1. Chapter at a Glance
If sin θ = ½, what is θ? It could be 30°, 150°, 390°, … — infinitely many answers, because sine is not one-one. To "undo" sine we must restrict its domain to an interval where it is one-one and onto, and then define the inverse. This gives the inverse trigonometric functions sin⁻¹, cos⁻¹, tan⁻¹ and the others, each with a carefully chosen principal value branch. They are essential in calculus (integration produces them often) and in engineering for finding angles from measurements.
- Why trigonometric functions need restricted domains to have inverses.
- Domain, range (principal value branch) of the six inverse functions.
- Graphs of sin⁻¹x, cos⁻¹x, tan⁻¹x.
- Principal values.
- Key properties: sin⁻¹(−x) = −sin⁻¹x, cos⁻¹(−x) = π − cos⁻¹x, etc.
- Simplifying expressions like sin(sin⁻¹x), sin⁻¹(sin x).
2. Principal Value Branches
sin x on [−π/2, π/2] is one-one and takes every value in [−1, 1] exactly once, so we define sin⁻¹ : [−1, 1] → [−π/2, π/2]: y = sin⁻¹x ⇔ sin y = x. The others are defined similarly.
| Function | Domain | Range (principal value branch) |
|---|---|---|
| sin⁻¹x | [−1, 1] | [−π/2, π/2] |
| cos⁻¹x | [−1, 1] | [0, π] |
| tan⁻¹x | ℝ | (−π/2, π/2) |
| cot⁻¹x | ℝ | (0, π) |
| sec⁻¹x | ℝ − (−1, 1) | [0, π] − {π/2} |
| cosec⁻¹x | ℝ − (−1, 1) | [−π/2, π/2] − {0} |
sin⁻¹x means the inverse sine (arcsin x), not 1/sin x. The reciprocal is (sin x)⁻¹ = cosec x.
3. Principal Values
Find the principal values of (a) sin⁻¹(1/√2) (b) cos⁻¹(−½) (c) tan⁻¹(−√3) (d) cosec⁻¹(2).
Solution: (a) π/4. (b) In [0, π], cos θ = −½ → 2π/3. (c) In (−π/2, π/2) → −π/3. (d) sin θ = ½ → π/6.
Evaluate tan⁻¹(1) + cos⁻¹(−½) + sin⁻¹(−½).
Solution: π/4 + 2π/3 − π/6 = (3π + 8π − 2π)/12 = 3π/4.
4. Properties
- sin⁻¹(−x) = −sin⁻¹x; tan⁻¹(−x) = −tan⁻¹x; cosec⁻¹(−x) = −cosec⁻¹x (odd functions).
- cos⁻¹(−x) = π − cos⁻¹x; sec⁻¹(−x) = π − sec⁻¹x; cot⁻¹(−x) = π − cot⁻¹x.
- sin⁻¹(1/x) = cosec⁻¹x, cos⁻¹(1/x) = sec⁻¹x (|x| ≥ 1).
- tan⁻¹(1/x) = cot⁻¹x for x > 0.
- sin⁻¹x + cos⁻¹x = π/2 (|x| ≤ 1); tan⁻¹x + cot⁻¹x = π/2.
- sin(sin⁻¹x) = x for x ∈ [−1, 1] — always true on the domain.
- sin⁻¹(sin x) = x only for x ∈ [−π/2, π/2]; otherwise find the angle in the principal branch with the same sine.
Find sin⁻¹(sin 3π/5) and cos⁻¹(cos 7π/6).
Solution: 3π/5 ∉ [−π/2, π/2]; sin(3π/5) = sin(π − 3π/5) = sin(2π/5), and 2π/5 is in the branch → 2π/5. cos(7π/6) = cos(2π − 7π/6) = cos(5π/6), and 5π/6 ∈ [0, π] → 5π/6.
Find tan(cos⁻¹(3/5)).
Solution: Let θ = cos⁻¹(3/5) ∈ [0, π/2): cos θ = 3/5, sin θ = 4/5 → tan θ = 4/3.
5. Solved Problems (Board Pattern)
Evaluate cos⁻¹(cos 13π/6).
Solution: 13π/6 = 2π + π/6; cos(13π/6) = cos(π/6), and π/6 ∈ [0, π] → π/6.
Find the domain of f(x) = sin⁻¹(2x − 1).
Solution: −1 ≤ 2x − 1 ≤ 1 → 0 ≤ x ≤ 1. Domain [0, 1].
Show that sin⁻¹x + cos⁻¹x = π/2 for x ∈ [−1, 1].
