Precalculus Quiz: Proving The Pythagorean Identity
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Proving The Pythagorean IdentityQuestion 1 of 20

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if cos⁡(θ)=45\cos(\theta)=\frac{4}{5} and θ\theta is in Quadrant I, what is sin⁡(θ)\sin(\theta)?

35\frac{3}{5}
−35-\frac{3}{5}
±35\pm\frac{3}{5}
45\frac{4}{5}
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Precalculus Quiz

Precalculus Quiz: Proving The Pythagorean Identity

Practice Proving The Pythagorean Identity 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 Proving The Pythagorean Identity, 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

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if cos⁡(θ)=45\cos(\theta)=\frac{4}{5} and θ\theta is in Quadrant I, what is sin⁡(θ)\sin(\theta)?

  1. 35\frac{3}{5} (correct answer)
  2. −35-\frac{3}{5}
  3. ±35\pm\frac{3}{5}
  4. 45\frac{4}{5}
Explanation: This question tests understanding of the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 for any angle θ\theta, which means that if you know one of these trig functions, you can find the other using the rearranged form sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta) = 1 - \cos^2(\theta) or cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta). Given cos⁡(θ)=45\cos(\theta) = \frac{4}{5}, we rearrange the identity to sin⁡2(θ)=1−cos⁡2(θ)=1−(45)2=1−1625=925\sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \left( \frac{4}{5} \right)^2 = 1 - \frac{16}{25} = \frac{9}{25}, so sin⁡(θ)=±925=±35\sin(\theta) = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}. The quadrant information tells us sine is positive in Quadrant I, giving sin⁡(θ)=35\sin(\theta) = \frac{3}{5}. Choice A is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the positive sign. Choice C provides both ±\pm solutions when the quadrant information specifies a unique sign. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to find the other by rearranging to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 2

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if tan⁡(θ)=43\tan(\theta)=\frac{4}{3} and θ\theta is in Quadrant III, what is cos⁡(θ)\cos(\theta)?

  1. 35\frac{3}{5}
  2. −35-\frac{3}{5} (correct answer)
  3. 45\frac{4}{5}
  4. −45-\frac{4}{5}
Explanation: This question tests understanding of the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta) = 1 - \cos^2(\theta) or cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta). Given tan⁡(θ)=43\tan(\theta) = \frac{4}{3}, we can use a right triangle where opposite = 4, adjacent = 3, hypotenuse = 5, so cos⁡(θ)=adjacenthypotenuse=35\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{5} in magnitude; since θ is in Quadrant III, where cosine is negative, cos⁡(θ)=−35\cos(\theta) = -\frac{3}{5}. Choice B is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the negative sign. Choice A uses the wrong sign for cosine, forgetting that in Quadrant III, cosine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to find the other by rearranging to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 3

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if sin⁡(θ)=−513\sin(\theta)=-\frac{5}{13} and θ\theta is in Quadrant III, what is cos⁡(θ)\cos(\theta)?

  1. 1213\frac{12}{13}
  2. −1213-\frac{12}{13} (correct answer)
  3. ±1213\pm\frac{12}{13}
  4. −513-\frac{5}{13}
Explanation: This question tests understanding of the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 for any angle θ\theta, which means that if you know one of these trig functions, you can find the other using the rearranged form sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta) = 1 - \cos^2(\theta) or cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta). Given sin⁡(θ)=−513\sin(\theta) = -\frac{5}{13}, we use the identity to find cos⁡2(θ)=1−sin⁡2(θ)=1−(−513)2=1−25169=144169\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - (-\frac{5}{13})^2 = 1 - \frac{25}{169} = \frac{144}{169}. Taking the square root gives cos⁡(θ)=±144169=±1213\cos(\theta) = \pm \sqrt{\frac{144}{169}} = \pm \frac{12}{13}, and since θ\theta is in Quadrant III, where cosine is negative, we choose cos⁡(θ)=−1213\cos(\theta) = -\frac{12}{13}. Choice B is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine sign. Choice A uses the wrong sign for cosine, forgetting that in Quadrant III, cosine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to rearrange to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 4

Given sin⁡(θ)=817\sin(\theta)=\frac{8}{17} and θ\theta is in Quadrant I, using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, what is cos⁡(θ)\cos(\theta)?

