Exercise 2.2 answers: Inverse Trigonometric Functions

Open NCERT p. 29One way to think about it

1. Let x = sin θ, so θ = sin⁻¹ x; since x ∈ [−1/2, 1/2], θ lies in [−π/6, π/6].
2. Use the triple-angle identity: sin 3θ = 3 sin θ − 4 sin³θ = 3x − 4x³.
3. Since θ ∈ [−π/6, π/6], 3θ ∈ [−π/2, π/2], so sin⁻¹(sin 3θ) = 3θ.
4. So sin⁻¹(3x − 4x³) = 3θ = 3 sin⁻¹ x.

In shortProved: 3 sin⁻¹ x = sin⁻¹(3x − 4x³) for x ∈ [−1/2, 1/2].

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Open NCERT p. 29One way to think about it

1. Let x = cos θ, so θ = cos⁻¹ x; since x ∈ [1/2, 1], θ lies in [0, π/3].
2. Use the triple-angle identity: cos 3θ = 4 cos³θ − 3 cos θ = 4x³ − 3x.
3. Since θ ∈ [0, π/3], 3θ ∈ [0, π], so cos⁻¹(cos 3θ) = 3θ.
4. So cos⁻¹(4x³ − 3x) = 3θ = 3 cos⁻¹ x.

In shortProved: 3 cos⁻¹ x = cos⁻¹(4x³ − 3x) for x ∈ [1/2, 1].

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Open NCERT p. 29Matches NCERT’s answer

1. Let x = tan θ, so θ = tan⁻¹ x. Then θ lies in (−π/2, π/2), and θ ≠ 0 because x ≠ 0.
2. In that range cos θ > 0, so √(1 + x²) = √(1 + tan²θ) = sec θ.
3. (sec θ − 1)/tan θ = (1 − cos θ)/sin θ = 2 sin²(θ/2) / (2 sin(θ/2) cos(θ/2)) = tan(θ/2).
4. θ/2 lies in (−π/4, π/4), which is inside (−π/2, π/2), so tan⁻¹[tan(θ/2)] = θ/2.
5. So the expression equals θ/2 = (1/2) tan⁻¹ x.

Answer(1/2) tan⁻¹ x

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Open NCERT p. 29Matches NCERT’s answer

1. Simplify: tan⁻¹ √((1 − cos x)/(1 + cos x)), 0 < x < π.
2. (1 − cos x)/(1 + cos x) = tan²(x/2), by the half-angle identities.
3. For 0 < x < π, x/2 lies in (0, π/2), so tan(x/2) is positive.
4. The square root of tan²(x/2) is therefore tan(x/2) itself.
5. So the expression is tan⁻¹[tan(x/2)] = x/2.

Answerx/2

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Open NCERT p. 29Matches NCERT’s answer

1. Simplify: tan⁻¹ ((cos x − sin x)/(cos x + sin x)), −π/4 < x < 3π/4.
2. When cos x ≠ 0, divide the numerator and denominator by cos x: (1 − tan x)/(1 + tan x).
3. Since tan(π/4) = 1, this is (tan(π/4) − tan x)/(1 + tan(π/4) tan x) = tan(π/4 − x).
4. At x = π/2, where cos x = 0, the fraction is (0 − 1)/(0 + 1) = −1 = tan(−π/4) = tan(π/4 − π/2), so the same result holds there too.
5. From −π/4 < x < 3π/4 we get −π/2 < π/4 − x < π/2, which is the range of tan⁻¹.
6. So tan⁻¹[tan(π/4 − x)] = π/4 − x.

Answerπ/4 − x

Watch this explained “The same move, five more times”, 14:36 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 29Matches NCERT’s answer

1. |x| < a means a > 0 and −1 < x/a < 1.
2. Let x = a sin θ, so θ = sin⁻¹(x/a). Then θ lies in (−π/2, π/2), so cos θ > 0.
3. √(a² − x²) = √(a² − a² sin²θ) = a cos θ.
4. x/√(a² − x²) = (a sin θ)/(a cos θ) = tan θ.
5. θ lies in (−π/2, π/2), so tan⁻¹(tan θ) = θ = sin⁻¹(x/a).

Answersin⁻¹(x/a)

Watch this explained “The same move, five more times”, 14:36 into Substituting an angle for the variable to collapse a messy expression

Open NCERT p. 29Matches NCERT’s answer

1. Simplify: tan⁻¹ ((3a²x − x³)/(a³ − 3ax²)), a > 0; −a/√3 < x < a/√3.
2. Let x = a tan θ, so θ = tan⁻¹(x/a).
3. 3a²x − x³ = a³(3 tanθ − tan³θ) and a³ − 3ax² = a³(1 − 3 tan²θ).
4. (3tanθ − tan³θ)/(1 − 3tan²θ) is the triple-angle formula for tan 3θ.
5. So the expression is tan⁻¹(tan 3θ) = 3θ = 3 tan⁻¹(x/a), since the given range keeps 3θ inside (−π/2, π/2).

