ACCUPLACER Advanced Algebra & Functions Quiz: Radian Measure And Unit Circle
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Radian Measure And Unit CircleQuestion 1 of 20

On the unit circle, the terminal side of angle α\alpha is in Quadrant IV and its y-coordinate is −12-\frac{1}{2}. What is the radian measure of α\alpha?

5π6\frac{5\pi}{6}
7π6\frac{7\pi}{6}
5π3\frac{5\pi}{3}
11π6\frac{11\pi}{6}
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Radian Measure And Unit Circle

Practice Radian Measure And Unit Circle in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Radian Measure And Unit Circle, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

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Question 1

On the unit circle, the terminal side of angle α\alpha is in Quadrant IV and its y-coordinate is −12-\frac{1}{2}. What is the radian measure of α\alpha?

  1. 5π6\frac{5\pi}{6}
  2. 7π6\frac{7\pi}{6}
  3. 5π3\frac{5\pi}{3}
  4. 11π6\frac{11\pi}{6} (correct answer)
Explanation: The y-coordinate on the unit circle represents sin⁡(α)\sin(\alpha). So, we have sin⁡(α)=−12\sin(\alpha) = -\frac{1}{2}. The reference angle for which sine is 12\frac{1}{2} is π6\frac{\pi}{6}. Since the angle α\alpha is in Quadrant IV, where sine is negative, the angle is found by 2π−reference angle2\pi - \text{reference angle}. Therefore, α=2π−π6=12π6−π6=11π6\alpha = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6}.

Question 2

Given that tan⁡(θ)=−23\tan(\theta) = -\frac{2}{3} and π2<θ<π\frac{\pi}{2} < \theta < \pi, what is the value of sin⁡(θ)\sin(\theta)?

  1. −31313-\frac{3\sqrt{13}}{13}
  2. −21313-\frac{2\sqrt{13}}{13}
  3. 21313\frac{2\sqrt{13}}{13} (correct answer)
  4. 31313\frac{3\sqrt{13}}{13}
Explanation: The angle θ\theta is in Quadrant II. For tan⁡(θ)=yx=−23\tan(\theta) = \frac{y}{x} = -\frac{2}{3}, and in Quadrant II, xx is negative and yy is positive, so we can set x=−3x = -3 and y=2y = 2. We find the radius rr using the Pythagorean theorem: r=x2+y2=(−3)2+22=9+4=13r = \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}. The value of sin⁡(θ)\sin(\theta) is yr\frac{y}{r}. Therefore, sin⁡(θ)=213\sin(\theta) = \frac{2}{\sqrt{13}}. Rationalizing the denominator gives 21313\frac{2\sqrt{13}}{13}.

Question 3

A central angle θ\theta in a circle with a radius of 6 cm subtends an arc of length 5π5\pi cm. What is the measure of θ\theta in radians?

  1. 5π12\frac{5\pi}{12}
  2. 5π6\frac{5\pi}{6} (correct answer)
  3. 65π\frac{6}{5\pi}
  4. 30π30\pi
Explanation: The relationship between arc length (s), radius (r), and central angle (θ\theta in radians) is given by the formula s=rθs = r\theta. We are given s=5πs = 5\pi cm and r=6r = 6 cm. To find θ\theta, we rearrange the formula to θ=sr\theta = \frac{s}{r}. Plugging in the values, we get θ=5π6\theta = \frac{5\pi}{6} radians.

Question 4

If sin⁡(θ)<0\sin(\theta) < 0 and cot⁡(θ)>0\cot(\theta) > 0, which of the following could be the value of θ\theta?

  1. 11π6\frac{11\pi}{6}
  2. 7π6\frac{7\pi}{6} (correct answer)
  3. 5π6\frac{5\pi}{6}
  4. π6\frac{\pi}{6}
Explanation: The condition sin⁡(θ)<0\sin(\theta) < 0 means the terminal side of θ\theta is in Quadrant III or Quadrant IV. The condition cot⁡(θ)>0\cot(\theta) > 0 (which means tan⁡(θ)>0\tan(\theta) > 0) means the terminal side of θ\theta is in Quadrant I or Quadrant III. The only quadrant that satisfies both conditions is Quadrant III. We now check which of the given angles lies in Quadrant III (between π\pi and 3π2\frac{3\pi}{2}). 7π6\frac{7\pi}{6} is in Quadrant III because π=6π6<7π6<9π6=3π2\pi = \frac{6\pi}{6} < \frac{7\pi}{6} < \frac{9\pi}{6} = \frac{3\pi}{2}.

