Answer :
To determine which expression is equivalent to [tex]\(\sin \frac{7 \pi}{6}\)[/tex], we need to use our knowledge of the unit circle and the properties of the sine function.
### Step-by-Step Solution:
1. Identify the Initial Angle:
[tex]\[ \theta = \frac{7 \pi}{6} \][/tex]
This angle is more than [tex]\(\pi\)[/tex] (180 degrees), so it is in the third quadrant of the unit circle.
2. Reference Angle:
To find the reference angle, we subtract [tex]\(\pi\)[/tex] from [tex]\(\theta\)[/tex]:
[tex]\[ \theta_{\text{ref}} = \theta - \pi = \frac{7 \pi}{6} - \pi = \frac{7 \pi}{6} - \frac{6 \pi}{6} = \frac{\pi}{6} \][/tex]
The reference angle is [tex]\(\frac{\pi}{6}\)[/tex].
3. Sine in the Third Quadrant:
In the third quadrant, sine values are negative. Therefore:
[tex]\[ \sin \frac{7 \pi}{6} = -\sin \frac{\pi}{6} \][/tex]
Since [tex]\(\sin \frac{\pi}{6} = \frac{1}{2}\)[/tex], we have:
[tex]\[ \sin \frac{7 \pi}{6} = -\frac{1}{2} \][/tex]
4. Compare to Other Expressions:
- [tex]\(\sin \frac{\pi}{6} = \frac{1}{2}\)[/tex]
- [tex]\(\sin \frac{5 \pi}{6} = \sin \left(\pi - \frac{\pi}{6}\right)\)[/tex]:
- [tex]\(\sin \)[/tex] in the second quadrant is positive, and
- [tex]\(\sin \frac{5 \pi}{6} = \sin \frac{\pi}{6} = \frac{1}{2}\)[/tex]
- [tex]\(\sin \frac{5 \pi}{3} = \sin \left(2\pi - \frac{\pi}{3}\right)\)[/tex]:
- [tex]\(\sin \)[/tex] in the fourth quadrant is negative, and
- [tex]\(\sin \frac{5 \pi}{3} = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}\)[/tex]
- [tex]\(\sin \frac{11 \pi}{6} = \sin \left(2\pi - \frac{\pi}{6}\right)\)[/tex]:
- [tex]\(\sin \)[/tex] in the fourth quadrant is negative, and
- [tex]\(\sin \frac{11 \pi}{6} = -\sin \frac{\pi}{6} = -\frac{1}{2}\)[/tex]
5. Conclusion:
The expression equivalent to [tex]\(\sin \frac{7 \pi}{6}\)[/tex] should have the same value of [tex]\(-\frac{1}{2}\)[/tex].
Only [tex]\(\sin \frac{11 \pi}{6}\)[/tex] matches this value:
[tex]\[ \sin \frac{11 \pi}{6} = -\frac{1}{2} \][/tex]
Thus, the expression equivalent to [tex]\(\sin \frac{7 \pi}{6}\)[/tex] is:
[tex]\[ \boxed{\sin \frac{11 \pi}{6}} \][/tex]
### Step-by-Step Solution:
1. Identify the Initial Angle:
[tex]\[ \theta = \frac{7 \pi}{6} \][/tex]
This angle is more than [tex]\(\pi\)[/tex] (180 degrees), so it is in the third quadrant of the unit circle.
2. Reference Angle:
To find the reference angle, we subtract [tex]\(\pi\)[/tex] from [tex]\(\theta\)[/tex]:
[tex]\[ \theta_{\text{ref}} = \theta - \pi = \frac{7 \pi}{6} - \pi = \frac{7 \pi}{6} - \frac{6 \pi}{6} = \frac{\pi}{6} \][/tex]
The reference angle is [tex]\(\frac{\pi}{6}\)[/tex].
3. Sine in the Third Quadrant:
In the third quadrant, sine values are negative. Therefore:
[tex]\[ \sin \frac{7 \pi}{6} = -\sin \frac{\pi}{6} \][/tex]
Since [tex]\(\sin \frac{\pi}{6} = \frac{1}{2}\)[/tex], we have:
[tex]\[ \sin \frac{7 \pi}{6} = -\frac{1}{2} \][/tex]
4. Compare to Other Expressions:
- [tex]\(\sin \frac{\pi}{6} = \frac{1}{2}\)[/tex]
- [tex]\(\sin \frac{5 \pi}{6} = \sin \left(\pi - \frac{\pi}{6}\right)\)[/tex]:
- [tex]\(\sin \)[/tex] in the second quadrant is positive, and
- [tex]\(\sin \frac{5 \pi}{6} = \sin \frac{\pi}{6} = \frac{1}{2}\)[/tex]
- [tex]\(\sin \frac{5 \pi}{3} = \sin \left(2\pi - \frac{\pi}{3}\right)\)[/tex]:
- [tex]\(\sin \)[/tex] in the fourth quadrant is negative, and
- [tex]\(\sin \frac{5 \pi}{3} = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2}\)[/tex]
- [tex]\(\sin \frac{11 \pi}{6} = \sin \left(2\pi - \frac{\pi}{6}\right)\)[/tex]:
- [tex]\(\sin \)[/tex] in the fourth quadrant is negative, and
- [tex]\(\sin \frac{11 \pi}{6} = -\sin \frac{\pi}{6} = -\frac{1}{2}\)[/tex]
5. Conclusion:
The expression equivalent to [tex]\(\sin \frac{7 \pi}{6}\)[/tex] should have the same value of [tex]\(-\frac{1}{2}\)[/tex].
Only [tex]\(\sin \frac{11 \pi}{6}\)[/tex] matches this value:
[tex]\[ \sin \frac{11 \pi}{6} = -\frac{1}{2} \][/tex]
Thus, the expression equivalent to [tex]\(\sin \frac{7 \pi}{6}\)[/tex] is:
[tex]\[ \boxed{\sin \frac{11 \pi}{6}} \][/tex]