Questions on how to find domain and range of arccosine functions.
## Theorem
1.
**y = arccos x is equivalent to cos y = x**
with **-1 ≤ x ≤ 1 and 0 ≤ y ≤ pi **## Questione with Detailed Solutions
## Question 1
Find the domain and range of y = arccos(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 + ) according to the domain of arccos(x) which is -1 ≤ x ≤ 1 . Hence
-1 ≤ (x + 1) ≤ 1
solve to obtain domain as: - 2 ≤ x ≤ 0
which as expected means that the graph of y = arccos(x + 1) is that of y = arccos(x) shifted one unit to the left.
2. Range: A shift to the left does not affect the range. Hence the range of y = arccos(x + 1) is the same as the range of arcsin(x) which is 0 ≤ y ≤ pi
## Question 2
Find the domain and range of y = - arccos(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 arccos(x) which is -1 ≤ x ≤ 1 . Hence
-1 ≤ (x - 2) ≤ 1
solve to obatain domain as: 1 ≤ x ≤ 3
which as expected means that the graph of y = arccos(x - 2) is that of y = arcsin(x) shifted two units to the right.
2. Range: The range of arccos(x - 2) is the same as the range of arccos(x) which is 0 ≤ y ≤ pi. Hence we can write
0 ≤ arcsin(x + 2) ≤ pi
We now multiply all terms of the above inequality by - 1 and invert the inequality symbols
0 ≥ - arcsin(x + 2) ≥ pi
which gives the range of y = - arccos(x + 2) as the interval [0 , pi]
## Question 3
Find the domain and range of y = -2 arccos(- 2 x + 1)
__Solution to question 3__
1. Domain: To find the domain, we need to impose the following condition
-1 ≤ (- 2 x + 1) ≤ 1
solve to obtain domain as: 0 ≤ x ≤ 1
2. Range: The range of arccos(- 2x + 1) is the same as the range of arccos(x) which is 0 ≤ y ≤ pi. Hence we can write
0 ≤ arccos(- 2x + 1) ≤ pi
We now multiply all terms of the above inequality by - 2 and invert the inequality symbols
0 ≥ - 2 arccos(3x - 1) ≥ - 2 pi
which gives the range of y = - 2 arccos(- 2x + 1) as the interval [- 2 pi]
## Question 4
Find the domain and range of y = 2 arccos( 2 x ) + pi/2
__Solution to question 4__
1. Domain: To find the domain, we need to impose the following condition
-1 ≤ 2x ≤ 1
solve to obtain domain as: - 1 / 2 ≤ x ≤ 1 / 2
2. Range: The range of arccos(2x) is the same as the range of arccos(x) which is 0 ≤ y ≤ pi. Hence we can write
0 ≤ arccos(2x) ≤ pi
We now multiply all terms of the above inequality by 2
0 ≥ 2 arccos(2x) ≥ 2 pi
We now subtract pi/2 to all terms of the above inequality.
pi / 2 ≥ 2 arccos(2x) + pi / 2 ≥ 5 pi / 2
which gives the range of 2 arccos(2x) + pi / 2 as the interval [ pi / 2 , 5 pi / 2]
## More References and Links to Inverse Trigonometric FunctionsInverse Trigonometric Functions
Graph, Domain and Range of Arcsin function
Graph, Domain and Range of Arctan function
Find Domain and Range of Arccosine Functions
Find Domain and Range of Arcsine Functions
Solve Inverse Trigonometric Functions Questions |