Principal Value Quiz: Principal Value And Restricted Domains

1) Which restriction makes sin x one-to-one and symmetric about the origin?

[−π/2, π/2]

[0, π]

(−π, 0]

[π/2, 3π/2]

On [−π/2, π/2], sin x is strictly increasing and odd, giving a one-to-one mapping with symmetry around 0.

Explanation

On [−π/2, π/2], sin x is strictly increasing and odd, giving a one-to-one mapping with symmetry around 0.

2) The principal range of y = arctan(x) is (−π/2, π/2).

True

False

arctan never returns ±π/2 because tanθ is undefined at those angles. Hence the open interval (−π/2, π/2).

Explanation

arctan never returns ±π/2 because tanθ is undefined at those angles. Hence the open interval (−π/2, π/2).

3) A function must be ____ on its restricted interval to have a well-defined inverse.

Injectivity ensures each output corresponds to exactly one input, making the inverse a function.

Explanation

Injectivity ensures each output corresponds to exactly one input, making the inverse a function.

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4) Why do inverse trigonometric functions use principal values?

To make the original trig function one-to-one on a restricted interval

To change the period to 1

To avoid using radians

To eliminate asymptotes

Trig functions are periodic and not one-to-one on ℝ. Restricting the domain to an interval where each y occurs once makes an inverse well-defined (each x maps to exactly one angle).

Explanation

Trig functions are periodic and not one-to-one on ℝ. Restricting the domain to an interval where each y occurs once makes an inverse well-defined (each x maps to exactly one angle).

5) Tan x is strictly increasing on (−π/2, π/2), which guarantees arctan is well-defined.

True

False

Strict monotonicity ensures a unique θ in (−π/2, π/2) for each real x, enabling the inverse.

Explanation

Strict monotonicity ensures a unique θ in (−π/2, π/2) for each real x, enabling the inverse.

6) Select all valid composition identities (with indicated domains).

Sin(arcsin x) = x for x ∈ [−1,1]

Arcsin(sin θ) = θ for θ ∈ [−π/2, π/2]

Arctan(tan θ) = θ for θ ∈ (−π/2, π/2)

Tan(arctan x) = x for all real x

Arccos(cos θ) = θ for θ ∈ [0, π]

Each equality holds when the inner angle lies in the principal interval or when the input lies in the inverse’s domain.

Explanation

Each equality holds when the inner angle lies in the principal interval or when the input lies in the inverse’s domain.

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7) Which is the principal range of y = arccos(x)?

[−π/2, π/2]

(−π/2, π/2)

[0, π]

(0, π)

arccos returns angles in [0, π], a closed interval, since cos reaches ±1 at 0 and π.

Explanation

arccos returns angles in [0, π], a closed interval, since cos reaches ±1 at 0 and π.

8) The principal value convention assigns exactly one angle to each valid input of an inverse trig function.

True

False

By restricting the original trig function to a one-to-one interval, each input to the inverse has a single, unique output angle.

Explanation

By restricting the original trig function to a one-to-one interval, each input to the inverse has a single, unique output angle.

9) What goes wrong if we define arcsin to return values in [0, π] instead of [−π/2, π/2]?

It would not be injective on that interval

It would not cover outputs near 0

It would no longer be a function

It would change sin’s period

On [0, π], sin is not one-to-one (it increases then decreases), so the inverse would fail to be a well-defined function.

Explanation

On [0, π], sin is not one-to-one (it increases then decreases), so the inverse would fail to be a well-defined function.

10) Select all true statements about inverse trig domains (real-valued).

Domain(arcsin) = [−1, 1]

Domain(arccos) = [−1, 1]

Domain(arctan) = ℝ

Domain(arccos) = ℝ

Domain(arcsin) = ℝ

sin and cos outputs are in [−1,1], so arcsin/arccos accept only x ∈ [−1,1]. tan outputs all reals, so arctan accepts any real x.

Explanation

sin and cos outputs are in [−1,1], so arcsin/arccos accept only x ∈ [−1,1]. tan outputs all reals, so arctan accepts any real x.

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11) The principal range of y = arcsin(x) is [−π/2, π/2].

True

False

By definition, arcsin returns the unique angle θ with sinθ = x and θ ∈ [−π/2, π/2]. Endpoints are included because sin(±π/2)=±1.

Explanation

By definition, arcsin returns the unique angle θ with sinθ = x and θ ∈ [−π/2, π/2]. Endpoints are included because sin(±π/2)=±1.

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13) Choose the principal values (if defined).

