Inverse Trigonometric
Functions
A complete introductory guide for students
You already know that sin(30°) = ½. But what if someone asks: "Which angle has a sine of ½?" — that's exactly what inverse trig functions answer.
If y = sin(x), then the inverse is written as x = sin⁻¹(y) or x = arcsin(y). It gives us the angle whose trig ratio equals the given value.
sin⁻¹(x) does NOT mean 1/sin(x). The superscript ⁻¹ here denotes the inverse function, not a reciprocal. For reciprocal, we write cosec(x).
Corresponding to the six trigonometric functions, we have six inverse functions:
Remember: sin⁻¹ and tan⁻¹ have the same range [−π/2, π/2], while cos⁻¹ lives in [0, π]. Think: sine and tangent are "symmetric about origin."
A function must be one-to-one to have an inverse. Since sin(x) repeats its values (e.g., sin(30°) = sin(150°) = ½), we restrict its domain to a chosen interval called the Principal Value Branch.
The value of an inverse trig function that lies in its principal value branch is called the principal value. This is always the answer you report unless the question asks for general solutions.
| Function | Principal Value Branch | In Degrees |
|---|---|---|
| sin⁻¹(x) | [−π/2, π/2] | [−90°, 90°] |
| cos⁻¹(x) | [0, π] | [0°, 180°] |
| tan⁻¹(x) | (−π/2, π/2) | (−90°, 90°) |
| cosec⁻¹(x) | [−π/2, π/2] \ {0} | [−90°, 90°] \ {0°} |
| sec⁻¹(x) | [0, π] \ {π/2} | [0°, 180°] \ {90°} |
| cot⁻¹(x) | (0, π) | (0°, 180°) |
| Value | sin⁻¹ | cos⁻¹ | tan⁻¹ |
|---|---|---|---|
| 0 | 0 | π/2 | 0 |
| 1/2 | π/6 | π/3 | — |
| 1/√2 | π/4 | π/4 | — |
| √3/2 | π/3 | π/6 | — |
| 1 | π/2 | 0 | π/4 (approx) |
| √3 | — | — | π/3 |
| −1/2 | −π/6 | 2π/3 | — |
| −1 | −π/2 | π | −π/4 |
For sin⁻¹ and tan⁻¹: negative values give negative angles (odd functions). For cos⁻¹: a negative input gives an angle greater than π/2 (neither odd nor even).
These identities are frequently tested. Learn them carefully.
A. Composition with trig functions:
- sin(sin⁻¹ x) = x, x ∈ [−1, 1]
- cos(cos⁻¹ x) = x, x ∈ [−1, 1]
- sin⁻¹(sin x) = x, x ∈ [−π/2, π/2]
- cos⁻¹(cos x) = x, x ∈ [0, π]
- tan(tan⁻¹ x) = x, x ∈ ℝ
- tan⁻¹(tan x) = x, x ∈ (−π/2, π/2)
B. Complementary angle identities:
- sin⁻¹ x + cos⁻¹ x = π/2
- tan⁻¹ x + cot⁻¹ x = π/2
- sec⁻¹ x + cosec⁻¹ x = π/2
C. Negative argument (odd/even) identities:
- sin⁻¹(−x) = −sin⁻¹(x)
- tan⁻¹(−x) = −tan⁻¹(x)
- cosec⁻¹(−x) = −cosec⁻¹(x)
- cos⁻¹(−x) = π − cos⁻¹(x)
- sec⁻¹(−x) = π − sec⁻¹(x)
- cot⁻¹(−x) = π − cot⁻¹(x)
D. Reciprocal identities:
- sin⁻¹(1/x) = cosec⁻¹(x), |x| ≥ 1
- cos⁻¹(1/x) = sec⁻¹(x), |x| ≥ 1
- tan⁻¹(1/x) = cot⁻¹(x), x > 0
- tan⁻¹(1/x) = −π + cot⁻¹(x), x < 0
E. Addition formulas for tan⁻¹:
tan⁻¹ x + tan⁻¹ y = tan⁻¹[(x+y)/(1−xy)] (if xy < 1)
tan⁻¹ x − tan⁻¹ y = tan⁻¹[(x−y)/(1+xy)]
2 tan⁻¹ x = tan⁻¹[2x/(1−x²)] (if |x| < 1)
- We need angle θ such that sin θ = −1/2 and θ ∈ [−π/2, π/2].
- We know sin(π/6) = 1/2, and since sin⁻¹ is an odd function:
- sin⁻¹(−1/2) = −sin⁻¹(1/2) = −π/6 ✓
- Note that 2π/3 ∉ [−π/2, π/2], so we cannot directly say the answer is 2π/3.
- Rewrite: sin(2π/3) = sin(π − π/3) = sin(π/3).
- Now π/3 ∈ [−π/2, π/2], so: sin⁻¹(sin 2π/3) = π/3 ✓
- Use the identity: sin⁻¹ x + cos⁻¹ x = π/2
- Therefore: cos⁻¹ x = π/2 − π/5 = 3π/10 ✓
- Here x = 1/2, y = 1/3, and xy = 1/6 < 1, so we use the addition formula.
- = tan⁻¹[(1/2 + 1/3) / (1 − 1/6)] = tan⁻¹[(5/6) / (5/6)] = tan⁻¹(1)
- = π/4 ✓