IDNLearn.com: Your go-to resource for finding precise and accurate answers. Ask your questions and receive prompt, detailed answers from our experienced and knowledgeable community members.
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]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.
