Answer :
To solve the problem of finding \(\sin\left(\frac{7\pi}{6}\right)\), let's follow these steps:
1. Determine the Quadrant:
The angle \(\frac{7\pi}{6}\) is in radians, and it can be converted to degrees for better understanding:
[tex]\[ \frac{7\pi}{6} \times \frac{180^\circ}{\pi} = 210^\circ \][/tex]
The angle 210 degrees is located in the third quadrant of the unit circle.
2. Identify the Reference Angle:
To find the reference angle, we need to subtract 180 degrees (as 210 degrees is 30 degrees beyond 180 degrees):
[tex]\[ 210^\circ - 180^\circ = 30^\circ \][/tex]
So, the reference angle is 30 degrees (or \(\frac{\pi}{6}\) radians).
3. Sine in the Third Quadrant:
In the third quadrant, the sine function is negative. So, we need to consider the sine of the reference angle with a negative sign.
[tex]\[ \sin(210^\circ) = -\sin(30^\circ) \][/tex]
4. Value of \(\sin(30^\circ)\) or \(\sin\left(\frac{\pi}{6}\right)\):
We know from trigonometric values that:
[tex]\[ \sin(30^\circ) = \frac{1}{2} \][/tex]
5. Combine the Results:
Since \(\sin(210^\circ)\) is negative in the third quadrant:
[tex]\[ \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} \][/tex]
6. Choose the Correct Option:
From the given options:
[tex]\[ \begin{array}{ll} \text{A.} & \frac{1}{2} \\ \text{B.} & -\frac{\sqrt{3}}{2} \\ \text{C.} & -\frac{1}{2} \\ \text{D.} & \frac{\sqrt{3}}{2} \\ \end{array} \][/tex]
The correct answer is:
[tex]\[ \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} \][/tex]
Hence, the correct choice is:
[tex]\[ \boxed{C} \][/tex]
1. Determine the Quadrant:
The angle \(\frac{7\pi}{6}\) is in radians, and it can be converted to degrees for better understanding:
[tex]\[ \frac{7\pi}{6} \times \frac{180^\circ}{\pi} = 210^\circ \][/tex]
The angle 210 degrees is located in the third quadrant of the unit circle.
2. Identify the Reference Angle:
To find the reference angle, we need to subtract 180 degrees (as 210 degrees is 30 degrees beyond 180 degrees):
[tex]\[ 210^\circ - 180^\circ = 30^\circ \][/tex]
So, the reference angle is 30 degrees (or \(\frac{\pi}{6}\) radians).
3. Sine in the Third Quadrant:
In the third quadrant, the sine function is negative. So, we need to consider the sine of the reference angle with a negative sign.
[tex]\[ \sin(210^\circ) = -\sin(30^\circ) \][/tex]
4. Value of \(\sin(30^\circ)\) or \(\sin\left(\frac{\pi}{6}\right)\):
We know from trigonometric values that:
[tex]\[ \sin(30^\circ) = \frac{1}{2} \][/tex]
5. Combine the Results:
Since \(\sin(210^\circ)\) is negative in the third quadrant:
[tex]\[ \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} \][/tex]
6. Choose the Correct Option:
From the given options:
[tex]\[ \begin{array}{ll} \text{A.} & \frac{1}{2} \\ \text{B.} & -\frac{\sqrt{3}}{2} \\ \text{C.} & -\frac{1}{2} \\ \text{D.} & \frac{\sqrt{3}}{2} \\ \end{array} \][/tex]
The correct answer is:
[tex]\[ \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2} \][/tex]
Hence, the correct choice is:
[tex]\[ \boxed{C} \][/tex]
