Principal Value Equations Quiz: Using Principal Values To Solve Equations

1) State the principal interval of arctan: ____.

By definition, arctan outputs are strictly between −π/2 and π/2 (open interval).

Explanation

By definition, arctan outputs are strictly between −π/2 and π/2 (open interval).

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3) Arctan is an odd, increasing function, so sign(arctan x) = sign(x).

True

False

Because tan is odd and strictly increasing on (−π/2,π/2), its inverse arctan is also odd and increasing, preserving the sign of x.

Explanation

Because tan is odd and strictly increasing on (−π/2,π/2), its inverse arctan is also odd and increasing, preserving the sign of x.

4) Find the principal value: cos θ = −√3/2.

π/6

5π/6

7π/6

11π/6

cosθ=−√3/2 at θ=5π/6 and 7π/6. The principal value for arccos is in [0,π], so θ=5π/6.

Explanation

cosθ=−√3/2 at θ=5π/6 and 7π/6. The principal value for arccos is in [0,π], so θ=5π/6.

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6) Arcsin(sin 5π/6) equals 5π/6.

True

False

sin(5π/6)=1/2, but arcsin returns the principal angle π/6 (in [−π/2, π/2]), not 5π/6.

Explanation

sin(5π/6)=1/2, but arcsin returns the principal angle π/6 (in [−π/2, π/2]), not 5π/6.

7) Which statements about principal ranges are true?

Arcsin returns QI or QIV angles

Arccos returns angles in QI or QII

Arctan returns QI or QIV angles

Arcsin can return angles in QII

Arctan can return ±π/2

arcsin range [−π/2,π/2] corresponds to QIV (negative) and QI (positive). arccos range [0,π] is QI–QII. arctan range (−π/2,π/2) is QIV–QI. arcsin does not return QII; arctan never equals ±π/2.

Explanation

arcsin range [−π/2,π/2] corresponds to QIV (negative) and QI (positive). arccos range [0,π] is QI–QII. arctan range (−π/2,π/2) is QIV–QI. arcsin does not return QII; arctan never equals ±π/2.

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8) Solve for the principal value: sin θ = −1/2. Choose θ in [−π/2, π/2].

−π/6

π/6

−π/3

5π/6

arcsin(−1/2)=−π/6 because sin(−π/6)=−1/2 and −π/6 lies in [−π/2, π/2]. Other listed angles are outside the principal range or give different sines.

Explanation

arcsin(−1/2)=−π/6 because sin(−π/6)=−1/2 and −π/6 lies in [−π/2, π/2]. Other listed angles are outside the principal range or give different sines.

9) Select all valid identities with proper domains/ranges.

Sin(arcsin x) = x for x ∈ [−1,1]

Arcsin(sin θ) = θ for θ ∈ [−π/2, π/2]

Tan(arctan x) = x for all real x

Arctan(tan θ) = θ for θ ∈ (−π/2, π/2)

Arccos(cos θ) = θ for all θ ∈ ℝ

Each identity holds when the inner value lies in the appropriate principal interval/domain. arccos(cos θ)=θ only for θ∈[0,π]; not for all real θ.

Explanation

Each identity holds when the inner value lies in the appropriate principal interval/domain. arccos(cos θ)=θ only for θ∈[0,π]; not for all real θ.

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10) Find the principal value: arctan(−√3) = ?

−π/6

−π/3

−π/4

π/3

arctan returns θ in (−π/2, π/2). Since tan(−π/3)=−√3 and −π/3 lies in (−π/2, π/2), arctan(−√3)=−π/3.

Explanation

arctan returns θ in (−π/2, π/2). Since tan(−π/3)=−√3 and −π/3 lies in (−π/2, π/2), arctan(−√3)=−π/3.

11) Compute cos(arccos(−0.2)) = ____.

Let θ=arccos(−0.2)∈[0,π]. Then cosθ=−0.2 by definition, so cos(arccos(−0.2))=−0.2.

Explanation

Let θ=arccos(−0.2)∈[0,π]. Then cosθ=−0.2 by definition, so cos(arccos(−0.2))=−0.2.

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12) Solve tan θ = 1. Choose the principal value of θ.

π/4

3π/4

−3π/4

5π/4

tanθ=1 occurs at θ=π/4+nπ. The principal range for arctan is (−π/2, π/2), so the principal value is π/4.

Explanation

tanθ=1 occurs at θ=π/4+nπ. The principal range for arctan is (−π/2, π/2), so the principal value is π/4.

13) Select all correct principal values.

Arcsin(√2/2) = π/4

Arccos(−1) = π

Arctan(1) = π/4

Arcsin(−√3/2) = −π/3

Arctan(0) = π

arcsin(√2/2)=π/4 (QI), arccos(−1)=π (endpoint in [0,π]), arctan(1)=π/4 (QI), arcsin(−√3/2)=−π/3 (QIV). arctan(0)=0, not π.

Explanation

arcsin(√2/2)=π/4 (QI), arccos(−1)=π (endpoint in [0,π]), arctan(1)=π/4 (QI), arcsin(−√3/2)=−π/3 (QIV). arctan(0)=0, not π.

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

True

False

arcsin outputs the unique θ with sinθ=x where θ ∈ [−π/2, π/2]. Endpoints included since sin(±π/2)=±1.

Explanation

arcsin outputs the unique θ with sinθ=x where θ ∈ [−π/2, π/2]. Endpoints included since sin(±π/2)=±1.

15) Select the correct principal solutions for each equation (choose all that apply).

Sin θ = 1/2 ⇒ θ = π/6

Sin θ = 1/2 ⇒ θ = 5π/6

Cos θ = −1 ⇒ θ = π

Tan θ = −1 ⇒ θ = −π/4

Tan θ = −1 ⇒ θ = 3π/4

Principal values: arcsin(1/2)=π/6 (not 5π/6 since that’s QII), arccos(−1)=π, arctan(−1)=−π/4. 3π/4 is outside arctan’s principal range.

Explanation

Principal values: arcsin(1/2)=π/6 (not 5π/6 since that’s QII), arccos(−1)=π, arctan(−1)=−π/4. 3π/4 is outside arctan’s principal range.

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16) Compute the principal value: arccos(−1/2) = ____.

cos(2π/3)=−1/2 and 2π/3 ∈ [0,π], which is the principal range for arccos.

Explanation

cos(2π/3)=−1/2 and 2π/3 ∈ [0,π], which is the principal range for arccos.

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17) Which angles solve tan θ = 1 within the principal range? Select all that apply.

θ = π/4

θ = −π/4

θ = 5π/4

θ = −3π/4

θ = 9π/4

Within (−π/2, π/2), tanθ=1 only at θ=π/4. −π/4 gives −1; the others are outside the principal interval.

Explanation

Within (−π/2, π/2), tanθ=1 only at θ=π/4. −π/4 gives −1; the others are outside the principal interval.

18) Given an angle x in QII with sin x = 1/2, what is arcsin(1/2)?

5π/6

π/6

11π/6

π − π/6

arcsin returns the principal value π/6 in [−π/2, π/2] regardless of the original quadrant of x. 5π/6 is a valid solution to sinθ=1/2 but not principal.

Explanation

arcsin returns the principal value π/6 in [−π/2, π/2] regardless of the original quadrant of x. 5π/6 is a valid solution to sinθ=1/2 but not principal.