Exercise 2.1
Find the principal values of the following (Q1 to Q10). Principal value ranges used: sin⁻¹ → [–π/2, π/2]; cos⁻¹ → [0, π]; tan⁻¹ → (–π/2, π/2); cot⁻¹ → (0, π); sec⁻¹ → [0, π] – {π/2}; cosec⁻¹ → [–π/2, π/2] – {0}.
sin⁻¹(–1/2)
Answer
Let sin⁻¹(–1/2) = y
⇒ sin y = –1/2 = –sin(π/6) = sin(–π/6). Since –π/6 ∈ [–π/2, π/2], the principal value is –π/6.
cos⁻¹(√3/2)
Answer
Let cos⁻¹(√3/2) = y
⇒ cos y = √3/2 = cos(π/6). Since π/6 ∈ [0, π], the principal value is π/6.
cosec⁻¹(2)
Answer
Let cosec⁻¹(2) = y
⇒ cosec y = 2 = cosec(π/6). Since π/6 ∈ [–π/2, π/2] – {0}, the principal value is π/6.
tan⁻¹(–√3)
Answer
Let tan⁻¹(–√3) = y
⇒ tan y = –√3 = –tan(π/3) = tan(–π/3). Since –π/3 ∈ (–π/2, π/2), the principal value is –π/3.
cos⁻¹(–1/2)
Answer
Let cos⁻¹(–1/2) = y
⇒ cos y = –1/2 = –cos(π/3) = cos(π – π/3) = cos(2π/3). Since 2π/3 ∈ [0, π], the principal value is 2π/3.
tan⁻¹(–1)
Answer
Let tan⁻¹(–1) = y
⇒ tan y = –1 = tan(–π/4). Since –π/4 ∈ (–π/2, π/2), the principal value is –π/4.
sec⁻¹(2/√3)
Answer
Let sec⁻¹(2/√3) = y
⇒ sec y = 2/√3, i.e. cos y = √3/2 = cos(π/6). Since π/6 ∈ [0, π] – {π/2}, the principal value is π/6.
cot⁻¹(√3)
Answer
Let cot⁻¹(√3) = y
⇒ cot y = √3 = cot(π/6). Since π/6 ∈ (0, π), the principal value is π/6.
cos⁻¹(–1/√2)
Answer
Let cos⁻¹(–1/√2) = y
⇒ cos y = –1/√2 = –cos(π/4) = cos(π – π/4) = cos(3π/4). Since 3π/4 ∈ [0, π], the principal value is 3π/4.
cosec⁻¹(–√2)
Answer
Let cosec⁻¹(–√2) = y
⇒ cosec y = –√2 = –cosec(π/4) = cosec(–π/4). Since –π/4 ∈ [–π/2, π/2] – {0}, the principal value is –π/4.
Find the values of the following:
tan⁻¹(1) + cos⁻¹(–1/2) + sin⁻¹(–1/2)
Answer
tan⁻¹(1) = π/4; cos⁻¹(–1/2) = π – π/3 = 2π/3; sin⁻¹(–1/2) = –π/6.
Sum = π/4 + 2π/3 – π/6 = (3π + 8π – 2π)/12 = 9π/12 = 3π/4.
cos⁻¹(1/2) + 2 sin⁻¹(1/2)
Answer
cos⁻¹(1/2) = π/3 and sin⁻¹(1/2) = π/6.
Value = π/3 + 2(π/6) = π/3 + π/3 = 2π/3.
If sin⁻¹ x = y, then (A) 0 ≤ y ≤ π (B) –π/2 ≤ y ≤ π/2 (C) 0 < y < π (D) –π/2 < y < π/2
Answer
The principal value branch of sin⁻¹ is [–π/2, π/2], so –π/2 ≤ y ≤ π/2.
Correct answer: (B).
tan⁻¹√3 – sec⁻¹(–2) is equal to (A) π (B) –π/3 (C) π/3 (D) 2π/3
Answer
tan⁻¹√3 = π/3. sec⁻¹(–2): cos y = –1/2
⇒ y = 2π/3 (in [0, π] – {π/2}).
Value = π/3 – 2π/3 = –π/3. Correct answer: (B).
