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Sagot :
Certainly! Let's find the reference angles for the given angles in radians: [tex]\(\frac{13 \pi}{5}\)[/tex], [tex]\(-\frac{8 \pi}{5}\)[/tex], [tex]\(\frac{2 \pi}{5}\)[/tex], [tex]\(-\frac{2 \pi}{5}\)[/tex], and [tex]\(\frac{8 \pi}{5}\)[/tex]. We will go through each angle step by step:
### 1. Reference Angle of [tex]\(\frac{13 \pi}{5}\)[/tex]
1.1. Normalize the angle to lie between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex]:
[tex]\[ \frac{13 \pi}{5} \mod 2 \pi = \frac{13 \pi}{5} - 2 \pi \left\lfloor \frac{\frac{13 \pi}{5}}{2 \pi} \right\rfloor = \frac{13 \pi}{5} - 2 \pi \cdot 1 = \frac{13 \pi}{5} - \frac{10 \pi}{5} = \frac{3 \pi}{5} \][/tex]
1.2. Identify the quadrant ([tex]\(\frac{3 \pi}{5}\)[/tex] is in the first quadrant):
Since it is already in the first quadrant, the reference angle is:
[tex]\[ \boxed{\frac{3 \pi}{5}} \][/tex]
### 2. Reference Angle of [tex]\(-\frac{8 \pi}{5}\)[/tex]
2.1. Normalize the angle to lie between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex]:
[tex]\[ -\frac{8 \pi}{5} \mod 2 \pi = -\frac{8 \pi}{5} + 2 \pi \left\lceil \frac{-\frac{8 \pi}{5}}{2 \pi} \right\rceil = -\frac{8 \pi}{5} + 2 \pi \cdot 2 = -\frac{8 \pi}{5} + \frac{10 \pi}{5} = \frac{2 \pi}{5} \][/tex]
2.2. Identify the quadrant ([tex]\(\frac{2 \pi}{5}\)[/tex] is in the first quadrant):
Since it is already in the first quadrant, the reference angle is:
[tex]\[ \boxed{\frac{2 \pi}{5}} \][/tex]
### 3. Reference Angle of [tex]\(\frac{2 \pi}{5}\)[/tex]
3.1. This angle is already between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex], and lies in the first quadrant.
The reference angle is:
[tex]\[ \boxed{\frac{2 \pi}{5}} \][/tex]
### 4. Reference Angle of [tex]\(-\frac{2 \pi}{5}\)[/tex]
4.1. Normalize the angle to lie between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex]:
[tex]\[ -\frac{2 \pi}{5} \mod 2 \pi = -\frac{2 \pi}{5} + 2 \pi \left\lceil \frac{-\frac{2 \pi}{5}}{2 \pi} \right\rceil = -\frac{2 \pi}{5} + 2 \pi \cdot 1 = -\frac{2 \pi}{5} + \frac{10 \pi}{5} = \frac{8 \pi}{5} \][/tex]
4.2. Identify the quadrant ([tex]\(\frac{8 \pi}{5}\)[/tex] is in the fourth quadrant):
For an angle in the fourth quadrant, the reference angle is:
[tex]\[ 2 \pi - \frac{8 \pi}{5} = \frac{10 \pi}{5} - \frac{8 \pi}{5} = \frac{2 \pi}{5} \][/tex]
So the reference angle is:
[tex]\[ \boxed{\frac{2 \pi}{5}} \][/tex]
### 5. Reference Angle of [tex]\(\frac{8 \pi}{5}\)[/tex]
5.1. Normalize the angle to lie between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex]:
[tex]\[ \frac{8 \pi}{5} \mod 2 \pi = \frac{8 \pi}{5} \][/tex]
Since it is already within the range.
