What an inverse trigonometric function is
sin x is not one-one on R, so we restrict its domain to [−π/2, π/2]. There it is a bijection onto [−1, 1], and its inverse is sin⁻¹ x.
The problem with reversing sin x
A function can be reversed (it has an inverse) only if it is one-one and onto: then each output comes from exactly one input. The sine function fails both tests on R. It is periodic: sin 0 = sin π = sin 2π = 0, so it is not one-one. Its values lie only in [−1, 1], so as a map from R to R it is not onto. The fix is to restrict the domain to an interval where sin x takes every value from −1 to 1 exactly once, and to use [−1, 1] as the codomain. On [−π/2, π/2], sin x rises steadily from −1 to 1, so sin: [−π/2, π/2] → [−1, 1] is a bijection. Its inverse is sin⁻¹: [−1, 1] → [−π/2, π/2]. The other five functions are handled in the same way.
Which intervals make sin x one-one?
| Interval for sin x | One-one there? | Values covered | Branch of sin⁻¹ |
|---|---|---|---|
| [−π/2, π/2] | Yes, rises from −1 to 1 | [−1, 1] | Principal value branch |
| [π/2, 3π/2] | Yes, falls from 1 to −1 | [−1, 1] | Another branch |
| [−3π/2, −π/2] | Yes, falls from 1 to −1 | [−1, 1] | Another branch |
| [0, π] | No: sin(π/6) = sin(5π/6) = 1/2 | [0, 1] only | Not usable |
sin⁻¹ x vs (sin x)⁻¹
The −1 in sin⁻¹ is not a power. Write (sin x)⁻¹, with brackets, when you mean the reciprocal.
Example 1: reading the definition
Find sin⁻¹(1/2).
- Rule: y = sin⁻¹ x means sin y = x and −π/2 ≤ y ≤ π/2.
- Let y = sin⁻¹(1/2). Then sin y = 1/2.
- sin(π/6) = 1/2, and π/6 lies in [−π/2, π/2].
- sin(5π/6) is also 1/2, but 5π/6 > π/2, so it is outside the branch.
sin⁻¹(1/2) = π/6.
Example 2: is it defined?
Which of these are defined: sin⁻¹(0.8), sin⁻¹(−1), sin⁻¹(√2), cos⁻¹(π/4)? Give the value where it is a standard angle.
- Rule: sin⁻¹ x and cos⁻¹ x are defined only for −1 ≤ x ≤ 1.
- 0.8 lies in [−1, 1], so sin⁻¹(0.8) is defined.
- −1 lies in [−1, 1], so sin⁻¹(−1) is defined. sin(−π/2) = −1, so sin⁻¹(−1) = −π/2.
- √2 ≈ 1.414 > 1, so sin⁻¹(√2) is not defined.
- π/4 ≈ 0.785 lies in [−1, 1], so cos⁻¹(π/4) is defined, even though it is not a standard angle.
sin⁻¹(0.8), sin⁻¹(−1) = −π/2 and cos⁻¹(π/4) are defined; sin⁻¹(√2) is not.
A wheelchair ramp at a railway station rises 1 m along a sloping surface 12 m long. Engineers use sin⁻¹(1/12) to find its slope angle, about 4.8°, and check that it is gentle enough.
1 marks: “sin⁻¹ x is defined only when x lies in:” [−1, 1], because sin y always lies between −1 and 1.
2 marks: “Why must the domain of sin x be restricted before sin⁻¹ x is defined? Name the interval used for the principal branch.” sin x is periodic, so it is not one-one on R: sin 0 = sin π = 0. On [−π/2, π/2] it rises from −1 to 1, so it is one-one and onto [−1, 1]. That interval gives the principal branch.
sin⁻¹ gives back the angle, not 1 over sine. Put in a number from [−1, 1]; get out an angle in [−π/2, π/2].