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

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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
$$\cot \left(\frac{7\pi}{6}\right)? $$

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3 Answers

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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.

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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: $$\cot \left(\frac{7\pi}{6}\right) = \frac{1}{\tan \left(\frac{7\pi}{6}\right)} = \frac{1}{\tan \left(\frac{7\pi}{6} - \pi\right)} = \frac{1}{\tan \left(\frac{\pi}{6}\right)} = \sqrt{3}$$

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

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

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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.

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