Double-Angle Applications with Pythagorean Identity Quiz

10th Grade

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Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.

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Quizzes Created: 8156 | Total Attempts: 9,588,805

| Attempts: 13 | Questions: 20 | Updated: Jan 22, 2026

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Question 1 / 21

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1) If sinθ = 3/5 and θ is in Quadrant II, find cosθ.

4/5

−4/5

−3/5

3/5

Given: sinθ = 3/5 (QII). Goal: cosθ.

Step 1: cosθ = −√(1 − (3/5)²) = −√(16/25) = −4/5.

So, the final answer is −4/5.

Explanation

Given: sinθ = 3/5 (QII). Goal: cosθ.

Step 1: cosθ = −√(1 − (3/5)²) = −√(16/25) = −4/5.

So, the final answer is −4/5.

Submit

About This Quiz

Double-angle Applications With Pythagorean Identity Quiz - Quiz

In this quiz, you’ll connect double-angle formulas to geometry, triangles, and the unit circle. You’ll apply sin(2θ) and cos(2θ) to find missing trig values, model angles on the coordinate plane, and explore how these identities describe real-world motion and rotation. By combining your knowledge of Pythagorean and double-angle relationships, you’ll... see morestrengthen your understanding of how trigonometric formulas work together in problem solving.
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2) If cosθ = −7/25 and θ is in Quadrant III, find sinθ.

24/25

7/25

−7/25

−24/25

Given: cosθ = −7/25 (QIII). Goal: sinθ.

Step 1: sinθ = −√(1 − 49/625) = −√(576/625) = −24/25.

So, the final answer is −24/25.

Explanation

Given: cosθ = −7/25 (QIII). Goal: sinθ.

Step 1: sinθ = −√(1 − 49/625) = −√(576/625) = −24/25.

So, the final answer is −24/25.

Submit

3) Given sinθ = −12/13 and θ is in Quadrant IV, compute cosθ.

5/13

−5/13

12/13

−12/13

Given: sinθ negative, QIV ⇒ cosθ > 0. Goal: cosθ.

Step 1: cosθ = √(1 − 144/169) = 5/13.

So, the final answer is 5/13.

Explanation

Given: sinθ negative, QIV ⇒ cosθ > 0. Goal: cosθ.

Step 1: cosθ = √(1 − 144/169) = 5/13.

So, the final answer is 5/13.

Submit

4) If cosθ = √3/2 and θ is in Quadrant I, find sinθ.

1/2

−1/2

√3/2

−√3/2

Given: cosθ = √3/2 (QI). Goal: sinθ.

Step 1: sinθ = √(1 − 3/4) = 1/2.

So, the final answer is 1/2.

Explanation

Given: cosθ = √3/2 (QI). Goal: sinθ.

Step 1: sinθ = √(1 − 3/4) = 1/2.

So, the final answer is 1/2.

Submit

5) If sinθ = 5/13, find cosθ assuming θ is in Quadrant I.

5/13

−12/13

12/13

−5/13

Given: sinθ = 5/13 (QI). Goal: cosθ.

Step 1: cosθ = √(1 − 25/169) = 12/13.

So, the final answer is 12/13.

Explanation

Given: sinθ = 5/13 (QI). Goal: cosθ.

Step 1: cosθ = √(1 − 25/169) = 12/13.

So, the final answer is 12/13.

Submit

6) A point P on the unit circle has y-coordinate −0.6. Find the possible x-coordinate(s).

±0.8

±0.6

±0.4

±0.3

Given: y = sinθ = −0.6. Goal: x = cosθ.

Step 1: x = ±√(1 − 0.36) = ±√0.64 = ±0.8.

So, the final answer is ±0.8.

Explanation

Given: y = sinθ = −0.6. Goal: x = cosθ.

Step 1: x = ±√(1 − 0.36) = ±√0.64 = ±0.8.

So, the final answer is ±0.8.

Submit

7) If sinθ = 0.8 and θ is in Quadrant II, compute cosθ.

0.6

−0.6

0.8

−0.8

Given: sinθ = 0.8 (QII ⇒ cosθ < 0). Goal: cosθ.

Step 1: cosθ = −√(1 − 0.64) = −0.6.