  1. 1517\frac{15}{17} (correct answer)
  2. −1517-\frac{15}{17}
  3. 817\frac{8}{17}
  4. 1715\frac{17}{15}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given sin(θ) = 8/17, we use the identity to find cos²(θ) = 1 - sin²(θ) = 1 - (8/17)² = 1 - 64/289 = 225/289. Taking the square root gives cos(θ) = ±√(225/289) = ±15/17, and since θ is in Quadrant I, where cosine is positive, we choose cos(θ) = 15/17. Choice A is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine sign. Choice B uses the wrong sign for cosine, forgetting that in Quadrant I, cosine is positive. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to rearrange to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 5

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if cos⁡(θ)=513\cos(\theta)=\frac{5}{13} and θ\theta is in Quadrant III, what is sin⁡(θ)\sin(\theta)?​​

  1. −1213-\frac{12}{13} (correct answer)
  2. 1213\frac{12}{13}
  3. ±1213\pm\frac{12}{13}
  4. −513-\frac{5}{13}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given cos(θ) = 5/13, we rearrange the identity to sin²(θ) = 1 - cos²(θ) = 1 - (5/13)² = 1 - 25/169 = 144/169, so sin(θ) = ±√(144/169) = ±12/13. The quadrant information tells us sine is negative in Quadrant III, giving sin(θ) = -12/13. Choice A is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the sign. Choice B uses the wrong sign for sine, forgetting that in Quadrant III, sine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 6

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if sin⁡(θ)=12\sin(\theta)=\frac{1}{2} and θ\theta is in Quadrant I, what is cos⁡(θ)\cos(\theta)?

  1. 32\frac{\sqrt{3}}{2} (correct answer)
  2. −32-\frac{\sqrt{3}}{2}
  3. 34\frac{3}{4}
  4. ±32\pm\frac{\sqrt{3}}{2}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given sin(θ) = 1/2, we use the identity to find cos²(θ) = 1 - sin²(θ) = 1 - (1/2)² = 1 - 1/4 = 3/4. Taking the square root gives cos(θ) = ±√(3/4) = ±√3/2, and since θ is in Quadrant I, where cosine is positive, we choose cos(θ) = √3/2. Choice A is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the positive sign. Choice B uses the wrong sign for cosine, forgetting that in Quadrant I, cosine is positive. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to find the other by rearranging to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 7

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II, what is cos⁡(θ)\cos(\theta)?

  1. 45\frac{4}{5}
  2. −45-\frac{4}{5} (correct answer)
  3. −35-\frac{3}{5}
  4. ±45\pm\frac{4}{5}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given sin(θ) = 3/5, we use the identity to find cos²(θ) = 1 - sin²(θ) = 1 - (3/5)² = 1 - 9/25 = 16/25. Taking the square root gives cos(θ) = ±√(16/25) = ±4/5, and since θ is in Quadrant II, where cosine is negative, we choose cos(θ) = -4/5. Choice B is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine sign. Choice A uses the wrong sign for cosine, forgetting that in Quadrant II, cosine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to rearrange to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 8

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if cos⁡(θ)=1213\cos(\theta)=\frac{12}{13} and θ\theta is in Quadrant IV, what is sin⁡(θ)\sin(\theta)?

  1. 513\frac{5}{13}
  2. −513-\frac{5}{13} (correct answer)
  3. −1213-\frac{12}{13}
  4. ±513\pm\frac{5}{13}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given cos(θ) = 12/13, we rearrange the identity to sin²(θ) = 1 - cos²(θ) = 1 - (12/13)² = 1 - 144/169 = 25/169, so sin(θ) = ±√(25/169) = ±5/13. The quadrant information tells us sine is negative, giving sin(θ) = -5/13. Choice B is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine sign. Choice A uses the wrong sign for sine, forgetting that in Quadrant IV, sine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to rearrange to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 9

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if sin⁡(θ)=817\sin(\theta)=\frac{8}{17} and θ\theta is in Quadrant I, what is cos⁡(θ)\cos(\theta)?