Exercise 2.2
Prove the following:
3sin⁻¹ x = sin⁻¹(3x – 4x³), x ∈ [–1/2, 1/2]
Answer
Put x = sin θ, so θ = sin⁻¹ x. Since x ∈ [–1/2, 1/2], θ ∈ [–π/6, π/6].
3x – 4x³ = 3 sin θ – 4 sin³θ = sin 3θ, and 3θ ∈ [–π/2, π/2].
RHS = sin⁻¹(sin 3θ) = 3θ = 3 sin⁻¹ x = LHS. Hence proved.
3cos⁻¹ x = cos⁻¹(4x³ – 3x), x ∈ [1/2, 1]
Answer
Put x = cos θ, so θ = cos⁻¹ x. Since x ∈ [1/2, 1], θ ∈ [0, π/3].
4x³ – 3x = 4 cos³θ – 3 cos θ = cos 3θ, and 3θ ∈ [0, π].
RHS = cos⁻¹(cos 3θ) = 3θ = 3 cos⁻¹ x = LHS. Hence proved.
Write the following functions in the simplest form:
tan⁻¹[(√(1 + x²) – 1)/x], x ≠ 0
Answer
Put x = tan θ, θ ∈ (–π/2, π/2), θ ≠ 0. Then √(1 + x²) = √(sec²θ) = sec θ (as cos θ > 0).
(sec θ – 1)/tan θ = (1 – cos θ)/sin θ
⇒ [2 sin²(θ/2)]/[2 sin(θ/2) cos(θ/2)]
⇒ tan(θ/2).
Since θ/2 ∈ (–π/4, π/4), tan⁻¹(tan(θ/2)) = θ/2.
Simplest form: (1/2) tan⁻¹ x.
tan⁻¹ √[(1 – cos x)/(1 + cos x)], 0 < x < π
Answer
(1 – cos x)/(1 + cos x) = [2 sin²(x/2)]/[2 cos²(x/2)] = tan²(x/2).
For 0 < x < π, x/2 ∈ (0, π/2), so tan(x/2) > 0 and √(tan²(x/2)) = tan(x/2).
tan⁻¹(tan(x/2)) = x/2. Simplest form: x/2.
tan⁻¹[(cos x – sin x)/(cos x + sin x)], –π/4 < x < 3π/4
Answer
cos x – sin x = √2 [cos x · (1/√2) – sin x · (1/√2)] = √2 sin(π/4 – x).
cos x + sin x = √2 [cos x · (1/√2) + sin x · (1/√2)] = √2 cos(π/4 – x).
So the expression = tan⁻¹[tan(π/4 – x)]. For –π/4 < x < 3π/4, π/4 – x ∈ (–π/2, π/2).
Simplest form: π/4 – x.
tan⁻¹[x/√(a² – x²)], |x| < a
Answer
Put x = a sin θ, θ ∈ (–π/2, π/2) (since |x| < a). Then √(a² – x²) = √(a² cos²θ) = a cos θ.
x/√(a² – x²) = a sin θ/(a cos θ) = tan θ
⇒ tan⁻¹(tan θ) = θ.
Simplest form: sin⁻¹(x/a).
tan⁻¹[(3a²x – x³)/(a³ – 3ax²)], a > 0; –a/√3 < x < a/√3
Answer
Put x = a tan θ. Since –a/√3 < x < a/√3, –1/√3 < tan θ < 1/√3, i.e. θ ∈ (–π/6, π/6).
(3a²·a tan θ – a³ tan³θ)/(a³ – 3a·a² tan²θ) = (3 tan θ – tan³θ)/(1 – 3 tan²θ) = tan 3θ.
3θ ∈ (–π/2, π/2), so tan⁻¹(tan 3θ) = 3θ.
Simplest form: 3 tan⁻¹(x/a).
Find the values of each of the following:
tan⁻¹[2 cos(2 sin⁻¹(1/2))]
Answer
sin⁻¹(1/2) = π/6
⇒ 2 sin⁻¹(1/2) = π/3
⇒ cos(π/3) = 1/2
⇒ 2 × 1/2 = 1.