Question 5

Which of the following is a solution to the equation 2sin⁡(θ)+2=02\sin(\theta) + \sqrt{2} = 0?

  1. π4\frac{\pi}{4}
  2. 3π4\frac{3\pi}{4}
  3. 5π4\frac{5\pi}{4} (correct answer)
  4. 11π6\frac{11\pi}{6}
Explanation: First, solve the equation for sin⁡(θ)\sin(\theta). 2sin⁡(θ)=−22\sin(\theta) = -\sqrt{2}, which gives sin⁡(θ)=−22\sin(\theta) = -\frac{\sqrt{2}}{2}. Sine is negative in Quadrants III and IV. The reference angle for sin⁡(θ)=22\sin(\theta) = \frac{\sqrt{2}}{2} is π4\frac{\pi}{4}. The solution in Quadrant III is π+π4=5π4\pi + \frac{\pi}{4} = \frac{5\pi}{4}. The solution in Quadrant IV is 2π−π4=7π42\pi - \frac{\pi}{4} = \frac{7\pi}{4}. Of the choices given, 5π4\frac{5\pi}{4} is a solution.

Question 6

For which of the following radian measures is the value of csc⁡(θ)−cot⁡(θ)\csc(\theta) - \cot(\theta) undefined?

  1. π2\frac{\pi}{2}
  2. 3π4\frac{3\pi}{4}
  3. 3π2\frac{3\pi}{2}
  4. 2π2\pi (correct answer)
Explanation: The expression can be written in terms of sine and cosine: csc⁡(θ)−cot⁡(θ)=1sin⁡(θ)−cos⁡(θ)sin⁡(θ)=1−cos⁡(θ)sin⁡(θ)\csc(\theta) - \cot(\theta) = \frac{1}{\sin(\theta)} - \frac{\cos(\theta)}{\sin(\theta)} = \frac{1 - \cos(\theta)}{\sin(\theta)}. This expression is undefined when the denominator, sin⁡(θ)\sin(\theta), is equal to zero. The sine function is zero at integer multiples of π\pi (e.g., 0,π,2π,−π,…0, \pi, 2\pi, -\pi, \dots). Of the given options, sin⁡(2π)=0\sin(2\pi) = 0, which makes the expression undefined.

Question 7

A circle is centered at the origin with a radius of 3. A point P starts at (3, 0) and moves counterclockwise along the circle for a distance of 2π2\pi units. What are the coordinates of the final position of P?

  1. (−32,332)(-\frac{3}{2}, \frac{3\sqrt{3}}{2}) (correct answer)
  2. (−12,32)(-\frac{1}{2}, \frac{\sqrt{3}}{2})
  3. (3,0)(3, 0)
  4. (−332,32)(-\frac{3\sqrt{3}}{2}, \frac{3}{2})
Explanation: The formula for arc length is s=rθs = r\theta, where ss is the arc length, rr is the radius, and θ\theta is the central angle in radians. We are given s=2πs = 2\pi and r=3r = 3. Plugging these in gives 2π=3θ2\pi = 3\theta, so θ=2π3\theta = \frac{2\pi}{3}. The coordinates of a point on a circle of radius rr are given by (rcos⁡(θ),rsin⁡(θ))(r\cos(\theta), r\sin(\theta)). Here, x=3cos⁡(2π3)=3(−12)=−32x = 3\cos(\frac{2\pi}{3}) = 3(-\frac{1}{2}) = -\frac{3}{2} and y=3sin⁡(2π3)=3(32)=332y = 3\sin(\frac{2\pi}{3}) = 3(\frac{\sqrt{3}}{2}) = \frac{3\sqrt{3}}{2}. The final coordinates are (−32,332)(-\frac{3}{2}, \frac{3\sqrt{3}}{2}).

