Use the unit circle to evaluate the function. tan 7pi/6
The Correct Answer and Explanation is:
To evaluate tan(7π/6) using the unit circle:
Correct Answer:
tan(7π6)=13\tan\left(\frac{7\pi}{6}\right) = \frac{1}{\sqrt{3}}tan(67π)=31
But because 7π/6 is in the third quadrant, where tangent is positive, and both sine and cosine are negative, the tangent value is:tan(7π6)=−12−32=13=33\tan\left(\frac{7\pi}{6}\right) = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}tan(67π)=−23−21=31=33
Detailed Explanation
To evaluate tan(7π/6), begin by understanding what this angle represents on the unit circle. The unit circle is a circle of radius 1 centered at the origin in the coordinate plane. Each angle on the unit circle corresponds to a point whose coordinates are given by (cos θ, sin θ). The tangent of an angle is defined as:tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}tan(θ)=cos(θ)sin(θ)
Now, consider 7π/6. This angle is slightly more than π (which is 6π/6). So:7π6=π+π6\frac{7\pi}{6} = \pi + \frac{\pi}{6}67π=π+6π
This tells us that 7π/6 is located in the third quadrant of the unit circle. In this quadrant, sine and cosine values are both negative, and since tangent is sine divided by cosine, tangent becomes positive in this quadrant.
To find the exact values of sine and cosine, use the reference angle. The reference angle for 7π/6 is:7π6−π=π6\frac{7\pi}{6} – \pi = \frac{\pi}{6}67π−π=6π
From known values on the unit circle:
- sin(π6)=12\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}sin(6π)=21
- cos(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}cos(6π)=23
So for 7π/6, since it’s in the third quadrant:
- sin(7π6)=−12\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}sin(67π)=−21
- cos(7π6)=−32\cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}cos(67π)=−23
Now compute tangent:tan(7π6)=−12−32=13=33\tan\left(\frac{7\pi}{6}\right) = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}tan(67π)=−23−21=31=33
Therefore, the final answer is √3⁄3.