Solution: Let sin⁻¹x = θ ∈ [−π/2, π/2]. Then x = sin θ = cos(π/2 − θ), and π/2 − θ ∈ [0, π], so cos⁻¹x = π/2 − θ. Adding gives π/2. ∎
6. Master Formula Table
| Result | Formula |
|---|---|
| sin⁻¹ branch | [−π/2, π/2], domain [−1, 1] |
| cos⁻¹ branch | [0, π], domain [−1, 1] |
| tan⁻¹ branch | (−π/2, π/2), domain ℝ |
| Odd | sin⁻¹(−x) = −sin⁻¹x, tan⁻¹(−x) = −tan⁻¹x |
| Supplement | cos⁻¹(−x) = π − cos⁻¹x |
| Complementary | sin⁻¹x + cos⁻¹x = π/2 |
| Reciprocal | cosec⁻¹x = sin⁻¹(1/x) |
| Composition | sin(sin⁻¹x) = x; sin⁻¹(sin x) = x only on the branch |
7. Master Summary Table
| Question type | Method | Watch out |
|---|---|---|
| Principal value | Find angle in the branch | cos⁻¹ of negatives → second quadrant |
| Sum of values | Evaluate each, then add | Signs of sin⁻¹, tan⁻¹ of negatives |
| sin⁻¹(sin x) | Bring angle into branch | Use sin(π − x) = sin x |
| trig(inverse trig) | Right-triangle method | Sign in the branch |
| Domain | Argument within allowed set | sec⁻¹, cosec⁻¹ need |x| ≥ 1 |
8. Mind Map
- Restrict domain
- Make bijective
- sin⁻¹ [−π/2, π/2]
- cos⁻¹ [0, π]
- Negative arguments
- sin⁻¹ + cos⁻¹ = π/2
- sin(sin⁻¹x) = x
- sin⁻¹(sin x): adjust
9. Tips and Tricks
- "sin⁻¹ and tan⁻¹ keep the sign; cos⁻¹ of a negative uses π − ."
- For tan(cos⁻¹x) type questions, draw a right triangle with the given ratio.
- Always check your final angle lies in the principal branch.
- Giving cos⁻¹(−½) = −π/3 (not in [0, π]).
- Writing sin⁻¹(sin 2π/3) = 2π/3 (correct answer π/3).
- Treating sin⁻¹x as 1/sin x.
10. Important One-liners
- Inverse trig functions exist only after restricting domains.
- The principal value of cos⁻¹x always lies in [0, π].
- sin⁻¹x + cos⁻¹x = π/2.
- tan⁻¹x is defined for every real x.
- The graph of an inverse function is the reflection in y = x.
11. Exam-pattern Questions with Detailed Answers
The principal value of cos⁻¹(−√3/2) is:
- −π/6
- π/6
- 5π/6
- 7π/6
Show answer
π − π/6 = 5π/6.
Answer: (C)
Find the principal value of sec⁻¹(2/√3).
Show answer
cos θ = √3/2 → π/6.
Answer: π/6
Find the value of cot⁻¹(−1).
Show answer
π − cot⁻¹(1) = π − π/4 = 3π/4.
Answer: 3π/4
Evaluate cos⁻¹(½) + 2 sin⁻¹(½).
Show answer
π/3 + 2(π/6) = 2π/3.
Answer: 2π/3
Find tan⁻¹(tan 3π/4).
Show answer
tan(3π/4) = −1 = tan(−π/4), and −π/4 ∈ (−π/2, π/2). Answer −π/4.
Answer: −π/4
Find sin(π/3 − sin⁻¹(−½)).
Show answer
sin⁻¹(−½) = −π/6 → sin(π/3 + π/6) = sin(π/2) = 1.
Answer: 1
Find the domain of cos⁻¹(x² − 4).
Show answer
−1 ≤ x² − 4 ≤ 1 → 3 ≤ x² ≤ 5 → x ∈ [−√5, −√3] ∪ [√3, √5].
Answer: [−√5, −√3] ∪ [√3, √5]
Find sin(2 tan⁻¹(3/4)).
Show answer
θ = tan⁻¹(3/4): sin θ = 3/5, cos θ = 4/5. sin 2θ = 2 × 3/5 × 4/5 = 24/25.
Answer: 24/25
Evaluate sin⁻¹(sin 10) (angle in radians).
Show answer
10 − 3π ≈ 0.575 and sin(10) = sin(3π + (10 − 3π)) = −sin(10 − 3π). So sin 10 = sin(3π − 10), and 3π − 10 ≈ −0.575 ∈ [−π/2, π/2]. Answer 3π − 10.
Answer: 3π − 10
Write cot⁻¹(1/√(x² − 1)), |x| > 1, in simplest form (take x > 1).
Show answer
Let x = sec θ, θ ∈ (0, π/2). √(x² − 1) = tan θ, so 1/√(x² − 1) = cot θ → expression = θ = sec⁻¹x.
Answer: sec⁻¹x
Show that tan⁻¹x + tan⁻¹(1/x) = π/2 for x > 0, and find its value for x < 0.
Show answer
For x > 0, tan⁻¹(1/x) = cot⁻¹x, and tan⁻¹x + cot⁻¹x = π/2. For x < 0, both terms are negative: the sum = −(tan⁻¹|x| + tan⁻¹(1/|x|)) = −π/2.
Answer: π/2 (x > 0); −π/2 (x < 0)
Find the value of cos(sec⁻¹x + cosec⁻¹x) for |x| ≥ 1.
Show answer
sec⁻¹x + cosec⁻¹x = cos⁻¹(1/x) + sin⁻¹(1/x) = π/2. cos(π/2) = 0.
Answer: 0
12. Quick Revision Summary
- Restrict domains to make trig functions bijective, then invert.
- Branches: sin⁻¹ [−π/2, π/2], cos⁻¹ [0, π], tan⁻¹ (−π/2, π/2), cot⁻¹ (0, π).
- sin⁻¹(−x) = −sin⁻¹x; cos⁻¹(−x) = π − cos⁻¹x.
- sin⁻¹x + cos⁻¹x = tan⁻¹x + cot⁻¹x = π/2.
- sin⁻¹(sin x) = x only within the branch.