Inverse Sine Quiz: Inverse Sine Fundamentals

1) Which statement best describes arcsin x?

It gives any angle with sine x

It gives the unique angle θ with sinθ = x in [−π/2, π/2]

It gives the smallest positive angle with sine x

It gives an angle in [0, π] with sine x

arcsin x is defined to return the unique principal angle θ ∈ [−π/2, π/2] such that sinθ = x.

Explanation

arcsin x is defined to return the unique principal angle θ ∈ [−π/2, π/2] such that sinθ = x.

2) Arcsin(sin θ) = θ for all real θ.

True

False

arcsin(sin θ)=θ only when θ ∈ [−π/2, π/2]; otherwise the principal angle with the same sine is returned.

Explanation

arcsin(sin θ)=θ only when θ ∈ [−π/2, π/2]; otherwise the principal angle with the same sine is returned.

3) Which of the following is the domain of y = arcsin(x)?

(−∞, ∞)

[−1, 1]

[0, 1]

[−1, 0]

arcsin is defined for inputs x with −1 ≤ x ≤ 1.

Explanation

arcsin is defined for inputs x with −1 ≤ x ≤ 1.

4) What is the principal range of y = arcsin(x)?

[−π, π]

[−π/2, π/2]

[0, π]

[0, 2π)

arcsin returns the unique angle θ with sinθ = x and θ constrained to [−π/2, π/2].

Explanation

arcsin returns the unique angle θ with sinθ = x and θ constrained to [−π/2, π/2].

5) Evaluate arcsin(0).

π/2

11/13

π

−π/2

sin0=0 with 0 in the principal range, so arcsin(0)=0.

Explanation

sin0=0 with 0 in the principal range, so arcsin(0)=0.

6) Arcsin is strictly increasing on [−1,1].

True

False

arcsin′(x)=1/√(1−x^2)>0 for x∈(−1,1), with monotone increase including endpoints.

Explanation

arcsin′(x)=1/√(1−x^2)>0 for x∈(−1,1), with monotone increase including endpoints.

7) If θ = arcsin(x), which identity is correct?

Cosθ = √(1 + x^2)

Cosθ = √(1 − x^2)

Cosθ = 1 − x

Cosθ = x

With sinθ=x and θ in [−π/2, π/2], cosθ ≥ 0 and cos^2θ = 1 − x^2 ⇒ cosθ = √(1 − x^2).

Explanation

With sinθ=x and θ in [−π/2, π/2], cosθ ≥ 0 and cos^2θ = 1 − x^2 ⇒ cosθ = √(1 − x^2).

8) If x = sin(5π/6), then arcsin(x) = 5π/6.

True

False

sin(5π/6)=1/2, but arcsin returns π/6 (principal value), not 5π/6.

Explanation

sin(5π/6)=1/2, but arcsin returns π/6 (principal value), not 5π/6.

9) For all x in [−1, 1], sin(arcsin x) = x.

True

False

Let θ = arcsin x ∈ [−π/2, π/2]. Then sinθ = x by definition.

Explanation

Let θ = arcsin x ∈ [−π/2, π/2]. Then sinθ = x by definition.

10) Select all x for which arcsin(x) is defined (real-valued).

X = −1

X = −3/2

X = 0

X = 5/4

X = 1

Domain of arcsin is [−1,1], so −1, 0, 1 are valid; −3/2 and 5/4 are not.

Explanation

Domain of arcsin is [−1,1], so −1, 0, 1 are valid; −3/2 and 5/4 are not.

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11) Select all exact evaluations.

Arcsin(−1) = −π/2

Arcsin(√2/2) = π/4

Arcsin(√3/2) = π/3

Arcsin(−√3/2) = −π/3

Arcsin(1) = π/2

Special angles: −1→−π/2, √2/2→π/4, √3/2→π/3, −√3/2→−π/3, and 1→π/2 (all within [−π/2, π/2]).

Explanation

Special angles: −1→−π/2, √2/2→π/4, √3/2→π/3, −√3/2→−π/3, and 1→π/2 (all within [−π/2, π/2]).

