Pythagorean Identity Quiz: Pythagorean Identity

1) On the unit circle, the coordinates are (cosθ, sinθ). What is cos^2θ + sin^2θ?

0

1

2

Depends on θ

The unit circle has radius 1, so the distance from the origin to (cosθ, sinθ) is √(cos^2θ + sin^2θ) = 1. Squaring both sides gives cos^2θ + sin^2θ = 1.

Explanation

The unit circle has radius 1, so the distance from the origin to (cosθ, sinθ) is √(cos^2θ + sin^2θ) = 1. Squaring both sides gives cos^2θ + sin^2θ = 1.

2) Select all pairs (sinθ, cosθ) that are possible for some real θ.

(3/5, 4/5)

(−12/13, 5/13)

(√2/2, √2/2)

(1.2, 0)

(−4/5, −3/5)

Check sin^2 + cos^2: A) 9/25 + 16/25 = 1 ✓; B) 144/169 + 25/169 = 1 ✓; C) 1/2 + 1/2 = 1 ✓; D) 1.44 + 0 > 1 and |sin| ≤ 1 is violated ✗; E) 16/25 + 9/25 = 1 ✓.

Explanation

Check sin^2 + cos^2: A) 9/25 + 16/25 = 1 ✓; B) 144/169 + 25/169 = 1 ✓; C) 1/2 + 1/2 = 1 ✓; D) 1.44 + 0 > 1 and |sin| ≤ 1 is violated ✗; E) 16/25 + 9/25 = 1 ✓.

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3) If sinθ = 7/25 and θ is acute, what is cosθ?

−24/25

24/25

7/24

−7/25

cos^2θ = 1 − sin^2θ = 1 − 49/625 = 576/625 ⇒ cosθ = √(576/625) = 24/25 (positive for acute angles).

Explanation

cos^2θ = 1 − sin^2θ = 1 − 49/625 = 576/625 ⇒ cosθ = √(576/625) = 24/25 (positive for acute angles).

4) Evaluate sin^2(π/6) + cos^2(π/6).

1/2

3/4

1

5/4

sin(π/6) = 1/2 ⇒ sin^2 = 1/4; cos(π/6) = √3/2 ⇒ cos^2 = 3/4; sum = 1/4 + 3/4 = 1.

Explanation

sin(π/6) = 1/2 ⇒ sin^2 = 1/4; cos(π/6) = √3/2 ⇒ cos^2 = 3/4; sum = 1/4 + 3/4 = 1.

5) Complete the unit circle equation: For (x, y) = (cosθ, sinθ), x^2 + y^2 = ________.

The unit circle has radius 1, so x^2 + y^2 = 1. Substituting x = cosθ, y = sinθ yields cos^2θ + sin^2θ = 1.

Explanation

The unit circle has radius 1, so x^2 + y^2 = 1. Substituting x = cosθ, y = sinθ yields cos^2θ + sin^2θ = 1.

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6) If cosθ = 0, what is sin^2θ?

0

1

-1

Depends on θ

Use sin^2θ + cos^2θ = 1. If cosθ = 0, then sin^2θ = 1 − 0 = 1, meaning sinθ = ±1 at θ = π/2 + kπ.

Explanation

Use sin^2θ + cos^2θ = 1. If cosθ = 0, then sin^2θ = 1 − 0 = 1, meaning sinθ = ±1 at θ = π/2 + kπ.

7) On the unit circle, the distance from (0,0) to (cosθ, sinθ) is 1 for every θ.

True

False

By definition of the unit circle, every point (cosθ, sinθ) lies at radius 1 from the origin, so √(cos^2θ + sin^2θ) = 1.

Explanation

By definition of the unit circle, every point (cosθ, sinθ) lies at radius 1 from the origin, so √(cos^2θ + sin^2θ) = 1.

8) Choose all angles for which sin^2θ + cos^2θ = 1 holds.

θ = 0

θ = π/3

θ = 11 (radians)

θ = −5π/4

Only angles in Quadrant I

sin^2θ + cos^2θ = 1 is an identity for all real θ. It is not restricted to any quadrant.

Explanation

sin^2θ + cos^2θ = 1 is an identity for all real θ. It is not restricted to any quadrant.

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9) Which expression is true for any real θ?

Cosθ = √(1 − sin^2θ)

Cosθ = −√(1 − sin^2θ)

Cosθ = ±√(1 − sin^2θ)

Cosθ = 1 − sinθ

From cos^2θ = 1 − sin^2θ, we take square roots to get cosθ = ±√(1 − sin^2θ). The sign depends on θ’s quadrant.

Explanation

From cos^2θ = 1 − sin^2θ, we take square roots to get cosθ = ±√(1 − sin^2θ). The sign depends on θ’s quadrant.

10) If sinθ = −0.8 and θ is in Quadrant III, what is cosθ?

0.8

0.6

-0.8

-0.6

Compute cos^2θ = 1 − sin^2θ = 1 − 0.64 = 0.36 ⇒ |cosθ| = 0.6. In Quadrant III, both sine and cosine are negative, so cosθ = −0.6.

Explanation

Compute cos^2θ = 1 − sin^2θ = 1 − 0.64 = 0.36 ⇒ |cosθ| = 0.6. In Quadrant III, both sine and cosine are negative, so cosθ = −0.6.

