How do you evaluate tan(-7pi/6)?

Question

Brooklyn Anderson · Accepted Answer

Evaluate# tan ((-7pi)/6)#

Ans: #-sqrt3/3# #tan ((-7pi)/6) = tan ((5pi)/6 - (12pi)/6) = tan ((5pi)/6 - 2pi) = # #= tan ((5pi)/6) = -sqrt3/3# (Trig Table)

Nolan Alexander · Answer

undefined To evaluate $ \tan\left(-\frac{7\pi}{6}\right) $, follow these steps:

1. Since the tangent function has a period of $ \pi $, you can add or subtract multiples of $ \pi $ to the given angle until it falls within the interval $ (-\pi/2, \pi/2) $ where the tangent function is defined.

2. $ -\frac{7\pi}{6} $ is equivalent to $ -\frac{6\pi}{6} - \frac{\pi}{6} $, which simplifies to $ -\pi - \frac{\pi}{6} $.

3. $ -\pi - \frac{\pi}{6} $ falls within the interval $ (-\pi, -\pi/2) $, where $ \tan $ is negative.

4. Use the periodicity of $ \tan $ to evaluate $ \tan\left(-\frac{\pi}{6}\right) $, which is $ -\tan\left(\frac{\pi}{6}\right) $.

5. $ \tan\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)} $.

6. Substitute the values of sine and cosine of $ \frac{\pi}{6} $, which are $ \frac{1}{2} $ and $ \frac{\sqrt{3}}{2} $, respectively.

7. $ \tan\left(\frac{\pi}{6}\right) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} $.

8. Therefore, $ \tan\left(-\frac{7\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} $.