Why trig functions need a principal value branch
A function has an inverse only if it is one-one and onto (bijective). sin x repeats every 2π, so many x give the same value. It is not one-one on R.
Fix: restrict the domain to a piece where the function takes every value exactly once. That piece is called the principal value branch. The value the inverse gives from this branch is the principal value.
For sin x we take [−π/2, π/2]. On this interval sin goes smoothly from −1 up to 1 and never repeats. So sin : [−π/2, π/2] → [−1, 1] is a bijection, and its inverse is sin⁻¹ : [−1, 1] → [−π/2, π/2].
Careful: sin⁻¹x does not mean 1/sin x. That is cosec x. The −1 here means inverse, not power.
Domain and range of all six inverse trig functions
| Function | Domain (x) | Principal range (answer) |
|---|---|---|
| sin⁻¹x | [−1, 1] | [−π/2, π/2] |
| cos⁻¹x | [−1, 1] | [0, π] |
| tan⁻¹x | R (all real numbers) | (−π/2, π/2) |
| cot⁻¹x | R | (0, π) |
| sec⁻¹x | (−∞, −1] ∪ [1, ∞) | [0, π] − {π/2} |
| cosec⁻¹x | (−∞, −1] ∪ [1, ∞) | [−π/2, π/2] − {0} |
Why the open ends and gaps? tan is undefined at ±π/2 and cot at 0 and π, so those ends are left open. sec is undefined at π/2 (cos = 0), and cosec at 0 (sin = 0), so those points are removed.
Remember: the domain of the inverse is the range of the original function, and the other way round.
Graphs of inverse trig functions
To draw the graph of an inverse, take the graph of the branch and reflect it in the line y = x. Each point (a, b) goes to (b, a).
- y = sin⁻¹x: an S-shaped piece from (−1, −π/2) to (1, π/2), passing through the origin, always rising.
- y = cos⁻¹x: from (−1, π) down to (1, 0), always falling, passing through (0, π/2).
- y = tan⁻¹x: defined for all x, rising, flattening towards the lines y = π/2 and y = −π/2 but never touching them.
- y = cot⁻¹x: falling from near π to near 0.
- y = sec⁻¹x and y = cosec⁻¹x: two separate parts, one for x ≤ −1 and one for x ≥ 1.
Open the last 3D step and tap each function to see its graph and branch.
How to find a principal value (method)
- Write y = the inverse value, e.g. y = cos⁻¹(−√3/2).
- Change it to the normal trig form: cos y = −√3/2.
- Find an angle with that value (use the standard angle table).
- Check the angle is inside the principal range. If not, use the negative-argument rules.
Negative-argument rules
- sin⁻¹(−x) = −sin⁻¹x, tan⁻¹(−x) = −tan⁻¹x, cosec⁻¹(−x) = −cosec⁻¹x
- cos⁻¹(−x) = π − cos⁻¹x, cot⁻¹(−x) = π − cot⁻¹x, sec⁻¹(−x) = π − sec⁻¹x
Board tip
This chapter often gives a 1–2 mark question: find the principal value, or evaluate an expression like tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2). Always write the range you used.
Try it: find angles with your phone
Lean a ruler on a stack of books. Measure the height h and the length along the ground d. Work out θ = tan⁻¹(h/d) using a calculator (set to degrees). Check with a phone's level/protractor app.
In the 3D free play: set sin⁻¹ and slide x to 0.5, then to −0.5. Notice the answers are π/6 and −π/6: the rule sin⁻¹(−x) = −sin⁻¹x. Now pick cos⁻¹ and do the same. You get π/3 and 2π/3, which add up to π.