Answer3 tan⁻¹(x/a)

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Open NCERT p. 29Matches NCERT’s answer

1. sin⁻¹(1/2) = π/6.
2. 2 × π/6 = π/3, and cos(π/3) = 1/2.
3. 2 × 1/2 = 1, so the expression becomes tan⁻¹(1).
4. tan⁻¹(1) = π/4.

Answerπ/4

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Open NCERT p. 29Matches NCERT’s answer

1. Find: tan (1/2)[sin⁻¹(2x/(1 + x²)) + cos⁻¹((1 − y²)/(1 + y²))], |x| < 1, y > 0 and xy < 1.
2. Since |x| < 1, write x = tan A with A between −π/4 and π/4. Then 2x/(1 + x²) = 2tan A/(1 + tan²A) = sin 2A, and 2A lies between −π/2 and π/2, the range of sin⁻¹. So sin⁻¹(2x/(1 + x²)) = 2A.
3. Since y > 0, write y = tan B with B between 0 and π/2. Then (1 − y²)/(1 + y²) = (1 − tan²B)/(1 + tan²B) = cos 2B, and 2B lies between 0 and π, the range of cos⁻¹. So cos⁻¹((1 − y²)/(1 + y²)) = 2B.
4. The expression becomes tan[(1/2)(2A + 2B)] = tan(A + B).
5. tan(A + B) = (tan A + tan B)/(1 − tan A·tan B) = (x + y)/(1 − xy). The condition xy < 1 keeps the denominator from being 0.

Answer(x + y)/(1 − xy)

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Open NCERT p. 30Matches NCERT’s answer

1. 2π/3 is more than π/2, so it is outside the range of sin⁻¹, [−π/2, π/2].
2. Write sin(2π/3) as sin(π − 2π/3) = sin(π/3), and π/3 does lie in [−π/2, π/2].
3. So sin⁻¹(sin(2π/3)) = sin⁻¹(sin(π/3)) = π/3.

Answerπ/3

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Open NCERT p. 30Matches NCERT’s answer

1. 3π/4 is more than π/2, so it is outside the range of tan⁻¹, (−π/2, π/2).
2. Write tan(3π/4) as tan(3π/4 − π) = tan(−π/4), and −π/4 does lie in (−π/2, π/2).
3. So tan⁻¹(tan(3π/4)) = tan⁻¹(tan(−π/4)) = −π/4.

Answer−π/4

Watch this explained “Five items, run the same way”, 12:37 into When the two cancellation rules hold, and the angles where the second one fails

Open NCERT p. 30Matches NCERT’s answer

1. Let A = sin⁻¹(3/5), so sinA = 3/5. In a 3-4-5 right triangle this gives cosA = 4/5 and tanA = 3/4.
2. Let B = cot⁻¹(3/2), so cotB = 3/2 and tanB = 2/3.
3. Use tan(A+B) = (tanA + tanB)/(1 − tanA·tanB) = (3/4 + 2/3)/(1 − (3/4)(2/3)).
4. Numerator = 17/12, denominator = 1/2, so tan(A+B) = (17/12) ÷ (1/2) = 17/6.

Answer17/6

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Open NCERT p. 30Matches NCERT’s answer

1. 7π/6 is more than π, so it is outside the range of cos⁻¹, [0, π].
2. Write cos(7π/6) as cos(2π − 7π/6) = cos(5π/6), and 5π/6 does lie in [0, π].
3. So cos⁻¹(cos(7π/6)) = cos⁻¹(cos(5π/6)) = 5π/6, option (B).

Answer5π/6, option (B)

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Open NCERT p. 30Matches NCERT’s answer

1. sin⁻¹(−1/2) = −π/6, since sin(−π/6) = −1/2 and −π/6 lies in [−π/2, π/2].
2. π/3 − (−π/6) = π/3 + π/6 = π/2.
3. sin(π/2) = 1, option (D).

Answer1, option (D)

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Open NCERT p. 30Matches NCERT’s answer

1. tan⁻¹√3 = π/3, since tan(π/3) = √3 and π/3 lies in (−π/2, π/2).
2. cot⁻¹(−√3): the range of cot⁻¹ is (0, π), and cot(5π/6) = −√3, so cot⁻¹(−√3) = 5π/6.
3. π/3 − 5π/6 = 2π/6 − 5π/6 = −3π/6 = −π/2, option (B).

Answer−π/2, option (B)

Watch this explained “Five items, run the same way”, 12:37 into When the two cancellation rules hold, and the angles where the second one fails