Let's solve this step by step:

Step 1: Use the unit circle to find the value of $\tan\left(\frac{\pi}{6}\right)$.

• On the unit circle, the angle $\frac{\pi}{6}$ (or 30°) corresponds to the point $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
• The tangent function is defined as $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.
• For $\frac{\pi}{6}$:
  • $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
  • $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

So, $\tan\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Thus, $\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$.

Step 2: Use even or odd properties of trigonometric functions to find $\tan\left(-\frac{\pi}{6}\right)$.

• Tangent is an odd function, which means that: $\tan(-\theta) = -\tan(\theta)$
• Therefore: $\tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right)$
• From Step 1, we know that $\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$, so: $\tan\left(-\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{3}$

Final Answer:

• $\tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}$
• $\tan\left(-\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{3}$

Would you like further details or have any questions?

5 Related Questions:

1. What is the value of $\tan\left(\frac{\pi}{4}\right)$ using the unit circle?
2. How do you find the value of $\tan\left(\frac{\pi}{3}\right)$?
3. What are the even and odd properties of the sine and cosine functions?
4. How do you calculate the value of $\sin(-\frac{\pi}{6})$ using the unit circle?
5. What is the relationship between tangent and cotangent?

Tip:

Remember that sine is also an odd function, while cosine is even. This can help you quickly evaluate trigonometric values for negative angles!