Question 8

A security camera on a tall building is programmed to scan a horizontal arc of length 150 feet. If the camera's beam reaches a distance of 60 feet from the building, what is the angle of its scan in radians?

  1. 0.4
  2. 2.5 (correct answer)
  3. 9000
  4. 2.5π2.5\pi
Explanation: This scenario can be modeled with a circle, where the distance the beam reaches is the radius (r=60r = 60 feet) and the arc length is s=150s = 150 feet. Using the arc length formula s=rθs = r\theta, we can solve for the angle θ\theta in radians: θ=sr=15060=156=52=2.5\theta = \frac{s}{r} = \frac{150}{60} = \frac{15}{6} = \frac{5}{2} = 2.5. The angle of the scan is 2.5 radians.

Question 9

On the unit circle, if point P corresponds to angle θ=7π4\theta = \frac{7\pi}{4}, and point Q corresponds to angle ϕ=11π6\phi = \frac{11\pi}{6}, what is the measure of the acute angle between the terminal sides of these two angles?

  1. π12\frac{\pi}{12} radians (correct answer)
  2. π6\frac{\pi}{6} radians
  3. π4\frac{\pi}{4} radians
  4. 5π12\frac{5\pi}{12} radians
Explanation: The difference between the angles is 11π6−7π4=22π12−21π12=π12\frac{11\pi}{6} - \frac{7\pi}{4} = \frac{22\pi}{12} - \frac{21\pi}{12} = \frac{\pi}{12}. Since this is less than π2\frac{\pi}{2}, it is already the acute angle between the terminal sides. Choice B incorrectly uses 2π12\frac{2\pi}{12}. Choice C uses the difference 3π12\frac{3\pi}{12}. Choice D incorrectly adds rather than subtracts the angles.

Question 10

An angle α\alpha in standard position has its terminal side in Quadrant III, and sin⁡α=−35\sin \alpha = -\frac{3}{5}. If α\alpha is between π\pi and 3π2\frac{3\pi}{2}, what is the exact radian measure of α\alpha?

  1. 2π−arcsin⁡(35)2\pi - \arcsin(\frac{3}{5})
  2. π−arcsin⁡(35)\pi - \arcsin(\frac{3}{5})
  3. π+arcsin⁡(35)\pi + \arcsin(\frac{3}{5}) (correct answer)
  4. 3π2+arcsin⁡(35)\frac{3\pi}{2} + \arcsin(\frac{3}{5})
Explanation: When you encounter a trigonometry problem about angles in specific quadrants, you need to understand both the sign patterns and how to construct angles using reference angles and quadrant locations. Since α\alpha is in Quadrant III where both sine and cosine are negative, and sin⁡α=−35\sin \alpha = -\frac{3}{5}, you first need to find the reference angle. The reference angle θr\theta_r satisfies sin⁡θr=35\sin \theta_r = \frac{3}{5}, so θr=arcsin⁡(35)\theta_r = \arcsin(\frac{3}{5}). In Quadrant III, angles are formed by starting at π\pi (the negative x-axis) and adding the reference angle. Therefore, α=π+arcsin⁡(35)\alpha = \pi + \arcsin(\frac{3}{5}). You can verify this makes sense: π<π+arcsin⁡(35)<3π2\pi < \pi + \arcsin(\frac{3}{5}) < \frac{3\pi}{2} since arcsin⁡(35)<π2\arcsin(\frac{3}{5}) < \frac{\pi}{2}. Choice A (2π−arcsin⁡(35)2\pi - \arcsin(\frac{3}{5})) gives an angle in Quadrant IV, where sine is negative but the angle would be between 3π2\frac{3\pi}{2} and 2π2\pi, outside our specified range. Choice B (π−arcsin⁡(35)\pi - \arcsin(\frac{3}{5})) places the angle in Quadrant II, where sine is positive, contradicting our given condition. Choice D (3π2+arcsin⁡(35)\frac{3\pi}{2} + \arcsin(\frac{3}{5})) creates an angle in Quadrant IV that exceeds 3π2\frac{3\pi}{2}, violating the given constraint. The correct answer is C. Study tip: For Quadrant III angles, always use the formula π+reference angle\pi + \text{reference angle}. Memorize the quadrant formulas: QI uses the reference angle directly, QII uses π−reference angle\pi - \text{reference angle}, QIII uses π+reference angle\pi + \text{reference angle}, and QIV uses 2π−reference angle2\pi - \text{reference angle}.