  1. 1517\frac{15}{17} (correct answer)
  2. −1517-\frac{15}{17}
  3. 817\frac{8}{17}
  4. 917\frac{9}{17}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given sin(θ) = 8/17, we use the identity to find cos²(θ) = 1 - sin²(θ) = 1 - (8/17)² = 1 - 64/289 = 225/289. Taking the square root gives cos(θ) = ±√(225/289) = ±15/17, and since θ is in Quadrant I, where cosine is positive, we choose cos(θ) = 15/17. Choice A is correct because it properly applies the identity with correct arithmetic and uses the correct positive sign for cosine in Quadrant I. Choice B uses the wrong sign for cosine, forgetting that in Quadrant I, cosine is positive. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 10

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if cos⁡(θ)=23\cos(\theta)=\frac{2}{3} and θ\theta is acute, what is sin⁡(θ)\sin(\theta)?

  1. 53\frac{\sqrt{5}}{3} (correct answer)
  2. −53-\frac{\sqrt{5}}{3}
  3. 59\frac{5}{9}
  4. ±53\pm\frac{\sqrt{5}}{3}
Explanation: This question tests understanding of the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 for any angle θ\theta, which means that if you know one of these trig functions, you can find the other using the rearranged form sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta) = 1 - \cos^2(\theta) or cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta). Given cos⁡(θ)=23\cos(\theta) = \frac{2}{3}, we rearrange the identity to sin⁡2(θ)=1−cos⁡2(θ)=1−(23)2=1−49=59\sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9}, so sin⁡(θ)=±59=±53\sin(\theta) = \pm \sqrt{\frac{5}{9}} = \pm \frac{\sqrt{5}}{3}. The quadrant information tells us sine is positive, giving sin⁡(θ)=53\sin(\theta) = \frac{\sqrt{5}}{3}. Choice A is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine sign. Choice B uses the wrong sign for sine, forgetting that for an acute angle, sine is positive. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to rearrange to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 11

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II, what is cos⁡(θ)\cos(\theta)?​

  1. 45\frac{4}{5}
  2. −45-\frac{4}{5} (correct answer)
  3. −35-\frac{3}{5}
  4. ±45\pm\frac{4}{5}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given sin(θ) = 3/5, we use the identity to find cos²(θ) = 1 - sin²(θ) = 1 - (3/5)² = 1 - 9/25 = 16/25. Taking the square root gives cos(θ) = ±√(16/25) = ±4/5, and since θ is in Quadrant II, where cosine is negative, we choose cos(θ) = -4/5. Choice B is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the negative sign. Choice A uses the wrong sign for cosine, forgetting that in Quadrant II, cosine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to find the other by rearranging to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 12

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if cos⁡(θ)=817\cos(\theta)=\frac{8}{17} and θ\theta is in Quadrant IV, what is sin⁡(θ)\sin(\theta)?

  1. ±1517\pm\frac{15}{17}
  2. 1517\frac{15}{17}
  3. −1517-\frac{15}{17} (correct answer)
  4. −817-\frac{8}{17}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given cos(θ) = 8/17, we rearrange the identity to sin²(θ) = 1 - cos²(θ) = 1 - (8/17)² = 1 - 64/289 = 225/289, so sin(θ) = ±√(225/289) = ±15/17. The quadrant information tells us sine is negative in Quadrant IV, giving sin(θ) = -15/17. Choice C is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the negative sign. Choice B uses the wrong sign for sine, forgetting that in Quadrant IV, sine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to find the other by rearranging to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 13

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if sin⁡(θ)=−513\sin(\theta)=-\frac{5}{13} and θ\theta is in Quadrant III, what is cos⁡(θ)\cos(\theta)?​

  1. 1213\frac{12}{13}
  2. −1213-\frac{12}{13} (correct answer)
  3. −513-\frac{5}{13}
  4. ±1213\pm\frac{12}{13}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given sin(θ) = -5/13, we use the identity to find cos²(θ) = 1 - sin²(θ) = 1 - (-5/13)² = 1 - 25/169 = 144/169. Taking the square root gives cos(θ) = ±√(144/169) = ±12/13, and since θ is in Quadrant III, where cosine is negative, we choose cos(θ) = -12/13. Choice B is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the negative sign. Choice C forgets to take the square root after finding cos²(θ) = 144/169, giving a wrong value. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to find the other by rearranging to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 14

Given that sin⁡(θ)=817\sin(\theta)=\frac{8}{17} and θ\theta is in Quadrant I, use the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1 to find cos⁡(θ)\cos(\theta).

