Using unit circle identities: tan⁡(π−β)=−tan⁡(β)=−1\tan(\pi - \beta) = -\tan(\beta) = -1 and sin⁡(2π−β)=−sin⁡(β)=−22\sin(2\pi - \beta) = -\sin(\beta) = -\frac{\sqrt{2}}{2}. Therefore, tan⁡(π−β)⋅sin⁡(2π−β)=(−1)⋅(−22)=22\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⁡(2π−β)\sin(2\pi - \beta) as −1-1. Choice D incorrectly computes the individual trigonometric values.

Question 3

In triangle ABCABC, angle C=π6C = \frac{\pi}{6} and the triangle is positioned so that vertex CC is at the origin, side CACA lies along the positive xx-axis, and vertex BB is in the first quadrant. If ∣CB∣=2|CB| = 2, what is cos⁡(π+π6)+sin⁡(π−π6)\cos(\pi + \frac{\pi}{6}) + \sin(\pi - \frac{\pi}{6})?

  1. 1−32\frac{1 - \sqrt{3}}{2} (correct answer)
  2. 3−12\frac{\sqrt{3} - 1}{2}
  3. 1+32\frac{1 + \sqrt{3}}{2}
  4. −1−32\frac{-1 - \sqrt{3}}{2}
Explanation: The triangle setup is meant to distract from the core calculation. We need cos⁡(π+π6)+sin⁡(π−π6)\cos(\pi + \frac{\pi}{6}) + \sin(\pi - \frac{\pi}{6}). Using reference angles from special triangles: cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} and sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2}. Then cos⁡(π+π6)=−cos⁡(π6)=−32\cos(\pi + \frac{\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} and sin⁡(π−π6)=sin⁡(π6)=12\sin(\pi - \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}. Therefore, −32+12=1−32-\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

An angle of 5π/65\pi/6 is drawn in standard position on the unit circle. Using the 30-60-90 reference triangle (reference angle π/6\pi/6) and applying quadrant signs, what is the exact value of cos⁡(5π/6)\cos(5\pi/6)?

  1. 3/2\sqrt{3}/2
  2. −3/2-\sqrt{3}/2 (correct answer)
  3. −1/2-1/2
  4. 1/21/2
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

Using the 30-60-90 triangle on the unit circle (hypotenuse 11), what is the exact value of cos⁡(π/3)\cos\left(\pi/3\right)?

  1. 32\frac{\sqrt{3}}{2}
  2. 22\frac{\sqrt{2}}{2}
  3. 12\frac{1}{2} (correct answer)
  4. 33\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 6

If PP is a point on the unit circle corresponding to the angle π6\frac{\pi}{6}, and QQ is the point corresponding to π−π6\pi - \frac{\pi}{6}, what is the distance between points PP and QQ?

  1. 62\frac{\sqrt{6}}{2}
  2. 2\sqrt{2}
  3. 11
  4. 3\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⁡θ,sin⁡θ)(\cos \theta, \sin \theta) for angle θ\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 PP corresponds to angle π6\frac{\pi}{6}, so P=(cos⁡π6,sin⁡π6)=(32,12)P = \left(\cos\frac{\pi}{6}, \sin\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right). Point QQ corresponds to angle π−π6=5π6\pi - \frac{\pi}{6} = \frac{5\pi}{6}, so Q=(cos⁡5π6,sin⁡5π6)=(−32,12)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=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} d=(−32−32)2+(12−12)2=(−3)2+02=3d = \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) 62\frac{\sqrt{6}}{2} might result from incorrectly adding the coordinates rather than using the distance formula. B) 2\sqrt{2} could come from mistakenly using 45°45° angles instead of 30°30° angles. C) 11 might seem reasonable since it's the radius, but these points aren't diametrically opposite. Key strategy: Notice that angles π6\frac{\pi}{6} and 5π6\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 7

A regular hexagon is inscribed in the unit circle with one vertex at (1,0)(1, 0). If a vertex of this hexagon corresponds to the angle π3\frac{\pi}{3}, what is the value of sin⁡(π+π3)−cos⁡(2π−π3)\sin(\pi + \frac{\pi}{3}) - \cos(2\pi - \frac{\pi}{3})?