Arcsin(√2/2) = π/4

Arccos(−1) = π

Arctan(1) = π/4

Arcsin(1.2) is undefined (real)

Arctan(∞) = π/2

All except E are correct: arctan has horizontal asymptotes at ±π/2 but never equals them; arcsin(1.2) is outside [−1,1] so undefined over ℝ.

Explanation

All except E are correct: arctan has horizontal asymptotes at ±π/2 but never equals them; arcsin(1.2) is outside [−1,1] so undefined over ℝ.

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14) Without restricting sine to [−π/2, π/2], the inverse would fail the ____ test for functions.

An inverse requires the original function to be one-to-one. Without restriction, many horizontal lines intersect y=sin x infinitely many times.

Explanation

An inverse requires the original function to be one-to-one. Without restriction, many horizontal lines intersect y=sin x infinitely many times.

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15) For every x ∈ [−1,1], there exists exactly one θ in [−π/2, π/2] such that sinθ = x.

True

False

Monotonicity of sin x on the restricted interval ensures a unique preimage for each output in [−1,1].

Explanation

Monotonicity of sin x on the restricted interval ensures a unique preimage for each output in [−1,1].

16) Why is the range of arctan open but the range of arcsin closed? Select all that apply.

Tan has vertical asymptotes at ±π/2 (values not attained)

Sin attains ±1 at ±π/2 (endpoints attained)

Arctan must exclude ±π/2 from outputs

Arcsin includes ±π/2 because sin equals ±1 there

Arctan includes ±π/2 to complete the interval

tan θ → ±∞ as θ→±π/2, so those outputs are not attained; arcsin includes endpoints since sin(±π/2)=±1.

Explanation

tan θ → ±∞ as θ→±π/2, so those outputs are not attained; arcsin includes endpoints since sin(±π/2)=±1.

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17) The principal value of an inverse trig function is the unique angle selected from a predetermined ____.

A principal value is defined by choosing a specific interval where the trig function is one-to-one.

Explanation

A principal value is defined by choosing a specific interval where the trig function is one-to-one.

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18) Which statement shows why unrestricted cos x cannot have an inverse on ℝ?

Cos(0)=1 and cos(2π)=1 (two different inputs give same output)

Cos(0)=0

Cos is unbounded

Cos has no period

Periodic repetition causes multiple inputs mapping to the same output, violating one-to-one.

Explanation

Periodic repetition causes multiple inputs mapping to the same output, violating one-to-one.

19) Why is [0, π] used for arccos?

Because cos x is increasing there

Because cos x is one-to-one and covers [−1,1] there

Because cos x equals sin x there

Because cos x is odd there

On [0, π], cos x is strictly decreasing and spans outputs from 1 to −1, giving a bijection with [−1,1].

Explanation

On [0, π], cos x is strictly decreasing and spans outputs from 1 to −1, giving a bijection with [−1,1].

20) Because arcsin uses [−π/2, π/2], arcsin(x) is an odd function.

True

False

On a symmetric interval with sin odd and one-to-one, the inverse arcsin is odd: arcsin(−x)=−arcsin(x).

Explanation

On a symmetric interval with sin odd and one-to-one, the inverse arcsin is odd: arcsin(−x)=−arcsin(x).

Principal Value Quiz: Principal Value and Restricted Domains

1) Which restriction makes sin x one-to-one and symmetric about the origin?

2)

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

2) The principal range of y = arctan(x) is (−π/2, π/2).

3) A function must be ____ on its restricted interval to have a well-defined inverse.

4) Why do inverse trigonometric functions use principal values?

5) Tan x is strictly increasing on (−π/2, π/2), which guarantees arctan is well-defined.

6) Select all valid composition identities (with indicated domains).

7) Which is the principal range of y = arccos(x)?

8) The principal value convention assigns exactly one angle to each valid input of an inverse trig function.

9) What goes wrong if we define arcsin to return values in [0, π] instead of [−π/2, π/2]?

10) Select all true statements about inverse trig domains (real-valued).

11) The principal range of y = arcsin(x) is [−π/2, π/2].

12) State the principal interval used for inverse tangent: ____.

13) Choose the principal values (if defined).

14) Without restricting sine to [−π/2, π/2], the inverse would fail the ____ test for functions.

15) For every x ∈ [−1,1], there exists exactly one θ in [−π/2, π/2] such that sinθ = x.

16) Why is the range of arctan open but the range of arcsin closed? Select all that apply.

17) The principal value of an inverse trig function is the unique angle selected from a predetermined ____.

18) Which statement shows why unrestricted cos x cannot have an inverse on ℝ?

19) Why is [0, π] used for arccos?

20) Because arcsin uses [−π/2, π/2], arcsin(x) is an odd function.