11) In a right triangle with hypotenuse 10, opposite 6, and adjacent 8, sin^2θ + cos^2θ = ________.

sinθ = opp/hyp = 6/10 = 3/5, cosθ = adj/hyp = 8/10 = 4/5. Then sin^2θ + cos^2θ = (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 25/25 = 1.

Explanation

sinθ = opp/hyp = 6/10 = 3/5, cosθ = adj/hyp = 8/10 = 4/5. Then sin^2θ + cos^2θ = (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 25/25 = 1.

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12) For any real θ, sin^2θ = 1 − cos^2θ.

True

False

Rearranging sin^2θ + cos^2θ = 1 gives sin^2θ = 1 − cos^2θ, valid for all real θ.

Explanation

Rearranging sin^2θ + cos^2θ = 1 gives sin^2θ = 1 − cos^2θ, valid for all real θ.

13) Using the identity, express cos^2θ in terms of sinθ.

Sin^2θ − 1

1 + sin^2θ

1 − sin^2θ

(1 − sinθ)^2

Rearrange sin^2θ + cos^2θ = 1 to get cos^2θ = 1 − sin^2θ.

Explanation

Rearrange sin^2θ + cos^2θ = 1 to get cos^2θ = 1 − sin^2θ.

14) If cosθ = −12/13 and θ is in Quadrant II, what is sinθ?

−5/13

−12/13

5/13

12/13

Compute sin^2θ = 1 − cos^2θ = 1 − (144/169) = 25/169 ⇒ |sinθ| = 5/13. In Quadrant II, sinθ > 0 and cosθ < 0, so sinθ = 5/13.

Explanation

Compute sin^2θ = 1 − cos^2θ = 1 − (144/169) = 25/169 ⇒ |sinθ| = 5/13. In Quadrant II, sinθ > 0 and cosθ < 0, so sinθ = 5/13.

15) For any real θ, sin^2θ + cos^2θ = 1 is always true.

True

False

This is an identity derived from x^2 + y^2 = 1 on the unit circle with x = cosθ and y = sinθ; it holds for all real θ.

Explanation

This is an identity derived from x^2 + y^2 = 1 on the unit circle with x = cosθ and y = sinθ; it holds for all real θ.

16) Which identity holds for every real angle θ?

Sin^2θ + cos^2θ = 1

Sin^2θ − cos^2θ = 1

Sin^2θ + cos^2θ = 2

Sinθ + cosθ = 1

On the unit circle a point at angle θ is (cosθ, sinθ). The radius is 1, so by the distance formula x^2 + y^2 = 1. Substituting x = cosθ and y = sinθ gives cos^2θ + sin^2θ = 1.

Explanation

On the unit circle a point at angle θ is (cosθ, sinθ). The radius is 1, so by the distance formula x^2 + y^2 = 1. Substituting x = cosθ and y = sinθ gives cos^2θ + sin^2θ = 1.

17) If a point on the unit circle is (−√3/2, 1/2), what is sin^2θ + cos^2θ?

1/2

3/2

0

1

Compute: cos^2θ + sin^2θ = ( (−√3/2)^2 ) + ( (1/2)^2 ) = (3/4) + (1/4) = 1.

Explanation

Compute: cos^2θ + sin^2θ = ( (−√3/2)^2 ) + ( (1/2)^2 ) = (3/4) + (1/4) = 1.

18) If sinθ = 3/5 and 0 ≤ θ ≤ π/2, then cosθ = ________.

Use sin^2θ + cos^2θ = 1. Compute cos^2θ = 1 − sin^2θ = 1 − (3/5)^2 = 1 − 9/25 = 16/25, so cosθ = √(16/25) = 4/5 (positive in Quadrant I).

Explanation

Use sin^2θ + cos^2θ = 1. Compute cos^2θ = 1 − sin^2θ = 1 − (3/5)^2 = 1 − 9/25 = 16/25, so cosθ = √(16/25) = 4/5 (positive in Quadrant I).

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19) There exists a real θ such that sin^2θ + cos^2θ = 0.

True

False

Since sin^2θ ≥ 0 and cos^2θ ≥ 0 for all θ and their sum is identically 1, it can never be 0. Therefore no real θ satisfies the equation.

Explanation

Since sin^2θ ≥ 0 and cos^2θ ≥ 0 for all θ and their sum is identically 1, it can never be 0. Therefore no real θ satisfies the equation.

20) Select all identities that follow from sin^2θ + cos^2θ = 1.

Sin^2θ + cos^2θ = 1

1 − cos^2θ = sin^2θ

1 − sin^2θ = cos^2θ

Tan^2θ + 1 = sec^2θ

Sinθ + cosθ = 1

From sin^2θ + cos^2θ = 1 we get 1 − cos^2θ = sin^2θ and 1 − sin^2θ = cos^2θ. Dividing by cos^2θ (cosθ ≠ 0) gives (sin^2θ/cos^2θ) + 1 = 1/cos^2θ ⇒ tan^2θ + 1 = sec^2θ. The equation sinθ + cosθ = 1 is not an identity.

Explanation

From sin^2θ + cos^2θ = 1 we get 1 − cos^2θ = sin^2θ and 1 − sin^2θ = cos^2θ. Dividing by cos^2θ (cosθ ≠ 0) gives (sin^2θ/cos^2θ) + 1 = 1/cos^2θ ⇒ tan^2θ + 1 = sec^2θ. The equation sinθ + cosθ = 1 is not an identity.

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