Cot Pi/6 : Exact Value Root 3 (Unit Circle Proof)

What Does Cot Pi/6 Mean?

Cotangent is the reciprocal of tangent, $\cot\theta = \dfrac{1}{\tan\theta}$, and from the core ratios it is $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$. Since its reciprocal partner tangent gives $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$, flipping it returns $\sqrt{3}$.

On the unit circle, cotangent is the $x$-coordinate divided by the $y$-coordinate. At $\frac{\pi}{6}$ the point is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, and because the $x$-value is larger than the $y$-value, the ratio comes out above $1$, at $\sqrt{3}$.

Where Does Cot Pi/6 Show Up?

The value shows up wherever a gentle $30^\circ$ incline is measured by its horizontal reach. Cotangent is run-over-rise, so a $30^\circ$ slope travels $\sqrt{3}$ units across for every $1$ unit up.

It sits inside the geometry of a regular hexagon, whose interior triangles carry $30^\circ$ angles, and in roof pitch or ramp calculations where a shallow grade needs a long horizontal run. Any structure that spreads out gently rather than rising sharply is trading on a cotangent near $\sqrt{3}$.

What Is The Value Of Cot Pi/6 Among The Standard Angles?

$\frac{\pi}{6}$ is the smallest of the common special angles, so its cotangent is the largest finite one, $\sqrt{3}$.

Cotangent is largest near $0$ and shrinks toward $\frac{\pi}{2}$, so the small angle $\frac{\pi}{6}$ carries the tall value $\sqrt{3}$. Notice $\cot\frac{\pi}{6}$ and $\cot\frac{\pi}{3}$ are reciprocals of each other, $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$, a symmetry that comes from the shared 30-60-90 triangle.

How Do You Find The Exact Value Of Cot Pi/6?

Three routes all reach $\sqrt{3}$.

Method 1: The 30-60-90 triangle.

Take an equilateral triangle of side $2$ and drop a perpendicular, splitting it into two right triangles with angles $30^\circ$, $60^\circ$, and $90^\circ$.

• the side opposite the $30^\circ$ angle is $1$,

• the side adjacent to the $30^\circ$ angle is $\sqrt{3}$ (from $\sqrt{2^2 - 1^2}$).

$$\cot\frac{\pi}{6} = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$

Method 2: The quotient $\cos\theta / \sin\theta$.

$$\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}, \qquad \sin\frac{\pi}{6} = \frac{1}{2}$$

$$\cot\frac{\pi}{6} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$$

Method 3: The unit circle.

Rotate a unit radius to $\frac{\pi}{6}$, that is $30^\circ$, above the positive $x$-axis. Its tip lands at $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, and $\frac{x}{y} = \sqrt{3}$. All three describe the same $30^\circ$ geometry, so they agree.

Examples Of Cot Pi/6

Example 1

Evaluate $4\cot\dfrac{\pi}{6}$.

$$4\cot\frac{\pi}{6} = 4 \times \sqrt{3} = 4\sqrt{3} \approx 6.928$$

Example 2

Find $\cot\dfrac{\pi}{6}$ from tangent.

Wrong attempt. A student recalls that $\frac{\pi}{6}$ is a "small angle" and writes $\cot\frac{\pi}{6} = \frac{1}{\sqrt{3}} \approx 0.577$, handing cotangent the small value.

That breaks: on the unit circle the point at $30^\circ$ has $x = \frac{\sqrt{3}}{2}$ larger than $y = \frac{1}{2}$, so $\frac{x}{y}$ must be greater than $1$, and $0.577$ is less than $1$.

Correct. The value $\frac{1}{\sqrt{3}}$ is $\tan\frac{\pi}{6}$; cotangent is its reciprocal, so $\cot\frac{\pi}{6} = \sqrt{3} \approx 1.732$.

Example 3

Evaluate $\cot\dfrac{\pi}{6} + \cot\dfrac{\pi}{4}$.

Using $\cot\frac{\pi}{4} = 1$ (the value at cot pi/4):

$$\cot\frac{\pi}{6} + \cot\frac{\pi}{4} = \sqrt{3} + 1 \approx 2.732$$

Example 4

Simplify $\cot\dfrac{\pi}{6} \times \sin\dfrac{\pi}{6}$.

$$\sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}$$

The result equals $\cos\frac{\pi}{6}$, which checks out because $\cot\theta \times \sin\theta = \cos\theta$.

Example 5

In a 30-60-90 right triangle, the side opposite the $30^\circ$ angle is $5$ cm. Find the side adjacent to it.

$$\cot 30^\circ = \frac{\text{adjacent}}{\text{opposite}} \implies \text{adjacent} = 5 \times \sqrt{3} = 5\sqrt{3} \approx 8.66 \text{ cm}$$

Where Students Trip Up On Cot Pi/6

Mistake 1: Swapping cot pi/6 with tan pi/6

Where it slips in: Handing the small-looking value to the small angle.

Don't do this: Writing $\cot\frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

The correct way: $\frac{1}{\sqrt{3}}$ is $\tan\frac{\pi}{6}$; cotangent flips it to $\sqrt{3}$. The first instinct is to give cotangent the smaller number because it feels like it belongs to $30^\circ$, but that value is tangent's.

Mistake 2: Confusing cot 30° with cot 60°

Where it slips in: Recall that attaches $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ to the wrong angle.

Don't do this: Writing $\cot\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\cot\frac{\pi}{3} = \sqrt{3}$.

The correct way: The smaller angle has the larger cotangent, so $\cot\frac{\pi}{6} = \sqrt{3}$ and $\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}$. Anchor on "small angle, tall cotangent."

Mistake 3: Giving a decimal when the exact value is asked

Where it slips in: Reading $1.732$ off a calculator and copying it.

Don't do this: Writing $\cot\frac{\pi}{6} = 1.732$ on a problem that wants the exact form.

The correct way: The exact value is the surd $\sqrt{3}$; $1.7321$ is only its rounded decimal, which never terminates.

Key Takeaways

• Cot pi/6 equals $\sqrt{3}$, because $\cot\frac{\pi}{6} = \dfrac{\cos(\pi/6)}{\sin(\pi/6)} = \dfrac{\sqrt{3}/2}{1/2}$.

• The 30-60-90 triangle gives $\frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{3}}{1} = \sqrt{3}$.

• It is the reciprocal of $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$, so do not swap the two.

• In radians or degrees the value is the same: $\cot\frac{\pi}{6} = \cot 30^\circ = \sqrt{3} \approx 1.7321$.

To take cot pi/6 and the special angles further with a teacher, explore Bhanzu's trigonometry tutor, high school math tutor, or online math classes.

Practice These To Solidify Your Understanding

1. Evaluate $2\cot\frac{\pi}{6} - \cot\frac{\pi}{4}$.

2. Show that $\cot\frac{\pi}{6} \times \tan\frac{\pi}{6} = 1$.

3. A ramp rises at $30^\circ$ and climbs $2$ m vertically. Use $\cot 30^\circ$ to find its horizontal run.

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