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12) Select all statements that are always true for arcsin.

Arcsin(0) = 0

Arcsin(1) = π/2

Arcsin(−1) = −π/2

Arcsin(x) is defined for all real x

Arcsin(x) ∈ [−π/2, π/2]

arcsin maps [−1,1] to [−π/2, π/2]; 0→0, 1→π/2, −1→−π/2. Not defined for |x|>1.

Explanation

arcsin maps [−1,1] to [−π/2, π/2]; 0→0, 1→π/2, −1→−π/2. Not defined for |x|>1.

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13) Select all correct simplifications (assume stated domains).

Sin(arcsin x) = x for x ∈ [−1,1]

Arcsin(sin θ) = θ for θ ∈ [−π/2, π/2]

Arcsin(sin θ) = π − θ for θ ∈ [π/2, 3π/2]

Arcsin(−x) = −arcsin(x) for x ∈ [−1,1]

Arcsin(sin θ) = −θ for θ ∈ [−π/2, π/2]

Compositions: sin∘arcsin is identity on [−1,1]; arcsin∘sin is identity only on [−π/2, π/2]; arcsin is odd.

Explanation

Compositions: sin∘arcsin is identity on [−1,1]; arcsin∘sin is identity only on [−π/2, π/2]; arcsin is odd.

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14) The equation arcsin(x) = 2 has no real solution.

True

False

arcsin(x) ∈ [−π/2, π/2] ≈ [−1.571, 1.571], so 2 is outside its range.

Explanation

arcsin(x) ∈ [−π/2, π/2] ≈ [−1.571, 1.571], so 2 is outside its range.

15) Select all equivalent statements to θ = arcsin(x).

Sinθ = x and θ ∈ [−π/2, π/2]

Sinθ = x for any θ

X ∈ [−1, 1]

θ is the unique angle in [−π/2, π/2] with sinθ = x

θ ∈ [0, π]

arcsin x is defined only for x∈[−1,1] and returns the unique θ in [−π/2, π/2] with sinθ=x.

Explanation

arcsin x is defined only for x∈[−1,1] and returns the unique θ in [−π/2, π/2] with sinθ=x.

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16) Solve θ = arcsin(−√3/2). Which principal value does θ equal?

−π/6

−π/3

π/3

5π/3

sin(−π/3)=−√3/2 and −π/3 ∈ [−π/2, π/2].

Explanation

sin(−π/3)=−√3/2 and −π/3 ∈ [−π/2, π/2].

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18) Solve for θ ∈ [−π/2, π/2]: sinθ = −1/2 ⇒ θ = ____.

Within the principal interval, the angle with sine −1/2 is −π/6.

Explanation

Within the principal interval, the angle with sine −1/2 is −π/6.

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Inverse Sine Quiz: Inverse Sine Fundamentals

1) Which statement best describes arcsin x?

2)

What first name or nickname would you like us to use?

2) Arcsin(sin θ) = θ for all real θ.

3) Which of the following is the domain of y = arcsin(x)?

4) What is the principal range of y = arcsin(x)?

5) Evaluate arcsin(0).

6) Arcsin is strictly increasing on [−1,1].

7) If θ = arcsin(x), which identity is correct?

8) If x = sin(5π/6), then arcsin(x) = 5π/6.

9) For all x in [−1, 1], sin(arcsin x) = x.

10) Select all x for which arcsin(x) is defined (real-valued).

11) Select all exact evaluations.

12) Select all statements that are always true for arcsin.

13) Select all correct simplifications (assume stated domains).

14) The equation arcsin(x) = 2 has no real solution.

15) Select all equivalent statements to θ = arcsin(x).

16) Solve θ = arcsin(−√3/2). Which principal value does θ equal?

17) Evaluate arcsin(1/2) = ____ (in radians).

18) Solve for θ ∈ [−π/2, π/2]: sinθ = −1/2 ⇒ θ = ____.

19) Evaluate sin(arcsin(0.3)) = ____.

20) Find all real x satisfying arcsin(x) = 0. ____

21) Compute sin(arcsin(−0.8)) = ____.