Lilia Mccarthy

2023-03-16

How to evaluate $\mathrm{cot}\left(-\frac{\pi }{6}\right)$?

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The following will be employed:
(1) $\mathrm{cot}\left(x\right)=\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$
(2) $\mathrm{sin}\left(-x\right)=-\mathrm{sin}\left(x\right)$
(3)$\mathrm{cos}\left(-x\right)=\mathrm{cos}\left(x\right)$
(4) $\mathrm{cos}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$
(5) $\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}$
Verifying these facts is a worthwhile exercise that may be carried out using the unit circle for units 1-3 and some basic definitions.
As a clue, divide an equilateral triangle with side length $1$ into two equal right triangles to solve problems 4 and 5. What are the right triangles' angles?
This has given us
$\mathrm{cot}\left(-\frac{\pi }{6}\right)=\frac{\mathrm{cos}\left(-\frac{\pi }{6}\right)}{\mathrm{sin}\left(-\frac{\pi }{6}\right)}$ (by 1)
$⇒\mathrm{cot}\left(-\frac{\pi }{6}\right)=\frac{\mathrm{cos}\left(\frac{\pi }{6}\right)}{-\mathrm{sin}\left(\frac{\pi }{6}\right)}$ (by 2 and 3)
$⇒\mathrm{cot}\left(-\frac{\pi }{6}\right)=-\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}$ (by 4 and 5)

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