5.2. Identify the quadrant ([tex]\(\frac{8 \pi}{5}\)[/tex] is in the fourth quadrant):
For an angle in the fourth quadrant, the reference angle is:
[tex]\[ 2 \pi - \frac{8 \pi}{5} = \frac{10 \pi}{5} - \frac{8 \pi}{5} = \frac{2 \pi}{5} \][/tex]
So the reference angle is:
[tex]\[ \boxed{\frac{2 \pi}{5}} \][/tex]
After computing all the reference angles:
- [tex]\(\frac{13 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
- [tex]\(-\frac{8 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
- [tex]\(\frac{2 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
- [tex]\(-\frac{2 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
- [tex]\(\frac{8 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
### 1. Reference Angle of [tex]\(\frac{13 \pi}{5}\)[/tex]
1.1. Normalize the angle to lie between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex]:
[tex]\[ \frac{13 \pi}{5} \mod 2 \pi = \frac{13 \pi}{5} - 2 \pi \left\lfloor \frac{\frac{13 \pi}{5}}{2 \pi} \right\rfloor = \frac{13 \pi}{5} - 2 \pi \cdot 1 = \frac{13 \pi}{5} - \frac{10 \pi}{5} = \frac{3 \pi}{5} \][/tex]
1.2. Identify the quadrant ([tex]\(\frac{3 \pi}{5}\)[/tex] is in the first quadrant):
Since it is already in the first quadrant, the reference angle is:
[tex]\[ \boxed{\frac{3 \pi}{5}} \][/tex]
### 2. Reference Angle of [tex]\(-\frac{8 \pi}{5}\)[/tex]
2.1. Normalize the angle to lie between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex]:
[tex]\[ -\frac{8 \pi}{5} \mod 2 \pi = -\frac{8 \pi}{5} + 2 \pi \left\lceil \frac{-\frac{8 \pi}{5}}{2 \pi} \right\rceil = -\frac{8 \pi}{5} + 2 \pi \cdot 2 = -\frac{8 \pi}{5} + \frac{10 \pi}{5} = \frac{2 \pi}{5} \][/tex]
2.2. Identify the quadrant ([tex]\(\frac{2 \pi}{5}\)[/tex] is in the first quadrant):
Since it is already in the first quadrant, the reference angle is:
[tex]\[ \boxed{\frac{2 \pi}{5}} \][/tex]
### 3. Reference Angle of [tex]\(\frac{2 \pi}{5}\)[/tex]
3.1. This angle is already between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex], and lies in the first quadrant.
The reference angle is:
[tex]\[ \boxed{\frac{2 \pi}{5}} \][/tex]
### 4. Reference Angle of [tex]\(-\frac{2 \pi}{5}\)[/tex]
4.1. Normalize the angle to lie between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex]:
[tex]\[ -\frac{2 \pi}{5} \mod 2 \pi = -\frac{2 \pi}{5} + 2 \pi \left\lceil \frac{-\frac{2 \pi}{5}}{2 \pi} \right\rceil = -\frac{2 \pi}{5} + 2 \pi \cdot 1 = -\frac{2 \pi}{5} + \frac{10 \pi}{5} = \frac{8 \pi}{5} \][/tex]
4.2. Identify the quadrant ([tex]\(\frac{8 \pi}{5}\)[/tex] is in the fourth quadrant):
For an angle in the fourth quadrant, the reference angle is:
[tex]\[ 2 \pi - \frac{8 \pi}{5} = \frac{10 \pi}{5} - \frac{8 \pi}{5} = \frac{2 \pi}{5} \][/tex]
So the reference angle is:
[tex]\[ \boxed{\frac{2 \pi}{5}} \][/tex]
### 5. Reference Angle of [tex]\(\frac{8 \pi}{5}\)[/tex]
5.1. Normalize the angle to lie between [tex]\(0\)[/tex] and [tex]\(2\pi\)[/tex]:
[tex]\[ \frac{8 \pi}{5} \mod 2 \pi = \frac{8 \pi}{5} \][/tex]
Since it is already within the range.
5.2. Identify the quadrant ([tex]\(\frac{8 \pi}{5}\)[/tex] is in the fourth quadrant):
For an angle in the fourth quadrant, the reference angle is:
[tex]\[ 2 \pi - \frac{8 \pi}{5} = \frac{10 \pi}{5} - \frac{8 \pi}{5} = \frac{2 \pi}{5} \][/tex]
So the reference angle is:
[tex]\[ \boxed{\frac{2 \pi}{5}} \][/tex]
After computing all the reference angles:
- [tex]\(\frac{13 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
- [tex]\(-\frac{8 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
- [tex]\(\frac{2 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
- [tex]\(-\frac{2 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
- [tex]\(\frac{8 \pi}{5}\)[/tex] has a reference angle of [tex]\(\boxed{1.2566370614359172}\)[/tex]
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