Special Triangles, Unit Circle in Trigonometry - Precalculus

An angle of $5\pi/6$ is drawn in standard position on the unit circle. Using the 30-60-90 reference triangle and quadrant signs, what is the exact value of $\sin\left(5\pi/6\right)$ ?
Answer: $\frac{1}{2}$ 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 5π/6, we find the reference angle by subtracting from π: π - 5π/6 = π/6, so we use a 30-60-90 triangle with reference angle π/6 in the second quadrant where sine is positive and cosine is negative. Choice C is correct because sin(5π/6) = sin(π/6) = 1/2, since 5π/6 is in the second quadrant where sine values are positive and the reference angle π/6 has sine value 1/2. Choice B incorrectly applies a negative sign to the sine value, when sine is positive in the second quadrant where 5π/6 is located. 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).
If $\cos(\beta) = \frac{\sqrt{2}}{2}$ where $\beta$ is a special angle in $[0, \frac{\pi}{2}]$ , what is the exact value of $\tan(\pi - \beta) \cdot \sin(2\pi - \beta)$ ?
Answer: $\frac{\sqrt{2}}{2}$ 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.
An angle of $7\pi/6$ is drawn in standard position on the unit circle. Using the 30-60-90 reference triangle (reference angle $\pi/6$ ) and the fact that $\sin\theta$ is the $y$ -coordinate, what is the exact value of $\sin(7\pi/6)$ ?
Answer: $-1/2$ 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).
A 45-45-90 triangle is used to locate the point on the unit circle at angle $\pi/4$ . Based on the special triangle ratios, what are the side ratios (leg : leg : hypotenuse) of a 45-45-90 triangle?
Answer: $1:1:\sqrt{2}$ 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 $1 : 1 : \sqrt{2}$ (leg : leg : hypotenuse), and when scaled to fit the unit circle (hypotenuse = 1), each leg has length $\sqrt{2}/2$ , giving sin( $\pi/4$ ) = cos( $\pi/4$ ) = $\sqrt{2}/2$ . In a 45-45-90 triangle, the legs are equal, and the hypotenuse is $\sqrt{2}$ times a leg, so the ratios are $1:1:\sqrt{2}$ . Choice A is correct because it directly states the standard ratios for a 45-45-90 triangle: equal legs and hypotenuse $\sqrt{2}$ times a leg. Choice B confuses the 30-60-90 triangle with the 45-45-90 triangle, using $1:\sqrt{3}:2$ when the correct ratio is $1:1:\sqrt{2}$ . For 45-45-90 triangles, the key insight is that the two legs are equal, so sin( $\pi/4$ ) = cos( $\pi/4$ ) = $\sqrt{2}/2$ , 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 ( $1:1:\sqrt{2}$ for 45-45-90, and $1:\sqrt{3}:2$ 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.
On the unit circle, $\theta=\pi/6$ forms a 30-60-90 triangle. Using special triangle ratios (scaled to radius $1$ ), what is the exact value of $\tan(\pi/6)$ ?
Answer: $\sqrt{3}/3$ 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(θ) = sin(θ)/cos(θ), with opposite side 1/2 and adjacent side √3/2, giving tan(π/6) = (1/2)/(√3/2) = 1/√3 = √3/3. Choice B is correct because the scaled ratios yield sin(π/6) = 1/2 and cos(π/6) = √3/2, so tan = (1/2)/(√3/2) = √3/3. Choice A swaps the values for 30° and 60°, using tan(π/3) = √3 when actually tan(π/6) = √3/3. 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).
Point $P$ on the unit circle corresponds to an angle of $\pi/4$ in standard position. Using the 45-45-90 triangle scaled to hypotenuse $1$ and the fact that $P=(\cos\theta,\sin\theta)$ , what is the exact value of $\sin(\pi/4)$ ?
Answer: $\sqrt{2}/2$ 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.
Consider the equation $\cos(x) = \frac{1}{2}$ where $x$ is a special angle. If $\alpha$ and $\beta$ are the two solutions in $[0, 2\pi)$ with $\alpha < \beta$ , what is the value of $\tan(\pi - \alpha) + \tan(\pi - \beta)$ ?
Answer: $0$ 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.
Compare $\sin\left(\pi/6\right)$ and $\sin\left(\pi/3\right)$ using the 30-60-90 triangle and their unit circle coordinates. Which statement is true?
Answer: $\sin\left(\pi/6\right)<\sin\left(\pi/3\right)$ 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).
Using the 30-60-90 triangle scaled to the unit circle (hypotenuse 1), what is the exact value of $\tan(\pi/6)$ ?
