Questions on how to find domain and range of arcsine functions.

**Theorems**

1.
**y = arcsin x is equivalent to sin y = x**

with **-1 ≤ x ≤ 1 and - pi / 2 ≤ y ≤ pi / 2**

**Question 1:** Find the domain and range of y = arcsin(x - 1)

**Solution to question 1:**

1. Domain: To find the domain of the above function, we need to impose a condition on the argument (x - 1) according to the domain of arcsin(x) which is -1 ≤ x ≤ 1 . Hence

-1 ≤ (x - 1) ≤ 1

solve to obtain domain as: 0 ≤ x ≤ 2

which as expected means that the graph of y = arcsin(x - 1) is that of y = arcsin(x) shifted one unit to the right.

2. Range: A shift to the right does not affect the range. Hence the range of y = arcsin(x - 1) is the same as the range of arcsin(x) which is - pi / 2 ≤ y ≤ pi / 2

**Question 2:** Find the domain and range of y = - arcsin(x + 2)

**Solution to question 2:**

1. Domain: To find the domain of the above function, we need to impose a condition on the argument (x + 2) according to the domain of arcsin(x) which is -1 ≤ x ≤ 1 . Hence

-1 ≤ (x + 2) ≤ 1

solve to obatain domain as: - 3 ≤ x ≤ - 1

which as expected means that the graph of y = arcsin(x + 2) is that of y = arcsin(x) shifted two units to the left.

2. Range: The range of arcsin(x + 2) is the same as the range of arcsin(x) which is - pi / 2 ≤ y ≤ pi / 2. Hence we can write

- pi / 2 ≤ arcsin(x + 2)
≤ pi / 2

We now multiply all terms of the above inequality by - 1 and invert the inequality symbols

pi / 2 ≥ - arcsin(x + 2) ≥ - pi / 2

Which is equivalent to

- pi / 2 ≤ - arcsin(x + 2)≤ pi / 2

which gives the range of y = - arcsin(x + 2) as the interval [- pi / 2 , pi / 2]

**Question 3:** Find the domain and range of y = -2 arcsin(3 x - 1)

**Solution to question 3:**

1. Domain: To find the domain, we need to impose the following condition

-1 ≤ (3 x - 1) ≤ 1

solve to obtain domain as: 0 ≤ x ≤ 2 / 3

2. Range: The range of arcsin(3x - 1) is the same as the range of arcsin(x) which is - pi / 2 ≤ y ≤ pi / 2. Hence we can write

- pi / 2 ≤ arcsin(3x - 1) ≤ pi / 2

We now multiply all terms of the above inequality by - 2 and invert the inequality symbols

pi ≥ - 2 arcsin(3x - 1) ≥ - pi

which gives the range of y = - 2 arcsin(3x - 1) as the interval [- pi , pi]

**Question 4:** Find the domain and range of y = 4 arcsin( -2(x - 1) ) - pi/2

**Solution to question 4:**

1. Domain: To find the domain, we need to impose the following condition

-1 ≤ -2(x - 1) ≤ 1

solve to obtain domain as: 1 / 2 ≤ x ≤ 3 / 2

2. Range: The range of arcsin(-2(x - 1)) is the same as the range of arcsin(x) which is - pi / 2 ≤ y ≤ pi / 2. Hence we can write

- pi / 2 ≤ arcsin(-2(x - 1)) ≤ pi / 2

We now multiply all terms of the above inequality by 4

-2 pi ≥ 4 arcsin(-2(x - 1)) ≥ 2 pi

We now subtract - pi/2 from all terms of the above inequality.

- 5 pi / 2 ≥ 4 arcsin(-2(x - 1)) ≥ 3 pi / 2
which gives the range of y = 4 arcsin(-2(x - 1)) - pi / 2 as the interval [- 5 pi / 2 , 3 pi / 2]