So, the final answer is −0.6.

Explanation

Given: sinθ = 0.8 (QII ⇒ cosθ < 0). Goal: cosθ.

Step 1: cosθ = −√(1 − 0.64) = −0.6.

So, the final answer is −0.6.

Submit

8) For cosθ = −√5/3 with θ in Quadrant II, determine sinθ.

√4/3

2/3

−2/3

√2/3

Given: cosθ negative, QII ⇒ sinθ > 0. Goal: sinθ.

Step 1: sinθ = √(1 − 5/9) = √(4/9) = 2/3.

So, the final answer is 2/3.

Explanation

Given: cosθ negative, QII ⇒ sinθ > 0. Goal: sinθ.

Step 1: sinθ = √(1 − 5/9) = √(4/9) = 2/3.

So, the final answer is 2/3.

Submit

9) If x = −9/10 and θ is in Quadrant II, find y.

1/10

3/10

√19/10

3/√10

Given: x = cosθ = −9/10 (QII ⇒ y > 0). Goal: y.

Step 1: y = √(1 − 81/100) = √(19/100) = √19/10.

So, the final answer is √19/10.

Explanation

Given: x = cosθ = −9/10 (QII ⇒ y > 0). Goal: y.

Step 1: y = √(1 − 81/100) = √(19/100) = √19/10.

So, the final answer is √19/10.

Submit

10) If cosθ = 4/5 and sinθ > 0, find sin(2θ).

24/25

−24/25

7/25

32/25

Given: cosθ = 4/5, sinθ > 0 ⇒ sinθ = 3/5. Goal: sin(2θ).

Step 1: sin(2θ) = 2·(3/5)·(4/5) = 24/25.

So, the final answer is 24/25.

Explanation

Given: cosθ = 4/5, sinθ > 0 ⇒ sinθ = 3/5. Goal: sin(2θ).

Step 1: sin(2θ) = 2·(3/5)·(4/5) = 24/25.

So, the final answer is 24/25.

Submit

11) Given sinθ = −3/5, determine cos(2θ).

−7/25

7/25

1/25

−1/25

Given: sinθ = −3/5. Goal: cos(2θ).

Step 1: cos(2θ) = 1 − 2·(9/25) = 1 − 18/25 = 7/25.

So, the final answer is 7/25.

Explanation

Given: sinθ = −3/5. Goal: cos(2θ).

Step 1: cos(2θ) = 1 − 2·(9/25) = 1 − 18/25 = 7/25.

So, the final answer is 7/25.

Submit

12) If cosθ = −5/13 and θ is in Quadrant II, evaluate sin(2θ).

120/169

5/169

−5/169

−120/169

Given: cosθ = −5/13 (QII ⇒ sinθ > 0). Goal: sin(2θ).

Step 1: sinθ = 12/13.

Step 2: sin(2θ) = 2·(12/13)·(−5/13) = −120/169.

So, the final answer is −120/169.

Explanation

Given: cosθ = −5/13 (QII ⇒ sinθ > 0). Goal: sin(2θ).

Step 1: sinθ = 12/13.

Step 2: sin(2θ) = 2·(12/13)·(−5/13) = −120/169.

So, the final answer is −120/169.

Submit

13) Right triangle with hypotenuse 13 and adjacent side to θ is 5. Find sinθ, then cos(2θ).

5/13 and 24/25

12/13 and −7/25

12/13 and −119/169

5/13 and −24/25

Given: adjacent = 5, hypotenuse = 13. Goal: sinθ, cos(2θ).

Step 1: opposite = √(13² − 5²) = 12 ⇒ sinθ = 12/13.

Step 2: cos(2θ) = 1 − 2 sin²θ = 1 − 2·(144/169) = −119/169.

So, the final answer is 12/13 and −119/169.

Explanation

Submit

14) On the unit circle, if sinθ = t (t ∈ [−1, 1]), which expression for cosθ is always valid?

1 − t

±√(1 − t²)

T²

√(1 + t²)

Given: sin²θ + cos²θ = 1. Goal: cosθ in terms of t.

Step 1: cosθ = ±√(1 − t²), sign by quadrant.