Answer: $\frac{\sqrt{3}}{3}$ 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).
An angle of $4\pi/3$ is drawn in standard position on the unit circle. Using the 30-60-90 reference triangle (reference angle $\pi/3$ ) and quadrant signs, which coordinates represent the terminal point $(\cos\theta,\sin\theta)$ for $\theta=4\pi/3$ ?
Answer: $(-1/2,-\sqrt{3}/2)$ 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).
A point $P$ on the unit circle corresponds to angle $\alpha$ where $\tan(\alpha) = \sqrt{3}$ and $\alpha$ is in the first quadrant. What are the coordinates of the point corresponding to angle $\pi + \alpha$ ?
Answer: $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$ 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.
On the unit circle, an angle of $\pi/3$ is drawn in standard position, creating a 30-60-90 triangle with radius $1$ . Based on the special triangle ratios and $\cos\theta$ being the $x$ -coordinate, what is the exact value of $\cos(\pi/3)$ ?
Answer: $1/2$ 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).
On the unit circle, the point corresponding to $\theta=\pi/4$ comes from a 45-45-90 triangle. Which coordinates represent the terminal point $(\cos\theta,\sin\theta)$ for $\theta=\pi/4$ ?
Answer: $(\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{2}}{2})$ 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.
Given the angle $\pi/3$ on the unit circle and using the 30-60-90 triangle (scaled to hypotenuse $1$ ), what is the exact value of $\sin\left(\pi/3\right)$ ?
Answer: $\frac{\sqrt{3}}{2}$ 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 sin(π/3) equals the opposite side to the 60° angle divided by the hypotenuse, which is the long leg (√3/2) in the 30-60-90 triangle scaled to the unit circle. Choice A swaps the values for 30° and 60°, using sin(π/6) = 1/2 when actually sin(π/6) = 1/2 and sin(π/3) = √3/2. 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).
If $P$ is a point on the unit circle corresponding to the angle $\frac{\pi}{6}$ , and $Q$ is the point corresponding to $\pi - \frac{\pi}{6}$ , what is the distance between points $P$ and $Q$ ?
Answer: $\sqrt{3}$ 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.
A point $P$ on the unit circle corresponds to an angle $\theta$ in standard position and has coordinates $P=(\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{2}}{2})$ . Based on the special right triangle connection to the unit circle, what is the exact value of $\theta$ ?
Answer: $\tfrac{\pi}{4}$ 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 coordinates (√2/2, √2/2), we use the 45-45-90 triangle where both legs are √2/2, corresponding to θ = π/4 in the first quadrant. Choice B is correct because the equal coordinates √2/2 match the scaled legs of the 45-45-90 triangle, directly identifying θ = π/4. Choice A confuses the 30-60-90 triangle with the 45-45-90 triangle, using π/6 when the coordinates would be (√3/2, 1/2) instead of equal values. 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.
On the unit circle, an angle of $\pi/3$ is drawn in standard position, creating a 30-60-90 triangle with hypotenuse 1. Based on special triangle ratios, what is the exact value of $\sin(\pi/3)$ ?
Answer: $\tfrac{\sqrt{3}}{2}$ 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 fit 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 (60°), sine is the opposite side over hypotenuse, which is the long leg √3/2 relative to 1. 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. 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).
A 30-60-90 triangle is used to determine unit circle values at $\theta=\pi/3$ . Since the terminal point is $(\cos\theta,\sin\theta)$ , what is the exact value of $\sin(\pi/3)$ ?
Answer: $\sqrt{3}/2$ 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.
On the unit circle, an angle of $\pi/6$ is drawn in standard position, forming a 30-60-90 triangle with hypotenuse $1$ . Using the special triangle ratios and the fact that the terminal point is $(\cos\theta,\sin\theta)$ , which coordinates represent the terminal point for $\theta=\pi/6$ ?
Answer: $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ 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 side adjacent to the 30° angle is the long leg (√3/2) and the side opposite is the short leg (1/2), giving coordinates (cos(π/6), sin(π/6)) = (√3/2, 1/2). Choice B is correct because it properly identifies that cos(π/6) = √3/2 (the x-coordinate) and sin(π/6) = 1/2 (the y-coordinate) based on the 30-60-90 triangle ratios. Choice C reverses sine and cosine, incorrectly placing 1/2 as the x-coordinate when it should be the y-coordinate for a 30° angle. 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.
Point $P$ on the unit circle corresponds to an angle of $7\pi/6$ in standard position. Using the 30-60-90 reference triangle and quadrant signs, what are the coordinates of $P$ ?
Answer: $\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$ 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.

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