Value = tan⁻¹(1) = π/4.
tan (1/2)[sin⁻¹(2x/(1 + x²)) + cos⁻¹((1 – y²)/(1 + y²))], |x| < 1, y > 0 and xy < 1
Answer
Put x = tan α with α ∈ (–π/4, π/4) (as |x| < 1). Then 2x/(1 + x²) = 2 tan α/(1 + tan²α) = sin 2α, and 2α ∈ (–π/2, π/2), so sin⁻¹(2x/(1 + x²)) = 2α = 2 tan⁻¹ x.
Put y = tan β with β ∈ (0, π/2) (as y > 0). Then (1 – y²)/(1 + y²) = cos 2β, and 2β ∈ (0, π), so cos⁻¹((1 – y²)/(1 + y²)) = 2β = 2 tan⁻¹ y.
Expression = tan[(1/2)(2α + 2β)] = tan(α + β) = (tan α + tan β)/(1 – tan α tan β).
Value = (x + y)/(1 – xy).
sin⁻¹(sin 2π/3)
Answer
2π/3 is not in [–π/2, π/2]. sin(2π/3) = sin(π – 2π/3) = sin(π/3), and π/3 ∈ [–π/2, π/2].
Value = sin⁻¹(sin π/3) = π/3.
tan⁻¹(tan 3π/4)
Answer
3π/4 is not in (–π/2, π/2). tan(3π/4) = tan(3π/4 – π) = tan(–π/4), and –π/4 ∈ (–π/2, π/2).
Value = –π/4.
tan(sin⁻¹(3/5) + cot⁻¹(3/2))
Answer
Let A = sin⁻¹(3/5): sin A = 3/5, cos A = 4/5, tan A = 3/4.
Let B = cot⁻¹(3/2): tan B = 2/3.
tan(A + B) = (tan A + tan B)/(1 – tan A tan B) = (3/4 + 2/3)/(1 – (3/4)(2/3))
⇒ (17/12)/(1/2)
⇒ 17/6.
Value = 17/6.
cos⁻¹(cos 7π/6) is equal to (A) 7π/6 (B) 5π/6 (C) π/3 (D) π/6
Answer
7π/6 ∉ [0, π]. cos(7π/6) = cos(2π – 7π/6) = cos(5π/6), and 5π/6 ∈ [0, π].
Value = 5π/6. Correct answer: (B).
sin(π/3 – sin⁻¹(–1/2)) is equal to (A) 1/2 (B) 1/3 (C) 1/4 (D) 1
Answer
sin⁻¹(–1/2) = –π/6
⇒ sin(π/3 + π/6) = sin(π/2) = 1.
Correct answer: (D).
tan⁻¹√3 – cot⁻¹(–√3) is equal to (A) π (B) –π/2 (C) 0 (D) 2√3
Answer
tan⁻¹√3 = π/3. cot⁻¹(–√3) = π – cot⁻¹(√3) = π – π/6 = 5π/6.
Value = π/3 – 5π/6 = –π/2. Correct answer: (B).
Miscellaneous Exercise on Chapter 2
Find the value of the following:
cos⁻¹(cos 13π/6)
Answer
cos(13π/6) = cos(2π + π/6) = cos(π/6), and π/6 ∈ [0, π].
Value = π/6.
tan⁻¹(tan 7π/6)
Answer
tan(7π/6) = tan(π + π/6) = tan(π/6), and π/6 ∈ (–π/2, π/2).
Value = π/6.
Prove that
2 sin⁻¹(3/5) = tan⁻¹(24/7)
Answer
Let θ = sin⁻¹(3/5): sin θ = 3/5, cos θ = 4/5, tan θ = 3/4. Since 3/5 < 1/√2, θ ∈ (0, π/4), so 2θ ∈ (0, π/2).
tan 2θ = 2 tan θ/(1 – tan²θ) = (3/2)/(1 – 9/16)
⇒ (3/2)/(7/16)
⇒ 24/7.