Question 11

On the unit circle, the coordinates of the point corresponding to angle 2π3\frac{2\pi}{3} are (−12,32)(-\frac{1}{2}, \frac{\sqrt{3}}{2}). What are the coordinates of the point corresponding to angle 8π3\frac{8\pi}{3}?

  1. (−32,−12)(-\frac{\sqrt{3}}{2}, -\frac{1}{2})
  2. (−12,−32)(-\frac{1}{2}, -\frac{\sqrt{3}}{2})
  3. (−32,12)(-\frac{\sqrt{3}}{2}, \frac{1}{2})
  4. (−12,32)(-\frac{1}{2}, \frac{\sqrt{3}}{2}) (correct answer)
Explanation: When working with angles on the unit circle, the key insight is recognizing that angles differing by multiples of 2π2\pi represent the same point, since one complete revolution brings you back to the starting position. To find the coordinates for 8π3\frac{8\pi}{3}, you need to determine its coterminal angle between 00 and 2π2\pi. Since 8π3=6π+2π3=6π3+2π3=2π+2π3\frac{8\pi}{3} = \frac{6\pi + 2\pi}{3} = \frac{6\pi}{3} + \frac{2\pi}{3} = 2\pi + \frac{2\pi}{3}, the angle 8π3\frac{8\pi}{3} is coterminal with 2π3\frac{2\pi}{3}. This means they have identical coordinates: (−12,32)(-\frac{1}{2}, \frac{\sqrt{3}}{2}), making choice D correct. Let's examine why the other answers are wrong. Choice A, (−32,−12)(-\frac{\sqrt{3}}{2}, -\frac{1}{2}), represents the coordinates for angle 7π6\frac{7\pi}{6} (third quadrant). Choice B, (−12,−32)(-\frac{1}{2}, -\frac{\sqrt{3}}{2}), corresponds to angle 4π3\frac{4\pi}{3} (also third quadrant). Choice C, (−32,12)(-\frac{\sqrt{3}}{2}, \frac{1}{2}), represents angle 5π6\frac{5\pi}{6} (second quadrant). These are all different standard angles on the unit circle, but none are coterminal with 8π3\frac{8\pi}{3}. Remember this strategy: when you encounter angles greater than 2π2\pi, subtract multiples of 2π2\pi to find the equivalent angle in the standard [0,2π)[0, 2\pi) range. This will always give you the same coordinates on the unit circle.

Question 12

A wheel rotates through an angle of 5π3\frac{5\pi}{3} radians. If the wheel then rotates an additional 7π6\frac{7\pi}{6} radians in the opposite direction, what is the wheel's net angular displacement from its starting position?

  1. π2\frac{\pi}{2} radians in the original direction (correct answer)
  2. π6\frac{\pi}{6} radians in the original direction
  3. π6\frac{\pi}{6} radians in the opposite direction
  4. π2\frac{\pi}{2} radians in the opposite direction
Explanation: The wheel first rotates 5π3\frac{5\pi}{3} radians, then −7π6-\frac{7\pi}{6} radians (negative because opposite direction). Net displacement = 5π3−7π6=10π6−7π6=3π6=π2\frac{5\pi}{3} - \frac{7\pi}{6} = \frac{10\pi}{6} - \frac{7\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} radians in the original direction. Choice B results from incorrectly using 2π6\frac{2\pi}{6}. Choice C gets the correct magnitude but wrong direction. Choice D uses incorrect arithmetic throughout.

Question 13

Which angle is coterminal with −5π3-\frac{5\pi}{3} and lies in the interval [π,3π][\pi, 3\pi]?