  1. 32−12\frac{\sqrt{3}}{2} - \frac{1}{2}
  2. −32+12-\frac{\sqrt{3}}{2} + \frac{1}{2}
  3. −32−12-\frac{\sqrt{3}}{2} - \frac{1}{2} (correct answer)
  4. −3-\sqrt{3}
Explanation: This problem combines properties of regular polygons inscribed in circles with trigonometric identities. When you see expressions like sin⁡(π+θ)\sin(\pi + \theta) or cos⁡(2π−θ)\cos(2\pi - \theta), think about using angle addition and reference angle relationships. Let's evaluate each term separately. For sin⁡(π+π3)\sin(\pi + \frac{\pi}{3}), use the identity sin⁡(π+θ)=−sin⁡(θ)\sin(\pi + \theta) = -\sin(\theta). So sin⁡(π+π3)=−sin⁡(π3)=−32\sin(\pi + \frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}. For cos⁡(2π−π3)\cos(2\pi - \frac{\pi}{3}), use the identity cos⁡(2π−θ)=cos⁡(−θ)=cos⁡(θ)\cos(2\pi - \theta) = \cos(-\theta) = \cos(\theta). So cos⁡(2π−π3)=cos⁡(π3)=12\cos(2\pi - \frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}. Therefore: sin⁡(π+π3)−cos⁡(2π−π3)=−32−12\sin(\pi + \frac{\pi}{3}) - \cos(2\pi - \frac{\pi}{3}) = -\frac{\sqrt{3}}{2} - \frac{1}{2} Choice A gives 32−12\frac{\sqrt{3}}{2} - \frac{1}{2}, which incorrectly applies sin⁡(π+θ)=sin⁡(θ)\sin(\pi + \theta) = \sin(\theta) instead of the negative identity. Choice B gives −32+12-\frac{\sqrt{3}}{2} + \frac{1}{2}, which correctly handles the sine term but treats the subtraction as addition, likely from misapplying cos⁡(2π−θ)=−cos⁡(θ)\cos(2\pi - \theta) = -\cos(\theta). Choice D gives −3-\sqrt{3}, which incorrectly combines the fractions or uses wrong trigonometric values. Choice C is correct: −32−12-\frac{\sqrt{3}}{2} - \frac{1}{2}. Key strategy: Master the fundamental trigonometric identities for angles in different quadrants, especially sin⁡(π+θ)=−sin⁡(θ)\sin(\pi + \theta) = -\sin(\theta) and cos⁡(2π−θ)=cos⁡(θ)\cos(2\pi - \theta) = \cos(\theta). These appear frequently in precalculus problems involving angle transformations.

Question 8

Point PP on the unit circle corresponds to an angle of 7π/67\pi/6 in standard position. Using the 30-60-90 reference triangle and quadrant signs, what are the coordinates of PP?

  1. (32,12)\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)
  2. (−32,12)\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)
  3. (−32,−12)\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right) (correct answer)
  4. (−12,−32)\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 9

Consider the equation cos⁡(x)=12\cos(x) = \frac{1}{2} where xx is a special angle. If α\alpha and β\beta are the two solutions in [0,2π)[0, 2\pi) with α<β\alpha < \beta, what is the value of tan⁡(π−α)+tan⁡(π−β)\tan(\pi - \alpha) + \tan(\pi - \beta)?

  1. 233\frac{2\sqrt{3}}{3}
  2. −233-\frac{2\sqrt{3}}{3}
  3. 00 (correct answer)
  4. −23-2\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)=12\cos(x) = \frac{1}{2}. The cosine function equals 12\frac{1}{2} at two angles in [0,2π)[0, 2\pi): x=π3x = \frac{\pi}{3} and x=5π3x = \frac{5\pi}{3}. So α=π3\alpha = \frac{\pi}{3} and β=5π3\beta = \frac{5\pi}{3}. Now calculate each tangent value using the identity tan⁡(π−θ)=−tan⁡(θ)\tan(\pi - \theta) = -\tan(\theta): For tan⁡(π−α)=tan⁡(π−π3)=tan⁡(2π3)=−tan⁡(π3)=−3\tan(\pi - \alpha) = \tan(\pi - \frac{\pi}{3}) = \tan(\frac{2\pi}{3}) = -\tan(\frac{\pi}{3}) = -\sqrt{3} For tan⁡(π−β)=tan⁡(π−5π3)=tan⁡(−2π3)\tan(\pi - \beta) = \tan(\pi - \frac{5\pi}{3}) = \tan(-\frac{2\pi}{3}) Since tangent has period π\pi, we have tan⁡(−2π3)=tan⁡(−2π3+π)=tan⁡(π3)=3\tan(-\frac{2\pi}{3}) = \tan(-\frac{2\pi}{3} + \pi) = \tan(\frac{\pi}{3}) = \sqrt{3} Therefore: tan⁡(π−α)+tan⁡(π−β)=−3+3=0\tan(\pi - \alpha) + \tan(\pi - \beta) = -\sqrt{3} + \sqrt{3} = 0 The answer is C. Choice A gives 233\frac{2\sqrt{3}}{3}, which you might get if you incorrectly used tan⁡(30°)\tan(30°) instead of tan⁡(60°)\tan(60°). Choice B gives −233-\frac{2\sqrt{3}}{3}, possibly from sign errors combined with the same angle mistake. Choice D gives −23-2\sqrt{3}, which could result from forgetting the cancellation and adding −3+(−3)-\sqrt{3} + (-\sqrt{3}). Study tip: When working with expressions like tan⁡(π−θ)\tan(\pi - \theta), always use the identity tan⁡(π−θ)=−tan⁡(θ)\tan(\pi - \theta) = -\tan(\theta) and watch for symmetric pairs that cancel out.