  1. 1517\frac{15}{17} (correct answer)
  2. −1517-\frac{15}{17}
  3. 817\frac{8}{17}
  4. ±1517\pm\frac{15}{17}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given sin(θ) = 8/17, we use the identity to find cos²(θ) = 1 - sin²(θ) = 1 - (8/17)² = 1 - 64/289 = 225/289. Taking the square root gives cos(θ) = ±√(225/289) = ±15/17, and since θ is in Quadrant I, where cosine is positive, we choose cos(θ) = 15/17. Choice A is correct because it properly applies the identity with correct arithmetic and uses the correct positive sign for cosine in Quadrant I. Choice B uses the wrong sign for cosine, forgetting that in Quadrant I, cosine is positive. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 15

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if cos⁡(θ)=1213\cos(\theta)=\frac{12}{13} and θ\theta is in Quadrant IV, what is sin⁡(θ)\sin(\theta)?​

  1. 513\frac{5}{13}
  2. −513-\frac{5}{13} (correct answer)
  3. −1213-\frac{12}{13}
  4. ±513\pm\frac{5}{13}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given cos(θ) = 12/13, we rearrange the identity to sin²(θ) = 1 - cos²(θ) = 1 - (12/13)² = 1 - 144/169 = 25/169, so sin(θ) = ±√(25/169) = ±5/13. The quadrant information tells us sine is negative in Quadrant IV, giving sin(θ) = -5/13. Choice B is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the negative sign. Choice A uses the wrong sign for sine, forgetting that in Quadrant IV, sine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to find the other by rearranging to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.

Question 16

A student incorrectly concludes that since sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, it follows that sin⁡(θ)+cos⁡(θ)=1\sin(\theta) + \cos(\theta) = 1 for all θ\theta. Which counterexample best demonstrates the flaw in this reasoning?

  1. θ=π4\theta = \frac{\pi}{4}, where sin⁡(π4)+cos⁡(π4)=22+22=2≠1\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \neq 1 (correct answer)
  2. θ=π6\theta = \frac{\pi}{6}, where sin⁡(π6)+cos⁡(π6)=12+32=1+32≠1\sin(\frac{\pi}{6}) + \cos(\frac{\pi}{6}) = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \neq 1
  3. θ=π3\theta = \frac{\pi}{3}, where sin⁡(π3)+cos⁡(π3)=32+12=1+32≠1\sin(\frac{\pi}{3}) + \cos(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 + \sqrt{3}}{2} \neq 1
  4. θ=π\theta = \pi, where sin⁡(π)+cos⁡(π)=0+(−1)=−1≠1\sin(\pi) + \cos(\pi) = 0 + (-1) = -1 \neq 1
Explanation: The student's error is assuming that a2+b2=a+b\sqrt{a^2 + b^2} = a + b, which is false. To find the best counterexample, we want a case where the difference between sin⁡(θ)+cos⁡(θ)\sin(\theta) + \cos(\theta) and 1 is most obvious. For θ=π4\theta = \frac{\pi}{4}: sin⁡(π4)+cos⁡(π4)=22+22=2≈1.414\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414. This gives the clearest counterexample because 2\sqrt{2} is a well-known irrational number distinctly different from 1. Choices B and C give 1+32≈1.366\frac{1 + \sqrt{3}}{2} \approx 1.366, which is also not 1, but the calculation is more complex. Choice D gives -1, which while clearly ≠ 1, uses a less intuitive angle for demonstrating the fundamental algebraic error.