So, the final answer is ±√(1 − t²).

Explanation

Given: sin²θ + cos²θ = 1. Goal: cosθ in terms of t.

Step 1: cosθ = ±√(1 − t²), sign by quadrant.

So, the final answer is ±√(1 − t²).

Submit

15) If sinθ = −√7/4 and θ is in Quadrant IV, find cosθ.

−3/4

3/4

√9/4

−√9/4

Given: sinθ negative, QIV ⇒ cosθ > 0. Goal: cosθ.

Step 1: cosθ = √(1 − 7/16) = √(9/16) = 3/4.

So, the final answer is 3/4.

Explanation

Given: sinθ negative, QIV ⇒ cosθ > 0. Goal: cosθ.

Step 1: cosθ = √(1 − 7/16) = √(9/16) = 3/4.

So, the final answer is 3/4.

Submit

16) If cosθ = 1/3 and θ is in Quadrant I, find sinθ and tanθ.

Sin = √8/3, tan = √8

Sin = 2√2/3, tan = 2/√2

Sin = √2/3, tan = √2/2

Sin = 2√2/3, tan = 2√2

Given: cosθ = 1/3 (QI). Goal: sinθ, tanθ.

Step 1: sinθ = √(1 − 1/9) = √(8/9) = 2√2/3.

Step 2: tanθ = sinθ/cosθ = (2√2/3)/(1/3) = 2√2.

So, the final answer is sinθ = 2√2/3, tanθ = 2√2.

Explanation

Submit

17) A unit direction vector has x-component 2/√5 (QI). Find sinθ.

1/√5

2/√5

√(1 − 4/5)

√(4/5)

Given: cosθ = 2/√5, QI. Goal: sinθ.

Step 1: sinθ = √(1 − 4/5) = √(1/5) = 1/√5.

So, the final answer is 1/√5.

Explanation

Given: cosθ = 2/√5, QI. Goal: sinθ.

Step 1: sinθ = √(1 − 4/5) = √(1/5) = 1/√5.

So, the final answer is 1/√5.

Submit

18) If sinθ = −4/5 and θ is in Quadrant III, find cosθ and then sin(2θ).

Cos = −3/5, sin(2θ) = −24/25

Cos = 3/5, sin(2θ) = −24/25

Cos = −3/5, sin(2θ) = 24/25

Cos = 3/5, sin(2θ) = 24/25

Step 1: cosθ = −3/5.

Step 2: sin(2θ) = 2·(−4/5)·(−3/5) = 24/25.

So, the final answer is cosθ = −3/5, sin(2θ) = 24/25.

Explanation

Step 1: cosθ = −3/5.

Step 2: sin(2θ) = 2·(−4/5)·(−3/5) = 24/25.

So, the final answer is cosθ = −3/5, sin(2θ) = 24/25.

Submit

19) A unit-circle point has (x, y) with y = 5/13 and x < 0. Compute cos(2θ).

119/169

120/169

−119/169

−120/169

Given: y = sinθ = 5/13, x < 0 ⇒ cosθ = −12/13. Goal: cos(2θ).

Step 1: cos(2θ) = cos²θ − sin²θ = (144 − 25)/169 = 119/169.

So, the final answer is 119/169.

Explanation

Given: y = sinθ = 5/13, x < 0 ⇒ cosθ = −12/13. Goal: cos(2θ).

Step 1: cos(2θ) = cos²θ − sin²θ = (144 − 25)/169 = 119/169.

So, the final answer is 119/169.

Submit

20) If cosθ = −√2/2 and θ is in Quadrant II, find sinθ and cos(2θ).

Sin = −√2/2, cos(2θ) = 0

Sin = √2/2, cos(2θ) = 0

Sin = √2/2, cos(2θ) = 1

Sin = −√2/2, cos(2θ) = −1

Given: cosθ = −√2/2, QII ⇒ sinθ > 0. Goal: sinθ, cos(2θ).

Step 1: sinθ = √2/2.

Step 2: cos(2θ) = cos²θ − sin²θ = 1/2 − 1/2 = 0.

So, the final answer is sinθ = √2/2, cos(2θ) = 0.

Explanation

Submit