Question 10

Compare sin⁡(π/6)\sin\left(\pi/6\right) and sin⁡(π/3)\sin\left(\pi/3\right) using the 30-60-90 triangle and their unit circle coordinates. Which statement is true?

  1. sin⁡(π/6)=sin⁡(π/3)\sin\left(\pi/6\right)=\sin\left(\pi/3\right)
  2. sin⁡(π/6)>sin⁡(π/3)\sin\left(\pi/6\right)>\sin\left(\pi/3\right)
  3. sin⁡(π/6)<sin⁡(π/3)\sin\left(\pi/6\right)<\sin\left(\pi/3\right) (correct answer)
  4. sin⁡(π/6)=cos⁡(π/6)\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 11

A point PP on the unit circle corresponds to angle α\alpha where tan⁡(α)=3\tan(\alpha) = \sqrt{3} and α\alpha is in the first quadrant. What are the coordinates of the point corresponding to angle π+α\pi + \alpha?

  1. (12,32)(\frac{1}{2}, \frac{\sqrt{3}}{2})
  2. (−12,−32)(-\frac{1}{2}, -\frac{\sqrt{3}}{2}) (correct answer)
  3. (−32,−12)(-\frac{\sqrt{3}}{2}, -\frac{1}{2})
  4. (32,12)(\frac{\sqrt{3}}{2}, \frac{1}{2})
Explanation: Since tan⁡(α)=3\tan(\alpha) = \sqrt{3} and α\alpha is in the first quadrant, we have α=π3\alpha = \frac{\pi}{3}. The coordinates of point PP are (cos⁡(π3),sin⁡(π3))=(12,32)(\cos(\frac{\pi}{3}), \sin(\frac{\pi}{3})) = (\frac{1}{2}, \frac{\sqrt{3}}{2}). For angle π+α\pi + \alpha, we use the identity that this places us in the third quadrant where both coordinates are negative: cos⁡(π+α)=−cos⁡(α)=−12\cos(\pi + \alpha) = -\cos(\alpha) = -\frac{1}{2} and sin⁡(π+α)=−sin⁡(α)=−32\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 12

An angle of 4π/34\pi/3 is drawn in standard position on the unit circle. Using the 30-60-90 reference triangle (reference angle π/3\pi/3) and quadrant signs, which coordinates represent the terminal point (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta) for θ=4π/3\theta=4\pi/3?

  1. (−1/2,−3/2)(-1/2,-\sqrt{3}/2) (correct answer)
  2. (1/2,−3/2)(1/2,-\sqrt{3}/2)
  3. (−3/2,−1/2)(-\sqrt{3}/2,-1/2)
  4. (1/2,3/2)(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 13

On the unit circle, an angle of π/6\pi/6 is drawn in standard position, forming a 30-60-90 triangle with hypotenuse (radius) 11. Using the special triangle ratios and the fact that the terminal point is (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta), which coordinates represent the terminal point for θ=π/6\theta=\pi/6?

  1. (2/2,2/2)(\sqrt{2}/2,\sqrt{2}/2)
  2. (1/2,3/2)(1/2,\sqrt{3}/2)
  3. (3/2,1/2)(\sqrt{3}/2,1/2) (correct answer)
  4. (3/2,3/2)(\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 14

A 4545-4545-9090 triangle is formed on the unit circle by drawing an angle of π/4\pi/4 in standard position. Based on the special triangle ratios scaled to hypotenuse 11, what is the exact value of sin⁡(π/4)\sin(\pi/4)?

  1. 12\tfrac{1}{2}
  2. 32\tfrac{\sqrt{3}}{2}
  3. 22\tfrac{\sqrt{2}}{2} (correct answer)
  4. 3\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 15

An angle of 7π/67\pi/6 is drawn in standard position on the unit circle. Using the 30-60-90 reference triangle (reference angle π/6\pi/6) and the fact that sin⁡θ\sin\theta is the yy-coordinate, what is the exact value of sin⁡(7π/6)\sin(7\pi/6)?