Key formulas and definitions
- y = sin⁻¹x ⇔ sin y = x, x ∈ [−1, 1], y ∈ [−π/2, π/2]
- y = cos⁻¹x ⇔ cos y = x, x ∈ [−1, 1], y ∈ [0, π]
- y = tan⁻¹x ⇔ tan y = x, x ∈ R, y ∈ (−π/2, π/2)
- cot⁻¹: x ∈ R, y ∈ (0, π); sec⁻¹: |x| ≥ 1, y ∈ [0, π] − {π/2}; cosec⁻¹: |x| ≥ 1, y ∈ [−π/2, π/2] − {0}
- sin⁻¹(−x) = −sin⁻¹x; tan⁻¹(−x) = −tan⁻¹x; cosec⁻¹(−x) = −cosec⁻¹x
- cos⁻¹(−x) = π − cos⁻¹x; cot⁻¹(−x) = π − cot⁻¹x; sec⁻¹(−x) = π − sec⁻¹x
- Graph of f⁻¹ = reflection of the graph of f (branch) in the line y = x
- sin⁻¹x ≠ 1/sin x (the −1 means inverse, not a power)
Worked examples
1. Find the principal value of sin⁻¹(√3/2).
Let y = sin⁻¹(√3/2). Then sin y = √3/2, with y in [−π/2, π/2]. sin(π/3) = √3/2 and π/3 is in the range. So the answer is π/3.
2. Find the principal value of cos⁻¹(−1/2).
cos⁻¹(−x) = π − cos⁻¹x. cos⁻¹(1/2) = π/3. So cos⁻¹(−1/2) = π − π/3 = 2π/3. Check: 2π/3 is in [0, π] and cos(2π/3) = −1/2.
3. Find the principal value of tan⁻¹(−1).
tan⁻¹(−x) = −tan⁻¹x, and tan⁻¹(1) = π/4. So tan⁻¹(−1) = −π/4, which lies in (−π/2, π/2).
4. Find the principal value of sec⁻¹(2).
Let y = sec⁻¹(2), so sec y = 2, which means cos y = 1/2. In [0, π] − {π/2}, cos(π/3) = 1/2. So the answer is π/3.
5. Find the principal value of cosec⁻¹(−√2).
cosec y = −√2 means sin y = −1/√2. In [−π/2, π/2] − {0}, sin(−π/4) = −1/√2. So cosec⁻¹(−√2) = −π/4.
6. Find the principal value of cot⁻¹(−√3).
cot⁻¹(−x) = π − cot⁻¹x. cot⁻¹(√3) = π/6 (since cot(π/6) = √3). So cot⁻¹(−√3) = π − π/6 = 5π/6, which is in (0, π).
7. Evaluate tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2).
tan⁻¹(1) = π/4. cos⁻¹(−1/2) = 2π/3. sin⁻¹(−1/2) = −π/6. Sum = π/4 + 2π/3 − π/6. Common denominator 12: 3π/12 + 8π/12 − 2π/12 = 9π/12 = 3π/4.
8. Find sin⁻¹(sin 2π/3).
Careful: 2π/3 is NOT in [−π/2, π/2], so the answer is not 2π/3. sin(2π/3) = sin(π − π/3) = sin(π/3). π/3 is in the range. So sin⁻¹(sin 2π/3) = π/3.
9. Find the domain of y = sin⁻¹(2x − 1).
sin⁻¹ accepts only inputs from −1 to 1. So −1 ≤ 2x − 1 ≤ 1. Add 1: 0 ≤ 2x ≤ 2. Divide by 2: 0 ≤ x ≤ 1. Domain = [0, 1].
Common mistakes
- Reading sin⁻¹x as 1/sin x. The reciprocal is cosec x; sin⁻¹x is an angle.
- Giving an answer outside the principal range, like cos⁻¹(−1/2) = −π/3 or sin⁻¹(sin 2π/3) = 2π/3.
- Using sin⁻¹(−x) = π − sin⁻¹x. That rule is for cos⁻¹, cot⁻¹ and sec⁻¹; for sin⁻¹, tan⁻¹ and cosec⁻¹ the minus just comes out.
- Forgetting domain limits: sin⁻¹(2) and cos⁻¹(−3) are not defined, and sec⁻¹(0.5) is not defined.