  1. π3\frac{\pi}{3}
  2. 4π3\frac{4\pi}{3}
  3. 7π3\frac{7\pi}{3} (correct answer)
  4. 11π3\frac{11\pi}{3}
Explanation: To find a coterminal angle, add or subtract multiples of 2π2\pi. First, find a positive coterminal angle: −5π3+2π=−5π3+6π3=π3-\frac{5\pi}{3} + 2\pi = -\frac{5\pi}{3} + \frac{6\pi}{3} = \frac{\pi}{3}. This angle is not in the specified interval [π,3π][\pi, 3\pi] because π3<π\frac{\pi}{3} < \pi. Add 2π2\pi again: π3+2π=π3+6π3=7π3\frac{\pi}{3} + 2\pi = \frac{\pi}{3} + \frac{6\pi}{3} = \frac{7\pi}{3}. To check if this is in the interval, note that π=3π3\pi = \frac{3\pi}{3} and 3π=9π33\pi = \frac{9\pi}{3}. Since 3π3≤7π3≤9π3\frac{3\pi}{3} \le \frac{7\pi}{3} \le \frac{9\pi}{3}, the angle 7π3\frac{7\pi}{3} is in the desired interval.

Question 14

If the terminal side of an angle θ\theta in standard position passes through the point (−817,1517)(-\frac{8}{17}, \frac{15}{17}) on the unit circle, what is the value of cot⁡(θ)\cot(\theta)?

  1. −158-\frac{15}{8}
  2. −815-\frac{8}{15} (correct answer)
  3. 815\frac{8}{15}
  4. 158\frac{15}{8}
Explanation: For a point (x,y)(x, y) on the unit circle, cos⁡(θ)=x\cos(\theta) = x and sin⁡(θ)=y\sin(\theta) = y. The definition of cotangent is cot⁡(θ)=cos⁡(θ)sin⁡(θ)=xy\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{x}{y}. Given the point (−817,1517)(-\frac{8}{17}, \frac{15}{17}), we have x=−817x = -\frac{8}{17} and y=1517y = \frac{15}{17}. Therefore, cot⁡(θ)=−8/1715/17=−815\cot(\theta) = \frac{-8/17}{15/17} = -\frac{8}{15}.

Question 15

An angle θ\theta has a measure of 4 radians. In which quadrant does the terminal side of θ\theta lie?

  1. Quadrant I
  2. Quadrant II
  3. Quadrant III (correct answer)
  4. Quadrant IV
Explanation: To determine the quadrant, we compare the angle measure to the radian values of the axes. We use the approximation π≈3.14\pi \approx 3.14. The quadrant boundaries are: Quadrant I is (0,π/2)(0, \pi/2) or (0,1.57)(0, 1.57); Quadrant II is (π/2,π)(\pi/2, \pi) or (1.57,3.14)(1.57, 3.14); Quadrant III is (π,3π/2)(\pi, 3\pi/2) or (3.14,4.71)(3.14, 4.71); Quadrant IV is (3π/2,2π)(3\pi/2, 2\pi) or (4.71,6.28)(4.71, 6.28). Since 3.14<4<4.713.14 < 4 < 4.71, the angle of 4 radians lies in Quadrant III.

Question 16

What is the value of cos⁡(17π3)\cos(\frac{17\pi}{3})?

  1. −32-\frac{\sqrt{3}}{2}
  2. −12-\frac{1}{2}
  3. 12\frac{1}{2} (correct answer)
  4. 32\frac{\sqrt{3}}{2}
Explanation: The angle 17π3\frac{17\pi}{3} is greater than 2π2\pi, so we find a coterminal angle within [0,2π)[0, 2\pi) by subtracting multiples of 2π=6π32\pi = \frac{6\pi}{3}. 17π3−2(6π3)=17π3−12π3=5π3\frac{17\pi}{3} - 2(\frac{6\pi}{3}) = \frac{17\pi}{3} - \frac{12\pi}{3} = \frac{5\pi}{3}. Thus, cos⁡(17π3)=cos⁡(5π3)\cos(\frac{17\pi}{3}) = \cos(\frac{5\pi}{3}). The angle 5π3\frac{5\pi}{3} is in Quadrant IV, where cosine is positive. The reference angle is 2π−5π3=π32\pi - \frac{5\pi}{3} = \frac{\pi}{3}. Therefore, cos⁡(5π3)=cos⁡(π3)=12\cos(\frac{5\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}.

Question 17

A circular saw blade makes 3000 revolutions per minute. Through how many radians does a point on the outer edge of the blade travel in 0.1 seconds?