  1. 1/21/2
  2. −1/2-1/2 (correct answer)
  3. −3/2-\sqrt{3}/2
  4. 2/2\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 7π/6, we use the 30-60-90 triangle with reference angle π/6; in quadrant III, sin is negative, so sin(7π/6) = -sin(π/6) = -1/2. Choice B is correct because the reference triangle gives sin(π/6) = 1/2, and the quadrant III sign makes it negative. Choice C confuses the values for π/6 and π/3, using -√3/2 which is sin(4π/3), not sin(7π/6). 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 16

Point PP on the unit circle corresponds to an angle of π/4\pi/4 in standard position. Using the 45-45-90 triangle scaled to hypotenuse 11 and the fact that P=(cos⁡θ,sin⁡θ)P=(\cos\theta,\sin\theta), what is the exact value of sin⁡(π/4)\sin(\pi/4)?

  1. 1/21/2
  2. 2/2\sqrt{2}/2 (correct answer)
  3. 3/2\sqrt{3}/2
  4. 3/3\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 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 B is correct because the 45-45-90 triangle's leg ratio, scaled by dividing by √2, gives the opposite side (sin) as √2/2 for θ=π/4. Choice C confuses the 30-60-90 triangle with the 45-45-90 triangle, using √3/2 when the correct ratio yields √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 17

An angle of 3π/43\pi/4 is drawn in standard position on the unit circle. Using the 45-45-90 triangle and quadrant signs, which coordinates represent the terminal point RR (where R=(cos⁡θ,sin⁡θ)R=(\cos\theta,\sin\theta))?

  1. (22,22)\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)
  2. (−22,22)\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) (correct answer)
  3. (22,−22)\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)
  4. (−32,12)\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 in quadrant II, using the 45-45-90 triangle with reference angle π/4, cosine is negative and sine is positive, so coordinates are (-√2/2, √2/2). Choice B is correct because it uses the 45-45-90 ratios scaled to unit circle and applies quadrant II signs: negative x (cos) and positive y (sin). Choice C reverses the quadrant signs, using positive cos and negative sin which would be for quadrant IV. 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). 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.

Question 18

On the unit circle, an angle of π/6\pi/6 is drawn in standard position, forming a 30-60-90 triangle with hypotenuse (radius) 11. Using the 30-60-90 triangle ratios and the fact that the terminal point is (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta), what are the coordinates of the terminal point for θ=π/6\theta=\pi/6?

  1. (12,32)(\tfrac{1}{2},\tfrac{\sqrt{3}}{2})
  2. (32,12)(\tfrac{\sqrt{3}}{2},\tfrac{1}{2}) (correct answer)
  3. (22,22)(\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{2}}{2})
  4. (12,22)(\tfrac{1}{2},\tfrac{\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 (30°), we use the 30-60-90 triangle where the angle at the origin is 30°, so the x-coordinate (cos(π/6)) is the adjacent side √3/2, and the y-coordinate (sin(π/6)) is the opposite side 1/2, giving coordinates (√3/2, 1/2). Choice B is correct because it uses the 30-60-90 ratios correctly scaled, with the longer leg adjacent to the 30° angle and the shorter leg opposite. Choice A reverses sine and cosine, giving the x-coordinate value as sin when the question asks for (cos, sin) order. 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 19

An angle of 3π/43\pi/4 is drawn in standard position on the unit circle. Using the 45-45-90 reference triangle and quadrant signs, which coordinates represent the terminal point (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta) for θ=3π/4\theta=3\pi/4?

  1. (−2/2,2/2)(-\sqrt{2}/2,\sqrt{2}/2) (correct answer)
  2. (2/2,2/2)(\sqrt{2}/2,\sqrt{2}/2)
  3. (−3/2,1/2)(-\sqrt{3}/2,1/2)
  4. (2/2,−2/2)(\sqrt{2}/2,-\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 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 for θ=3π/4 in quadrant II, the reference 45-45-90 triangle gives absolute values √2/2, with cos negative and sin positive. Choice C confuses the 30-60-90 triangle with the 45-45-90 triangle, using √3/2 and 1/2 when the correct ratios are both √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

In a 45-45-90 triangle used to find unit circle values at θ=π/4\theta=\pi/4, the hypotenuse is scaled to 11 (the unit circle radius). Based on the side ratios 1:1:21:1:\sqrt{2}, what is the length of each leg in this scaled triangle?

  1. 11
  2. 3/2\sqrt{3}/2
  3. 2/2\sqrt{2}/2 (correct answer)
  4. 1/21/2
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 C is correct because scaling the 1:1:√2 ratios by 1/√2 gives each leg as √2/2. 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.