Question 17

Which of the following correctly demonstrates why sin⁡4(x)+cos⁡4(x)+2sin⁡2(x)cos⁡2(x)=1\sin^4(x) + \cos^4(x) + 2\sin^2(x)\cos^2(x) = 1 using the Pythagorean identity?

  1. The expression equals (sin⁡2(x)+cos⁡2(x))2(\sin^2(x) + \cos^2(x))^2, and since sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1, the result is 12=11^2 = 1 (correct answer)
  2. The expression can be factored as sin⁡2(x)(sin⁡2(x)+2cos⁡2(x))+cos⁡4(x)=1\sin^2(x)(\sin^2(x) + 2\cos^2(x)) + \cos^4(x) = 1 using the identity
  3. Since sin⁡4(x)=(sin⁡2(x))2\sin^4(x) = (\sin^2(x))^2 and cos⁡4(x)=(cos⁡2(x))2\cos^4(x) = (\cos^2(x))^2, we apply the identity to each term separately
  4. The expression simplifies by substituting cos⁡2(x)=1−sin⁡2(x)\cos^2(x) = 1 - \sin^2(x) into each term and then expanding completely
Explanation: The key insight is recognizing that sin⁡4(x)+cos⁡4(x)+2sin⁡2(x)cos⁡2(x)\sin^4(x) + \cos^4(x) + 2\sin^2(x)\cos^2(x) is a perfect square: (sin⁡2(x)+cos⁡2(x))2=(sin⁡2(x))2+2sin⁡2(x)cos⁡2(x)+(cos⁡2(x))2=sin⁡4(x)+2sin⁡2(x)cos⁡2(x)+cos⁡4(x)(\sin^2(x) + \cos^2(x))^2 = (\sin^2(x))^2 + 2\sin^2(x)\cos^2(x) + (\cos^2(x))^2 = \sin^4(x) + 2\sin^2(x)\cos^2(x) + \cos^4(x). By the Pythagorean identity, sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1, so the expression equals 12=11^2 = 1. Choice B shows an incorrect factorization. Choice C incorrectly suggests applying the identity to individual terms. Choice D suggests a more complicated substitution method that, while possible, doesn't reveal the elegant structure.

Question 18

Using a right triangle derivation: in a right triangle with legs aa and bb and hypotenuse cc, a2+b2=c2a^2+b^2=c^2. Which statement correctly proves the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1 for an acute angle θ\theta?​

  1. Divide a2+b2=c2a^2+b^2=c^2 by cc to get (ac)2+(bc)2=c\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=c.
  2. Divide a2+b2=c2a^2+b^2=c^2 by c2c^2 to get (ac)2+(bc)2=1\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1, then use sin⁡(θ)=ac\sin(\theta)=\frac{a}{c} and cos⁡(θ)=bc\cos(\theta)=\frac{b}{c}. (correct answer)
  3. Use a2+b2=c2a^2+b^2=c^2 and substitute sin⁡(θ)=ca\sin(\theta)=\frac{c}{a} and cos⁡(θ)=cb\cos(\theta)=\frac{c}{b}.
  4. Add sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) to get sin⁡(θ)+cos⁡(θ)=1\sin(\theta)+\cos(\theta)=1.
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how it derives from right triangles. The Pythagorean identity comes from the Pythagorean theorem in a right triangle: starting with a² + b² = c² and dividing both sides by c², we get (a/c)² + (b/c)² = 1, which becomes sin²(θ) + cos²(θ) = 1 since sin(θ) = a/c and cos(θ) = b/c. In a right triangle with opposite side a, adjacent side b, and hypotenuse c, the Pythagorean theorem gives a² + b² = c². Dividing every term by c² yields (a/c)² + (b/c)² = 1. Since sin(θ) = a/c (opposite/hypotenuse) and cos(θ) = b/c (adjacent/hypotenuse), this becomes sin²(θ) + cos²(θ) = 1. Choice B is correct because it accurately states the identity. Choice A incorrectly derives the identity, failing to divide by c² in the Pythagorean theorem, leaving a² + b² = c² instead of the ratio form. The Pythagorean identity sin²(θ) + cos²(θ) = 1 is one of the most fundamental trig identities: it works for any angle, derives directly from either the unit circle or the Pythagorean theorem, and is essential for solving countless trig problems. Don't confuse the Pythagorean identity with similar-looking statements: sin(θ) + cos(θ) does NOT equal 1 (missing squares), and sin²(θ) + cos²(θ) always equals 1, not 0 or any other number.

