Inverse Trigonometric Functions · Lesson 2 of 4
The Remaining Principal Branches and Values
“Complete the six principal inverse functions and use their ranges to select exact angles with confidence.”
• Find the natural domains and ranges of cosecant, secant, tangent and cotangent. • Explain the restrictions that define their principal inverse branches. • Read the domains, excluded values, endpoints and asymptotes of the four inverse graphs. • Distinguish inverse functions from reciprocal expressions. • Evaluate individual principal values and combinations of inverse-trigonometric values.
The same inverse idea, four more functions
Sine and cosine showed why an inverse needs a selected angle interval. We now apply the same idea to cosecant, secant, tangent and cotangent. Their natural domains contain excluded angles, so we must track both the repeating values and the points where a denominator becomes zero.
Recall that cosec θ = 1/sin θ, sec θ = 1/cos θ, tan θ = sin θ/cos θ and cot θ = cos θ/sin θ. Here cosec is the abbreviation for cosecant, also written csc. The integer n in the exclusions below may be positive, negative or zero; it describes every repeating forbidden angle, not just those in one turn.
| Function | Natural angle domain | Natural range |
|---|---|---|
| sin θ and cos θ | ℝ | [−1, 1] |
| cosec θ | ℝ excluding θ = nπ, n ∈ ℤ | (−∞, −1] ∪ [1, ∞) |
| sec θ | ℝ excluding θ = (2n + 1)π/2, n ∈ ℤ | (−∞, −1] ∪ [1, ∞) |
| tan θ | ℝ excluding θ = (2n + 1)π/2, n ∈ ℤ | ℝ |
| cot θ | ℝ excluding θ = nπ, n ∈ ℤ | ℝ |
The reciprocal of a nonzero sine or cosine has absolute value at least 1. This explains the gap (−1, 1) in the ranges of cosecant and secant. The values −1 and 1 remain included, because the sine or cosine denominator can equal −1 or 1. Tangent and cotangent, in contrast, attain every real value on each complete interval between consecutive forbidden angles.
Inverse cosecant keeps the sine interval but removes zero
Cosecant is undefined where sine is zero. The sine principal interval [−π/2, π/2] contains just one such point, 0, so we remove it when defining the principal cosecant restriction. The resulting domain has two pieces, but their positive and negative outputs do not overlap.
On [−π/2, 0), cosecant moves from −1 towards arbitrarily large negative values. On (0, π/2], it moves from arbitrarily large positive values down to 1. Every number at most −1 or at least 1 occurs exactly once across those two pieces. This makes the restriction bijective onto ℝ excluding the open interval (−1, 1).
If y = cosec⁻¹ x, then cosec y = x, or equivalently sin y = 1/x. Because |x| ≥ 1, the number 1/x is in [−1, 1] and is nonzero. Its principal inverse sine therefore lies in exactly the required interval. This gives a convenient evaluation relationship rather than a new branch choice.
Other valid restrictions include [−3π/2, −π/2] with −π removed, and [π/2, 3π/2] with π removed. They give different inverse-cosecant branches. For the principal graph, the endpoint values are cosec⁻¹(−1) = −π/2 and cosec⁻¹(1) = π/2. As the absolute size of the input grows, the output approaches 0; no finite input gives output 0.
Problem
Find cosec⁻¹(−√2).
- 1.The input is allowed because |−√2| ≥ 1. Let y = cosec⁻¹(−√2), so sin y = −1/√2.
- 2.The principal range is [−π/2, π/2] excluding 0. A negative sine in this range needs a negative angle.
- 3.Since sin(−π/4) = −1/√2, that angle satisfies both the ratio condition and the range condition.
- 4.Therefore cosec⁻¹(−√2) = −π/4.
Inverse secant keeps the cosine interval but removes π/2
Secant is undefined where cosine is zero. On the cosine principal interval [0, π], that happens at π/2. Removing this angle gives two reversible pieces: one supplies the positive secant values, and the other supplies the negative values.
On [0, π/2), secant rises from 1 towards arbitrarily large positive values. On (π/2, π], it rises from arbitrarily large negative values to −1. Together these pieces attain every value outside (−1, 1) exactly once. Thus ±1 are allowed inverse inputs, while inputs strictly between −1 and 1 are not.
If y = sec⁻¹ x, then cos y = 1/x. Inverse cosine selects the same angle because its range is [0, π]; the nonzero input 1/x ensures that the output cannot be π/2. Therefore the principal inverse secant can be evaluated through inverse cosine.
