The exact value of $\cot\frac{7\pi}{6}$?

Question

I am working on a trigonometry question at the moment and am very stuck. I have looked through all the tips to solving it and I cant seem to come up with the right answer. The problem is

What is exact value of
$$\cot \left(\frac{7\pi}{6}\right)? $$

Draw the reference triangle. Read off the value ($\pi/6$ is $30^\circ$, you have a "special triangle"). — David Mitra, Jul 22 '15 at 21:17

Zain Patel · Answer 1 · 2015-07-22T21:20:33.337

6

We have $$\cot \left(\frac{7\pi}{6}\right) = \frac{1}{\tan \left(\frac{7\pi}{6}\right)} = \frac{\cos \left(\frac{7\pi}{6}\right)}{\sin \left(\frac{7\pi}{6}\right)} \equiv \frac{-\cos \left(\frac{\pi}{6}\right)}{-\sin \left(\frac{\pi}{6}\right)} = \sqrt{3}$$

You can easily see this using a "CAST" diagram to reduce $\cos \left(\frac{7\pi}{6}\right)$and $\sin \left(\frac{7\pi}{6}\right)$ to standard results.

As per @Scientifica's comment, an easier method would be to simply note that $\tan$ is $\pi$-periodic so that shifting its argument by $\pi$ will yield no change to it's value, or: $$\cot \left(\frac{7\pi}{6}\right) = \frac{1}{\tan \left(\frac{7\pi}{6}\right)} = \frac{1}{\tan \left(\frac{7\pi}{6} - \pi\right)} = \frac{1}{\tan \left(\frac{\pi}{6}\right)} = \sqrt{3}$$

edited Jul 22 '15 at 21:20

answered Jul 22 '15 at 21:07

Zain Patel

16,802

I didn't downvote. In fact I gave you +1. I comment to tell you that it is easier using the fact that $\tan$ is a $\pi$-periodic function (which the OP itself used but confused $\tan$ and $\cot$) – Scientifica Jul 22 '15 at 21:11
@Scientifica, thanks for that. I included your method (which was the one I originally thought of but for some reason didn't work out in my head), I credited you as well. :-) – Zain Patel Jul 22 '15 at 21:14
1

Thank you you too for being comprehensive and it's ok wether you credit me or not. It happens to me too that I think of something but then get the impression that it won't work hhh. One last remark: you forgot $\tan$ in $\tan\left( \frac{7\pi}{6}-\pi\right)$ – Scientifica Jul 22 '15 at 21:20

score 4 · Accepted Answer · answered Jul 22 '15 at 21:13

4

$$\cot \left(\frac{7\pi}{6}\right) = \cot \left(\frac{\pi}{6}\right) = \sqrt 3 $$

because $$ \tan ( \theta + \pi) = \tan ( \theta ) $$

answered Jul 22 '15 at 21:13

Narasimham

40,495

score 1 · Answer 3 · edited Jul 22 '15 at 21:14

1

Hint: $\cot x=\dfrac{\cos x}{\sin x}$ and

$$ \cos \left(\dfrac{7 \pi}{6}\right)=\cos \left(\pi+\dfrac{\pi}{6}\right) $$

$$ \sin \left(\dfrac{7 \pi}{6}\right)=\sin \left(\pi+\dfrac{\pi}{6}\right) $$

now you can use sum formulas or reduction to the first quadrant.

edited Jul 22 '15 at 21:14

Zain Patel

16,802

answered Jul 22 '15 at 21:11

Emilio Novati

62,675

The exact value of $\cot\frac{7\pi}{6}$?

3 Answers3