So 2θ = tan⁻¹(24/7), i.e. 2 sin⁻¹(3/5) = tan⁻¹(24/7). Hence proved.
sin⁻¹(8/17) + sin⁻¹(3/5) = tan⁻¹(77/36)
Answer
Let A = sin⁻¹(8/17): cos A = 15/17, tan A = 8/15. Let B = sin⁻¹(3/5): cos B = 4/5, tan B = 3/4.
tan(A + B) = (8/15 + 3/4)/(1 – (8/15)(3/4)) = (77/60)/(36/60) = 77/36.
A, B are acute and tan A tan B = 2/5 < 1, so A + B ∈ (0, π/2). Hence A + B = tan⁻¹(77/36). Proved.
cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65)
Answer
Let A = cos⁻¹(4/5): sin A = 3/5. Let B = cos⁻¹(12/13): sin B = 5/13.
cos(A + B) = cos A cos B – sin A sin B = (4/5)(12/13) – (3/5)(5/13) = 48/65 – 15/65 = 33/65.
A, B are acute, so A + B ∈ (0, π), the range of cos⁻¹. Hence A + B = cos⁻¹(33/65). Proved.
cos⁻¹(12/13) + sin⁻¹(3/5) = sin⁻¹(56/65)
Answer
Let A = cos⁻¹(12/13): sin A = 5/13. Let B = sin⁻¹(3/5): cos B = 4/5.
sin(A + B) = sin A cos B + cos A sin B = (5/13)(4/5) + (12/13)(3/5) = 20/65 + 36/65 = 56/65.
Also cos(A + B) = (12/13)(4/5) – (5/13)(3/5) = 33/65 > 0, so A + B ∈ (0, π/2). Hence A + B = sin⁻¹(56/65). Proved.
tan⁻¹(63/16) = sin⁻¹(5/13) + cos⁻¹(3/5)
Answer
Let A = sin⁻¹(5/13): cos A = 12/13, tan A = 5/12. Let B = cos⁻¹(3/5): sin B = 4/5, tan B = 4/3.
tan(A + B) = (5/12 + 4/3)/(1 – (5/12)(4/3)) = (21/12)/(16/36)
⇒ (21/12) × (36/16)
⇒ 63/16.
A, B acute and tan A tan B = 5/9 < 1, so A + B ∈ (0, π/2). Hence sin⁻¹(5/13) + cos⁻¹(3/5) = tan⁻¹(63/16). Proved.
Prove that
tan⁻¹ √x = (1/2) cos⁻¹[(1 – x)/(1 + x)], x ∈ [0, 1]
Answer
Put x = tan²θ, i.e. √x = tan θ and θ = tan⁻¹√x. Since x ∈ [0, 1], θ ∈ [0, π/4].
(1 – x)/(1 + x) = (1 – tan²θ)/(1 + tan²θ) = cos 2θ, with 2θ ∈ [0, π/2] ⊂ [0, π].
RHS = (1/2) cos⁻¹(cos 2θ) = (1/2)(2θ) = θ = tan⁻¹√x = LHS. Hence proved.
cot⁻¹[(√(1 + sin x) + √(1 – sin x))/(√(1 + sin x) – √(1 – sin x))] = x/2, x ∈ (0, π/4)
Answer
1 + sin x = cos²(x/2) + sin²(x/2) + 2 sin(x/2) cos(x/2) = (cos(x/2) + sin(x/2))².
1 – sin x = (cos(x/2) – sin(x/2))².
For x ∈ (0, π/4), x/2 ∈ (0, π/8), so cos(x/2) > sin(x/2) > 0. Hence √(1 + sin x) = cos(x/2) + sin(x/2) and √(1 – sin x) = cos(x/2) – sin(x/2).
Numerator = 2 cos(x/2); denominator = 2 sin(x/2); ratio = cot(x/2).
LHS = cot⁻¹(cot(x/2)) = x/2 (as x/2 ∈ (0, π)). Hence proved.
tan⁻¹[(√(1 + x) – √(1 – x))/(√(1 + x) + √(1 – x))] = π/4 – (1/2) cos⁻¹ x, –1/√2 ≤ x ≤ 1 [Hint: Put x = cos 2θ]
Answer
Put x = cos 2θ, so 2θ = cos⁻¹ x. Since –1/√2 ≤ x ≤ 1, 2θ ∈ [0, 3π/4], i.e. θ ∈ [0, 3π/8].
