9 min readConcept sheet1-mark MCQ; 2-mark 'write the range' or 'find the range of an expression' questions

Ranges and principal value branches

Each inverse function's range is its principal value branch, the interval chosen to make the original function one-one: [−π/2, π/2] for sin⁻¹ and [0, π] for cos⁻¹.

sin⁻¹, tan⁻¹, cosec⁻¹[−π/2, π/2], (−π/2, π/2) and [−π/2, π/2] − {0}

cos⁻¹, cot⁻¹, sec⁻¹[0, π], (0, π) and [0, π] − {π/2}

Principal valueThe value of an inverse trigonometric function that lies in its principal value branch

Negative inputssin⁻¹, tan⁻¹, cosec⁻¹ give an angle between −π/2 and 0; cos⁻¹, cot⁻¹, sec⁻¹ give an angle between π/2 and π.

How each branch is chosen

Each trigonometric function repeats, so many intervals make it one-one. The principal value branch is the one your textbook fixes as standard. It is chosen so that the function is one-one on it and takes every value of its range. It also includes the angles from 0 to π/2, so a positive input gives an angle from 0 to π/2. Unless a question names another branch, always use the principal one. The range of each inverse function is this branch.

Domain and principal value branch of all six

Function	Domain	Range (principal value branch)
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}

Two families of branches

sin⁻¹, tan⁻¹, cosec⁻¹Branch centred on 0, from −π/2 to π/2 (fourth and first quadrants). A negative input gives a negative angle.

cos⁻¹, cot⁻¹, sec⁻¹Branch from 0 to π (first and second quadrants). A negative input gives an angle between π/2 and π, never a negative one.

Open ends: tan⁻¹ never reaches ±π/2 and cot⁻¹ never reaches 0 or π, because tan(±π/2), cot 0 and cot π are not defined. For the same reason cosec⁻¹ leaves out 0 and sec⁻¹ leaves out π/2.

Example 1: can it be a principal value?

Problem

Which of the angles 3π/4, −π/4 and 5π/4 can be a value of (i) tan⁻¹ x, (ii) cos⁻¹ x?

Working

Rule: a principal value must lie in the principal value branch.
(i) tan⁻¹ has range (−π/2, π/2). Only −π/4 lies in it.
3π/4 and 5π/4 are bigger than π/2, so they are outside.
(ii) cos⁻¹ has range [0, π]. Only 3π/4 lies in it.
−π/4 < 0 and 5π/4 > π, so they are outside.

Answer

(i) Only −π/4. (ii) Only 3π/4.

Example 2: range of an expression (board style)

Problem

Find the range of f(x) = 2 sin⁻¹ x + π/2.

Working

Rule: −π/2 ≤ sin⁻¹ x ≤ π/2.
Multiply by 2: −π ≤ 2 sin⁻¹ x ≤ π.
Add π/2: −π + π/2 ≤ f(x) ≤ π + π/2.
So −π/2 ≤ f(x) ≤ 3π/2.

Answer

The range is [−π/2, 3π/2].

In real life

Your phone's map app turns east-west and north-south readings into a direction using inverse tangent. tan⁻¹ alone gives the same angle for north-east and south-west, so the app also checks the signs of the readings.

Exam angle

1 marks: “Write the principal value branch of sec⁻¹ x.” [0, π] − {π/2}

2 marks: “Find the range of f(x) = 3 cos⁻¹ x − π.” 0 ≤ cos⁻¹ x ≤ π, so 0 ≤ 3 cos⁻¹ x ≤ 3π. Subtract π: −π ≤ f(x) ≤ 2π. The range is [−π, 2π].

1 marks: “Why is the range of tan⁻¹ x an open interval?” tan(π/2) and tan(−π/2) are not defined, so tan⁻¹ x never equals ±π/2. Its range is (−π/2, π/2).

Remember it

Sine family sits around zero (−π/2 to π/2); cosine family sits from 0 to π. Negative input: the sine family goes negative, the cosine family goes past π/2.