Other secant branches can be obtained by restricting to [−π, 0] excluding −π/2, or to [π, 2π] excluding 3π/2. These have different output angle sets. On the principal inverse, the graph approaches the horizontal line y = π/2 as the input tends towards positive or negative infinity, but never reaches that line.
Problem
Find sec⁻¹(−2).
- 1.Let y = sec⁻¹(−2). Then sec y = −2 and cos y = −1/2.
- 2.The principal secant range is [0, π] excluding π/2. The angle −2π/3 has the correct cosine but is outside this range.
- 3.The angle 2π/3 lies in the permitted range and has cosine −1/2.
- 4.Therefore sec⁻¹(−2) = 2π/3.
Inverse tangent uses an open interval around zero
Tangent is continuous and strictly increasing between −π/2 and π/2. It takes every real value there, approaching large negative values near the left boundary and large positive values near the right. The boundary angles are excluded because tangent is undefined at them.
The principal values include tan⁻¹ 0 = 0, tan⁻¹ 1 = π/4 and tan⁻¹(−1) = −π/4. A positive input gives a positive angle, and a negative input gives a negative angle. The graph increases and becomes flatter towards the horizontal asymptotes y = −π/2 and y = π/2.
A tangent branch can instead use any interval (−π/2 + kπ, π/2 + kπ), where k is an integer. For example, (−3π/2, −π/2) and (π/2, 3π/2) each contain exactly one angle for every real tangent value. The principal choice is k = 0.
Problem
Find tan⁻¹(−√3).
- 1.We need an angle y in (−π/2, π/2) satisfying tan y = −√3.
- 2.Since tan(π/3) = √3 and tangent changes sign when its angle changes sign, tan(−π/3) = −√3.
- 3.The angle −π/3 is in the permitted open interval.
- 4.Thus tan⁻¹(−√3) = −π/3. The angle 2π/3 has the same tangent, but it is not the principal value.
Inverse cotangent uses the interval from zero to π
Cotangent decreases continuously from arbitrarily large positive values to arbitrarily large negative values on (0, π). It is undefined at the two boundary angles and equals zero at π/2. The convention in this chapter therefore makes the principal inverse cotangent a function with positive angle outputs, even for negative inputs.
The values cot⁻¹ 1 = π/4 and cot⁻¹ 0 = π/2 lie at familiar points. If x > 0, then cot⁻¹ x lies in (0, π/2); if x < 0, it lies in (π/2, π). Its graph decreases towards 0 for large positive inputs and towards π for large negative inputs. Other branches use intervals (kπ, (k + 1)π), such as (−π, 0) or (π, 2π).
Problem
Find cot⁻¹(−1/√3).
- 1.Let y = cot⁻¹(−1/√3). Then cot y = −1/√3 and 0 < y < π.
- 2.We know cot(π/3) = 1/√3. The supplementary angle has cot(π − π/3) = −cot(π/3).
- 3.Thus cot(2π/3) = −1/√3, and 2π/3 lies in (0, π).
- 4.Therefore cot⁻¹(−1/√3) = 2π/3. The angle −π/3 is not an allowed principal cotangent output.
For negative x, cot⁻¹ x is not tan⁻¹(1/x). Their tangent/cotangent ratios agree, but their principal output intervals differ. For example, cot⁻¹(−1) = 3π/4 while tan⁻¹(−1) = −π/4. Using a reciprocal does not remove the need to check the branch.
One comparison table, six precise choices
The six principal functions can now be viewed together. The input column tells you whether a real value is accepted, while the output column tells you which angle to choose. Brackets include an endpoint, parentheses exclude it, and a listed exception removes a particular output within the surrounding interval.
| Principal function | Input domain | Output range |
|---|---|---|
| sin⁻¹ x | [−1, 1] | [−π/2, π/2] |
| cos⁻¹ x | [−1, 1] | [0, π] |
| cosec⁻¹ x | (−∞, −1] ∪ [1, ∞) | [−π/2, π/2] excluding 0 |
| sec⁻¹ x | (−∞, −1] ∪ [1, ∞) | [0, π] excluding π/2 |
| tan⁻¹ x | ℝ | (−π/2, π/2) |
| cot⁻¹ x | ℝ | (0, π) |
The table also groups related functions. Inverse sine and inverse cosine accept the same bounded inputs but return different angle ranges. Inverse cosecant and inverse secant accept the same two unbounded input pieces; their ranges inherit the sine and cosine restrictions with one undefined angle removed. Inverse tangent and inverse cotangent accept all real inputs, but the former increases through zero while the latter decreases through π/2.
