Inverse Trigonometric Functions · Lesson 1 of 4
Why Inverse Trigonometry Needs Branches
“See how restricting an angle interval makes sine and cosine reversible, and learn to read their principal inverses.”
• Explain why repeated trigonometric values prevent an inverse on the full natural domain. • Describe how an inverse exchanges a function’s domain and range. • Distinguish a principal branch from other valid branches. • Use the principal domains and ranges of inverse sine and inverse cosine. • Interpret inverse graphs as reflections across y = x and use composition with the correct restrictions.
Recovering an angle from a ratio
A trigonometric function takes an angle and returns a number. For example, sine takes π/6 and returns 1/2. An inverse trigonometric function asks the reverse question: which angle has the given ratio? This is useful whenever a measurement gives a ratio and the unknown is an angle; it also provides functions used later in calculus, science and engineering.
Throughout this chapter, angles are in radians. An angle of π radians is 180°, so π/2, π/3, π/4 and π/6 correspond to 90°, 60°, 45° and 30°. A value such as sin θ is a real number, while sin⁻¹ x returns an angle. Keeping these two roles separate makes the domain and range restrictions easier to understand.
The difficulty is that a ratio does not determine a unique angle among all real angles. Both π/6 and 5π/6 have sine 1/2, and adding 2π to either angle does not change its sine. Cosine also repeats its values. If an inverse were allowed to return all these angles at once, it would not be a function: a function must assign exactly one output to each allowed input.
A function is one-to-one when different inputs have different outputs. It is onto a declared codomain when every element of that codomain occurs as an output. A function that has both properties is bijective; only then can it be reversed as a function between those declared sets.
Sine and cosine on ℝ are onto [−1, 1], but they are not one-to-one. Similarly, the other four trigonometric functions are onto their natural ranges but repeat values. The obstruction here is repetition, not a failure to attain their natural ranges. We therefore choose a smaller angle domain on which every output occurs once.
What is exchanged when a function is inverted?
An inverse reverses each input-output pair of a bijection. The numbers that were outputs become the allowed inputs, and the original inputs become the possible outputs. Restricting a trigonometric function changes the set of angles that the inverse may return.
If g = f⁻¹, then g is itself bijective, its inverse is f, and applying one function after the other returns the starting value. These conclusions refer to the chosen sets X and Y. For trigonometric functions, choosing X carefully is what makes the conclusion possible.
Problem
Why can sine on [−π/2, π/2] be inverted onto [−1, 1]?
- 1.At the left endpoint, sin(−π/2) = −1. At the right endpoint, sin(π/2) = 1.
- 2.Between those endpoints sine increases strictly: a larger angle in this interval has a larger sine. Thus no two distinct inputs share an output.
- 3.The continuous curve passes through every value from −1 to 1. Therefore it is onto [−1, 1].
- 4.Both requirements hold. Each number in [−1, 1] can be sent back to its unique angle in [−π/2, π/2].
Inverse sine and its principal branch
The interval [−π/2, π/2] provides a particularly useful reversible part of sine. It contains 0, allows both negative and positive angles, and covers every sine value. By convention, this is the branch used when sin⁻¹ is written without specifying another branch.
A branch is an inverse obtained after selecting a domain on which the original function is bijective onto its full natural range. The principal branch is the conventionally selected branch. A principal value is the angle returned within that branch’s specified range.
Both endpoints are included: sin⁻¹(−1) = −π/2 and sin⁻¹(1) = π/2. Also sin⁻¹(0) = 0. If the input is positive, the principal angle is positive; if it is negative, the principal angle is negative. An input such as 3/2 is not allowed for a real inverse sine, because no real angle has sine greater than 1.
Other choices are possible. Sine decreases from 1 to −1 on [π/2, 3π/2], and it also covers [−1, 1] one-to-one on [−3π/2, −π/2]. Each restriction produces a different inverse branch. A ratio such as 1/2 therefore gives π/6 on the principal branch but 5π/6 on the branch with range [π/2, 3π/2]. Choosing a branch selects one answer; it does not change the sine of that answer.
Problem
Find the principal inverse sine of 1/2, and compare it with the inverse whose output must lie in [π/2, 3π/2].
- 1.The principal range is [−π/2, π/2]. Since sin(π/6) = 1/2 and π/6 is in that range, sin⁻¹(1/2) = π/6.
- 2.On [π/2, 3π/2], the angle 5π/6 has sine 1/2 and belongs to the required interval.
- 3.The other branch returns 5π/6. Writing sin⁻¹(1/2) without a branch qualification means π/6.
- 4.Neither value is arbitrary: each is the unique answer allowed by its chosen range.
Seeing an inverse as a reflection
A point (a, b) on a graph records that input a produces output b. Reversing the pair gives (b, a), so an inverse graph is the reflection of the restricted original graph in the diagonal y = x. The principal restriction must be made before the reflected curve is treated as the graph of a function.
