Exercise 2.1 answers: Inverse Trigonometric Functions

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1. Let sin⁻¹(−1/2) = y, so sin y = −1/2.
2. sin(π/6) = 1/2, so sin(−π/6) = −1/2.
3. −π/6 lies in [−π/2, π/2], the range of sin⁻¹.

Answer−π/6

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1. Let cos⁻¹(√3/2) = y, so cos y = √3/2.
2. cos(π/6) = √3/2.
3. π/6 lies in [0, π], the range of cos⁻¹.

Answerπ/6

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1. Let cosec⁻¹(2) = y, so cosec y = 2, which means sin y = 1/2.
2. sin(π/6) = 1/2.
3. π/6 lies in [−π/2, π/2] and is not 0, the range of cosec⁻¹.

Answerπ/6

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1. Let tan⁻¹(−√3) = y, so tan y = −√3.
2. tan(π/3) = √3, so tan(−π/3) = −√3.
3. −π/3 lies in (−π/2, π/2), the range of tan⁻¹.

Answer−π/3

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1. Let cos⁻¹(−1/2) = y, so cos y = −1/2.
2. cos(π/3) = 1/2, and cos(π − π/3) = −cos(π/3) = −1/2.
3. So y = 2π/3, which lies in [0, π].

Answer2π/3

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1. Let tan⁻¹(−1) = y, so tan y = −1.
2. tan(π/4) = 1, so tan(−π/4) = −1.
3. −π/4 lies in (−π/2, π/2).

Answer−π/4

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1. Let sec⁻¹(2/√3) = y, so sec y = 2/√3, which means cos y = √3/2.
2. cos(π/6) = √3/2.
3. π/6 lies in [0, π] and is not π/2, the range of sec⁻¹.

Answerπ/6

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1. Let cot⁻¹(√3) = y, so cot y = √3, which means tan y = 1/√3.
2. tan(π/6) = 1/√3.
3. π/6 lies in (0, π), the range of cot⁻¹.

Answerπ/6

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1. Let cos⁻¹(−1/√2) = y, so cos y = −1/√2.
2. cos(π/4) = 1/√2, and cos(π − π/4) = −cos(π/4) = −1/√2.
3. So y = 3π/4, which lies in [0, π].

Answer3π/4

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1. Let cosec⁻¹(−√2) = y, so cosec y = −√2, which means sin y = −1/√2.
2. sin(π/4) = 1/√2, so sin(−π/4) = −1/√2.
3. −π/4 lies in [−π/2, π/2] and is not 0.

Answer−π/4

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1. tan⁻¹(1) = π/4, since tan(π/4) = 1.
2. cos⁻¹(−1/2) = 2π/3, since cos(2π/3) = −1/2 and 2π/3 lies in [0, π].
3. sin⁻¹(−1/2) = −π/6, since sin(−π/6) = −1/2 and −π/6 lies in [−π/2, π/2].
4. Add the three angles: π/4 + 2π/3 − π/6 = 3π/12 + 8π/12 − 2π/12 = 9π/12 = 3π/4.

Answer3π/4

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1. cos⁻¹(1/2) = π/3, since cos(π/3) = 1/2.
2. sin⁻¹(1/2) = π/6, since sin(π/6) = 1/2.
3. Add: π/3 + 2(π/6) = π/3 + π/3 = 2π/3.

Answer2π/3

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1. y = sin⁻¹ x is, by definition, the principal value — the one angle in the principal value branch of sin⁻¹.
2. The principal value branch of sin⁻¹ is [−π/2, π/2], with both end angles included.
3. So y always satisfies −π/2 ≤ y ≤ π/2.

AnswerOption (B): −π/2 ≤ y ≤ π/2

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1. tan⁻¹√3 = π/3, since tan(π/3) = √3 and π/3 lies in (−π/2, π/2).
2. sec⁻¹(−2) means cos y = −1/2 with y in [0, π]; cos(2π/3) = −1/2, so sec⁻¹(−2) = 2π/3.
3. π/3 − 2π/3 = −π/3, which is option (B).

Answer−π/3 (option B)

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