Why restrict the domain?
A function has an inverse only if it is a bijection (one-one and onto). sin x is not one-one on ℝ: sin 0 = sin π = 0. But if we restrict its domain to [−π/2, π/2], it becomes one-one and takes every value in [−1, 1] exactly once. On that restricted domain, it has an inverse, written sin⁻¹ x or arcsin x.
sin⁻¹ x means "the angle in [−π/2, π/2] whose sine is x". Note that sin⁻¹ x is not 1/sin x.
The chosen range for each inverse function is its principal value branch, and the value it gives is the principal value.
Domains and principal value ranges
Function | Domain | Principal value range |
|---|---|---|
sin⁻¹ x | [−1, 1] | [−π/2, π/2] |
cos⁻¹ x | [−1, 1] | [0, π] |
tan⁻¹ x | ℝ | (−π/2, π/2) |
cot⁻¹ x | ℝ | (0, π) |
sec⁻¹ x | ℝ − (−1, 1) | [0, π] − {π/2} |
cosec⁻¹ x | ℝ − (−1, 1) | [−π/2, π/2] − {0} |
These ranges are worth memorising: almost every mistake in this chapter comes from giving an answer outside the principal range.
Example. sin⁻¹(1/2) = π/6, since sin π/6 = 1/2 and π/6 lies in [−π/2, π/2]. (5π/6 also has sine 1/2, but it is outside the range.)
Example. cos⁻¹(−1/2) = 2π/3, since cos 2π/3 = −1/2 and 2π/3 lies in [0, π].
Example. tan⁻¹(−√3) = −π/3.
Example. cot⁻¹(−1) = 3π/4, which lies in (0, π).
Graphs
The graph of an inverse function is the reflection of the graph of the original (restricted) function in the line y = x.
sin⁻¹ x rises from (−1, −π/2) to (1, π/2).
cos⁻¹ x falls from (−1, π) to (1, 0).
tan⁻¹ x rises across all of ℝ, approaching the horizontal lines y = −π/2 and y = π/2 but never reaching them.
Properties
1. Composition:
sin(sin⁻¹ x) = x for x ∈ [−1, 1], and sin⁻¹(sin x) = x only for x ∈ [−π/2, π/2].
Similar results hold for the other functions within their principal ranges.
Example. sin⁻¹(sin 2π/3). It is tempting to say 2π/3, but that is outside [−π/2, π/2]. Since sin 2π/3 = sin π/3, the answer is π/3.
2. Reciprocal relations:
sin⁻¹(1/x) = cosec⁻¹ x
cos⁻¹(1/x) = sec⁻¹ x
tan⁻¹(1/x) = cot⁻¹ x, for x > 0
3. Negative arguments:
sin⁻¹(−x) = −sin⁻¹ x, and similarly tan⁻¹ and cosec⁻¹ (odd functions)
cos⁻¹(−x) = π − cos⁻¹ x, and similarly sec⁻¹ and cot⁻¹
4. Complementary pairs:
sin⁻¹ x + cos⁻¹ x = π/2, x ∈ [−1, 1]
tan⁻¹ x + cot⁻¹ x = π/2
sec⁻¹ x + cosec⁻¹ x = π/2, |x| ≥ 1
5. Sum of inverse tangents (for xy < 1):
tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y) ⁄ (1 − xy))
and tan⁻¹ x − tan⁻¹ y = tan⁻¹((x − y)/(1 + xy)).
6. Double angle forms: 2 tan⁻¹ x = tan⁻¹(2x/(1 − x²)) for |x| < 1; also 2 tan⁻¹ x = sin⁻¹(2x/(1 + x²)) for |x| ≤ 1 and cos⁻¹((1 − x²)/(1 + x²)) for x ≥ 0.
Worked examples
Example. Find tan⁻¹ ½ + tan⁻¹ ⅓.
Using the sum formula: tan⁻¹((½ + ⅓)/(1 − ⅙)) = tan⁻¹((5/6)/(5/6)) = tan⁻¹ 1 = π/4.
Example. Find cos(sin⁻¹ ⅗).
Let θ = sin⁻¹ ⅗, so sin θ = 3/5 with θ in [−π/2, π/2], where cosine is non-negative. Then cos θ = √(1 − 9/25) = 4/5.
Example. Simplify tan⁻¹(cos x ⁄ (1 − sin x)) for −π/2 < x < π/2.
Use cos x = sin(π/2 − x) and 1 − sin x = 1 − cos(π/2 − x). Writing θ = π/2 − x, the fraction becomes sin θ/(1 − cos θ) = cot(θ/2) = tan(π/2 − θ/2). The result is π/2 − θ/2 = π/4 + x/2.
Summary
Restrict each trigonometric function to a principal branch to make it invertible.
Principal ranges: sin⁻¹ [−π/2, π/2], cos⁻¹ [0, π], tan⁻¹ (−π/2, π/2).
sin⁻¹ x + cos⁻¹ x = π/2; cos⁻¹(−x) = π − cos⁻¹ x; sin⁻¹(−x) = −sin⁻¹ x.
tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y)/(1 − xy)) when xy < 1.