For sine, the point (π/6, 1/2) becomes (1/2, π/6) on inverse sine. The endpoints (−π/2, −1) and (π/2, 1) become (−1, −π/2) and (1, π/2). This coordinate exchange explains the inverse’s horizontal domain [−1, 1] and vertical range [−π/2, π/2].
Reflecting the entire periodic sine graph gives an inverse relation with many angle outputs for some inputs. It is not a single inverse function. Use one bijective branch before labelling the reflected curve y = sin⁻¹ x.
Inverse cosine chooses a different angle interval
Cosine is strictly decreasing from 1 to −1 on [0, π]. This interval gives exactly one angle for every number in [−1, 1], so it can be used to define inverse cosine. Its principal range is different from that of inverse sine, even though the two inverses accept the same inputs.
Here cos⁻¹(1) = 0, cos⁻¹(0) = π/2 and cos⁻¹(−1) = π. A negative input produces an angle between π/2 and π, rather than a negative output. As x increases, cos⁻¹ x decreases; compare this with the increasing graph of sin⁻¹ x.
The intervals [−π, 0] and [π, 2π] give other valid cosine branches. For example, cosine rises from −1 to 1 on [−π, 0], and its inverse on that branch has range [−π, 0]. Unless explicitly stated otherwise, cos⁻¹ always refers here to the branch with range [0, π].
Problem
Find sin⁻¹(−1/2) and cos⁻¹(−1/2).
- 1.For inverse sine, the output must lie in [−π/2, π/2]. The angle −π/6 is in that interval and has sine −1/2.
- 2.Therefore sin⁻¹(−1/2) = −π/6.
- 3.For inverse cosine, the output must lie in [0, π]. The angle 2π/3 is in that interval and has cosine −1/2.
- 4.Therefore cos⁻¹(−1/2) = 2π/3. The input number is the same, but the function and principal range are different.
Composition needs the declared range
There are two possible orders for composing a trigonometric function with its inverse. Starting with an allowed ratio, inverse sine finds an angle and sine returns that ratio. Starting with an angle, sine loses information about which repeating branch the angle came from; inverse sine can return the starting angle only when it belongs to the selected interval.
Problem
Evaluate sin(sin⁻¹(3/5)) and cos⁻¹(cos(2π/3)).
- 1.The input 3/5 is in [−1, 1]. Let α = sin⁻¹(3/5). By definition, sin α = 3/5, so the first expression equals 3/5.
- 2.The angle 2π/3 belongs to [0, π], the principal cosine range.
- 3.Thus cos⁻¹(cos(2π/3)) = 2π/3. The two outputs have different roles: one is a ratio and the other is an angle.
- 4.The argument for the second expression would need an extra principal-angle check if its inner angle lay outside [0, π].
The superscript −1 in sin⁻¹ x names an inverse function. It does not mean 1/sin x. Also, cancelling inverse symbols is justified only after checking the allowed input or principal angle interval.
Check your understanding
For each question, identify whether the input is a ratio or an angle. Then use the definition of the inverse and its specified range to decide which answer can be correct.
Quiz
Why does sine on ℝ fail to have a single-valued inverse onto all of ℝ?
If f : X → Y is bijective, which statement describes f⁻¹?
Which is the principal range of sin⁻¹ x?
What is cos⁻¹(−1/2)?
Which point lies on inverse sine because (π/6, 1/2) lies on the selected sine graph?
Which expression is undefined as a real principal inverse?
Which statement has the necessary condition for sin⁻¹(sin θ) = θ?
On the inverse-sine branch with range [π/2, 3π/2], which angle is returned for input 1/2?
Practice Problems
- Explain why adding 2π to an angle prevents the unrestricted sine function from being one-to-one.
- Find sin⁻¹(−√3/2), sin⁻¹(1), and cos⁻¹(−1). State the range used for each answer.
- List three points on restricted cosine on [0, π] and exchange their coordinates to obtain points on inverse cosine.
- A student writes cos⁻¹(−1/2) = −2π/3. Explain why the cosine value is correct but the inverse value is not.
- Evaluate sin(sin⁻¹(−4/5)) and cos⁻¹(cos(π/4)), explaining why each composition is valid.
- Compare the principal inverse sine of −1/2 with the inverse on [π/2, 3π/2]. Which output lies in each range?
- Decide whether sin⁻¹ 0 and (sin 0)⁻¹ are defined. Explain the difference in notation.
- Sketch the principal inverse-sine and inverse-cosine graphs, labelling their endpoints and whether each increases or decreases.
Key Takeaways
• Trigonometric values repeat, so a bijective restriction is needed before an inverse can return one angle. • An inverse exchanges the chosen domain and range and swaps each graph point’s coordinates. • Inverse sine accepts [−1, 1] and returns [−π/2, π/2]; inverse cosine accepts [−1, 1] and returns [0, π]. • Other branch choices give other valid inverses, but unqualified inverse notation uses the principal branch. • Inverse graphs reflect restricted original graphs across y = x. • Composition identities require the correct ratio domain or principal angle range; inverse notation is not reciprocal notation.
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Next · Lesson 2
The Remaining Principal Branches and Values