√(1 + x) = √(2 cos²θ) = √2 cos θ and √(1 – x) = √(2 sin²θ) = √2 sin θ (both cos θ, sin θ ≥ 0 here).
Ratio = (cos θ – sin θ)/(cos θ + sin θ) = (1 – tan θ)/(1 + tan θ) = tan(π/4 – θ).
π/4 – θ ∈ [–π/8, π/4] ⊂ (–π/2, π/2), so LHS = π/4 – θ = π/4 – (1/2) cos⁻¹ x = RHS. Hence proved.
Solve the following equations:
2 tan⁻¹(cos x) = tan⁻¹(2 cosec x)
Answer
Let θ = tan⁻¹(cos x), so tan θ = cos x. Taking tan of both sides: tan 2θ = 2 cosec x.
tan 2θ = 2 tan θ/(1 – tan²θ) = 2 cos x/(1 – cos²x) = 2 cos x/sin²x.
2 cos x/sin²x = 2/sin x
⇒ cos x = sin x (sin x ≠ 0)
⇒ tan x = 1.
x = π/4 (in general x = nπ + π/4, n ∈ Z).
tan⁻¹[(1 – x)/(1 + x)] = (1/2) tan⁻¹ x, (x > 0)
Answer
Let θ = tan⁻¹ x, θ ∈ (0, π/2) since x > 0. Then (1 – x)/(1 + x) = (1 – tan θ)/(1 + tan θ) = tan(π/4 – θ), with π/4 – θ ∈ (–π/4, π/4).
So LHS = π/4 – θ. Equation: π/4 – θ = θ/2
⇒ 3θ/2 = π/4
⇒ θ = π/6.
x = tan(π/6) = 1/√3.
sin(tan⁻¹ x), |x| < 1 is equal to (A) x/√(1 – x²) (B) 1/√(1 – x²) (C) 1/√(1 + x²) (D) x/√(1 + x²)
Answer
Let tan⁻¹ x = θ, θ ∈ (–π/4, π/4), tan θ = x. Then cos θ > 0 and sec θ = √(1 + x²).
sin θ = tan θ · cos θ = x/√(1 + x²).
Correct answer: (D).
sin⁻¹(1 – x) – 2 sin⁻¹ x = π/2, then x is equal to (A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2
Answer
Let sin⁻¹ x = θ, so x = sin θ. Then sin⁻¹(1 – x) = π/2 + 2θ.
1 – x = sin(π/2 + 2θ) = cos 2θ = 1 – 2 sin²θ = 1 – 2x²
⇒ 2x² – x = 0
⇒ x = 0 or x = 1/2.
Check x = 1/2: sin⁻¹(1/2) – 2 sin⁻¹(1/2) = π/6 – π/3 = –π/6 ≠ π/2, rejected.
Check x = 0: sin⁻¹(1) – 0 = π/2 ✓.
x = 0. Correct answer: (C).
Important Extra Questions
Find the principal value of sec⁻¹(–√2).
Answer
cos y = –1/√2
⇒ y = π – π/4 = 3π/4, which lies in [0, π] – {π/2}. Principal value = 3π/4.
Find the value of sin⁻¹(sin 3π/5).
Answer
sin(3π/5) = sin(π – 3π/5) = sin(2π/5), and 2π/5 ∈ [–π/2, π/2]. Value = 2π/5.
Write the domain and range (principal branch) of cos⁻¹ x and tan⁻¹ x.
Answer
cos⁻¹: domain [–1, 1], range [0, π]. tan⁻¹: domain R, range (–π/2, π/2).
Write cot⁻¹[1/√(x² – 1)], x > 1 in simplest form.
Answer
Put x = sec θ, θ ∈ (0, π/2). Then √(x² – 1) = tan θ, so 1/√(x² – 1) = cot θ
⇒ cot⁻¹(cot θ) = θ = sec⁻¹ x.
Find the value of cos(sec⁻¹ x + cosec⁻¹ x), |x| ≥ 1.
Answer
sec⁻¹ x + cosec⁻¹ x = π/2 (for |x| ≥ 1), so the value = cos(π/2) = 0.