  1. 5π5\pi
  2. 10π10\pi (correct answer)
  3. 50π50\pi
  4. 100π100\pi
Explanation: First, convert the rotational speed from revolutions per minute to revolutions per second: 3000revmin×1 min60 sec=50revsec3000 \frac{\text{rev}}{\text{min}} \times \frac{1 \text{ min}}{60 \text{ sec}} = 50 \frac{\text{rev}}{\text{sec}}. Next, find the number of revolutions in 0.1 seconds: 50revsec×0.1 sec=550 \frac{\text{rev}}{\text{sec}} \times 0.1 \text{ sec} = 5 revolutions. Since one full revolution is 2π2\pi radians, the total angle in radians is 5 rev×2πradrev=10π5 \text{ rev} \times 2\pi \frac{\text{rad}}{\text{rev}} = 10\pi radians.

Question 18

What are the coordinates of the point on the unit circle that corresponds to an angle of 9π2\frac{9\pi}{2} radians?

  1. (1,0)(1, 0)
  2. (0,1)(0, 1) (correct answer)
  3. (−1,0)(-1, 0)
  4. (0,−1)(0, -1)
Explanation: To find the coordinates, we first find an angle coterminal with 9π2\frac{9\pi}{2} that is in the interval [0,2π)[0, 2\pi). We subtract multiples of 2π=4π22\pi = \frac{4\pi}{2}. 9π2−2(2π)=9π2−8π2=π2\frac{9\pi}{2} - 2(2\pi) = \frac{9\pi}{2} - \frac{8\pi}{2} = \frac{\pi}{2}. The angle π2\frac{\pi}{2} corresponds to the point at the top of the unit circle on the positive y-axis. The coordinates of this point are (0,1)(0, 1).

Question 19

Suppose cos⁡(θ)=−14\cos(\theta) = -\frac{1}{4} and π<θ<3π2\pi < \theta < \frac{3\pi}{2}. What is the value of sin⁡(θ)\sin(\theta)?

  1. −174-\frac{\sqrt{17}}{4}
  2. −154-\frac{\sqrt{15}}{4} (correct answer)
  3. 154\frac{\sqrt{15}}{4}
  4. 174\frac{\sqrt{17}}{4}
Explanation: We use the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1. Substituting the given value, we get sin⁡2(θ)+(−14)2=1\sin^2(\theta) + (-\frac{1}{4})^2 = 1, which simplifies to sin⁡2(θ)+116=1\sin^2(\theta) + \frac{1}{16} = 1. Subtracting 116\frac{1}{16} from both sides gives sin⁡2(θ)=1516\sin^2(\theta) = \frac{15}{16}. Taking the square root, sin⁡(θ)=±1516=±154\sin(\theta) = \pm\sqrt{\frac{15}{16}} = \pm\frac{\sqrt{15}}{4}. The given interval π<θ<3π2\pi < \theta < \frac{3\pi}{2} is Quadrant III, where sine is negative. Therefore, sin⁡(θ)=−154\sin(\theta) = -\frac{\sqrt{15}}{4}.

Question 20

What is the exact value of tan⁡(5π)+sin⁡(7π2)\tan(5\pi) + \sin(\frac{7\pi}{2})?

  1. -1 (correct answer)
  2. 0
  3. 1
  4. Undefined
Explanation: We evaluate each term separately. The angle 5π5\pi is coterminal with π\pi (since 5π=4π+π5\pi = 4\pi + \pi). Thus, tan⁡(5π)=tan⁡(π)=sin⁡(π)cos⁡(π)=0−1=0\tan(5\pi) = \tan(\pi) = \frac{\sin(\pi)}{\cos(\pi)} = \frac{0}{-1} = 0. The angle 7π2\frac{7\pi}{2} is coterminal with 3π2\frac{3\pi}{2} (since 7π2=4π2+3π2=2π+3π2\frac{7\pi}{2} = \frac{4\pi}{2} + \frac{3\pi}{2} = 2\pi + \frac{3\pi}{2}). Thus, sin⁡(7π2)=sin⁡(3π2)=−1\sin(\frac{7\pi}{2}) = \sin(\frac{3\pi}{2}) = -1. The sum is 0+(−1)=−10 + (-1) = -1.