The principal value of an inverse trigonometric expression is its unique angle output in the specified principal range. When no other branch is named, the principal range is understood.
There is also a notation distinction. In sin⁻¹ x, the superscript −1 names the inverse sine function. In (sin x)⁻¹, the parentheses make it the reciprocal of the number sin x. The latter equals 1/sin x = cosec x wherever sin x is nonzero. The same distinction applies to the other trigonometric functions.
Evaluating principal values systematically
An inverse-trigonometric evaluation is a short reasoning problem. First check the input domain, then name the output angle and write the corresponding ordinary trigonometric equation. Choose the angle only after using the principal range; standard trigonometric values then supply exact answers.
Problem
Find sin⁻¹(1/√2).
- 1.The input 1/√2 is in [−1, 1], so the real inverse sine is defined.
- 2.Let y = sin⁻¹(1/√2). Then sin y = 1/√2 and −π/2 ≤ y ≤ π/2.
- 3.The angle π/4 has sine 1/√2 and lies in the required interval.
- 4.Therefore sin⁻¹(1/√2) = π/4. Although 3π/4 has the same sine, it is outside the principal range.
Problem
Evaluate tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
- 1.For inverse tangent, the angle with tangent 1 in (−π/2, π/2) is π/4.
- 2.For inverse cosine, the angle with cosine −1/2 in [0, π] is 2π/3.
- 3.For inverse sine, the angle with sine −1/2 in [−π/2, π/2] is −π/6.
- 4.Add these chosen values: π/4 + 2π/3 − π/6 = (3π + 8π − 2π)/12 = 3π/4.
- 5.The final sum need not lie in any one inverse function’s principal range. Each separate inverse value has already satisfied its own range condition.
Problem
Evaluate tan⁻¹(√3) − sec⁻¹(−2).
- 1.The tangent principal range gives tan⁻¹(√3) = π/3.
- 2.The secant principal range gives sec⁻¹(−2) = 2π/3, since cos(2π/3) = −1/2.
- 3.Subtract the angles: π/3 − 2π/3 = −π/3.
- 4.Do not replace the secant value with a negative angle merely because its input is negative.
Check your understanding
These questions compare the allowed input sets, graph features and exact values. An attractive-looking angle is only correct if both its trigonometric ratio and its principal range agree with the question.
Quiz
Which is the real input domain of sec⁻¹ x?
Which principal inverse-cosecant output is excluded?
Which value equals cot⁻¹(−√3)?
What is sec⁻¹(2/√3)?
What is tan⁻¹(−1)?
Which inverse has principal value π/2 at input 0?
What does (sin x)⁻¹ mean when sin x ≠ 0?
Evaluate cos⁻¹(1/2) + 2 sin⁻¹(1/2).
Practice Problems
- Find the principal values of cos⁻¹(√3/2), cosec⁻¹(2), cos⁻¹(−1/2), and tan⁻¹(−1).
- Find sec⁻¹(2/√3), cot⁻¹(√3), cos⁻¹(−1/√2), and cosec⁻¹(−√2). State the output range for each.
- Classify sin⁻¹(2), sec⁻¹(1/2), tan⁻¹(10), and cot⁻¹(−10) as defined or undefined over the reals.
- List all six inverse functions that are defined at input 1, and find their principal values.
- Evaluate tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2) by choosing each value separately.
- Evaluate cos⁻¹(1/2) + 2 sin⁻¹(1/2), and explain why the sum is not restricted to the inverse-sine range.
- A student uses cot⁻¹(−1/√3) = −π/3. Identify the failed condition and correct the angle.
- Explain why the inverse-tangent graph never reaches its horizontal lines y = ±π/2.
- Write the relationship between principal inverse cosecant and inverse sine of a reciprocal, and give its input restriction.
- Compare sec⁻¹(−1), cosec⁻¹(−1), and cot⁻¹(0), including the endpoints that are allowed.
Key Takeaways
• Cosecant and secant attain only values with absolute value at least 1; both endpoint values ±1 are included. • Inverse cosecant returns [−π/2, π/2] excluding 0; inverse secant returns [0, π] excluding π/2. • Inverse tangent returns (−π/2, π/2), while inverse cotangent returns (0, π); both accept every real input. • A negative input gives a negative principal sine, cosecant or tangent angle, but a positive cosine, secant or cotangent angle. • Principal values require both a correct ordinary trigonometric ratio and membership in the specified output interval. • Evaluate each inverse separately before combining its angle with other values; the resulting sum has no automatic principal-range restriction. • Inverse-function notation and reciprocal notation have different meanings and different domains.