Lilia Mccarthy

2023-03-16

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

coguarsbq2q

Beginner2023-03-17Added 10 answers

The following will be employed:

(1) $\mathrm{cot}\left(x\right)=\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$

(2) $\mathrm{sin}(-x)=-\mathrm{sin}\left(x\right)$

(3)$\mathrm{cos}(-x)=\mathrm{cos}\left(x\right)$

(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

$\mathrm{cot}(-\frac{\pi}{6})=\frac{\mathrm{cos}(-\frac{\pi}{6})}{\mathrm{sin}(-\frac{\pi}{6})}$ (by 1)

$\Rightarrow \mathrm{cot}(-\frac{\pi}{6})=\frac{\mathrm{cos}\left(\frac{\pi}{6}\right)}{-\mathrm{sin}\left(\frac{\pi}{6}\right)}$ (by 2 and 3)

$\Rightarrow \mathrm{cot}(-\frac{\pi}{6})=-\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}$ (by 4 and 5)

(1) $\mathrm{cot}\left(x\right)=\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}$

(2) $\mathrm{sin}(-x)=-\mathrm{sin}\left(x\right)$

(3)$\mathrm{cos}(-x)=\mathrm{cos}\left(x\right)$

(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

$\mathrm{cot}(-\frac{\pi}{6})=\frac{\mathrm{cos}(-\frac{\pi}{6})}{\mathrm{sin}(-\frac{\pi}{6})}$ (by 1)

$\Rightarrow \mathrm{cot}(-\frac{\pi}{6})=\frac{\mathrm{cos}\left(\frac{\pi}{6}\right)}{-\mathrm{sin}\left(\frac{\pi}{6}\right)}$ (by 2 and 3)

$\Rightarrow \mathrm{cot}(-\frac{\pi}{6})=-\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}$ (by 4 and 5)

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