Question 19

Based on the unit circle, a point P(x,y)P(x,y) lies on the circle x2+y2=1x^2+y^2=1 and corresponds to an angle θ\theta in standard position where x=cos⁡(θ)x=\cos(\theta) and y=sin⁡(θ)y=\sin(\theta). How is the identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1 derived from the unit circle?

  1. Substitute x=sin⁡(θ)x=\sin(\theta) and y=cos⁡(θ)y=\cos(\theta) into x2+y2=2x^2+y^2=2.
  2. Substitute x=cos⁡(θ)x=\cos(\theta) and y=sin⁡(θ)y=\sin(\theta) into x2+y2=1x^2+y^2=1 to get cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta)+\sin^2(\theta)=1. (correct answer)
  3. Use x+y=1x+y=1 and set x=cos⁡(θ)x=\cos(\theta), y=sin⁡(θ)y=\sin(\theta) to get sin⁡(θ)+cos⁡(θ)=1\sin(\theta)+\cos(\theta)=1.
  4. Differentiate x2+y2=1x^2+y^2=1 to obtain sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1.
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how it derives from the unit circle. The Pythagorean identity derives from the unit circle equation x² + y² = 1: since a point at angle θ on the unit circle has coordinates (cos(θ), sin(θ)), substituting gives cos²(θ) + sin²(θ) = 1. On the unit circle with radius 1, any point satisfies x² + y² = 1. Since the coordinates at angle θ are (cos(θ), sin(θ)), substituting x = cos(θ) and y = sin(θ) into the circle equation gives cos²(θ) + sin²(θ) = 1. Choice B is correct because it correctly describes the unit circle derivation. Choice A uses the wrong equation x² + y² = 2 instead of 1. The Pythagorean identity sin²(θ) + cos²(θ) = 1 is one of the most fundamental trig identities: it works for any angle, derives directly from either the unit circle or the Pythagorean theorem, and is essential for solving countless trig problems. Don't confuse the Pythagorean identity with similar-looking statements: sin(θ) + cos(θ) does NOT equal 1 (missing squares), and sin²(θ) + cos²(θ) always equals 1, not 0 or any other number.

Question 20

Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1, if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II, what is cos⁡(θ)\cos(\theta)?​​

  1. 45\frac{4}{5}
  2. −45-\frac{4}{5} (correct answer)
  3. −35-\frac{3}{5}
  4. ±45\pm\frac{4}{5}
Explanation: This question tests understanding of the Pythagorean identity sin²(θ) + cos²(θ) = 1 and how to use it to find missing trig values. The Pythagorean identity states that sin²(θ) + cos²(θ) = 1 for any angle θ, which means that if you know one of these trig functions, you can find the other using the rearranged form sin²(θ) = 1 - cos²(θ) or cos²(θ) = 1 - sin²(θ). Given sin(θ) = 3/5, we use the identity to find cos²(θ) = 1 - sin²(θ) = 1 - (3/5)² = 1 - 9/25 = 16/25. Taking the square root gives cos(θ) = ±√(16/25) = ±4/5, and since θ is in Quadrant II, where cosine is negative, we choose cos(θ) = -4/5. Choice B is correct because it properly applies the identity with correct arithmetic and uses the right quadrant to determine the sign. Choice A uses the wrong sign for cosine, forgetting that in Quadrant II, cosine is negative. Key to using the Pythagorean identity: when given one trig value (sin or cos), use the identity to isolate the unknown, then take the square root and determine the correct sign based on which quadrant the angle is in. Remember the quadrant sign rules: Quadrant I (both positive), Quadrant II (sin positive, cos negative), Quadrant III (both negative), Quadrant IV (sin negative, cos positive)—use these to choose the correct sign after taking the square root.