Tan 7pi/6 : Exact Value 1/√3 Explained

What Does Tan 7pi/6 Mean?

Tangent is one of the three core trigonometric ratios: for an angle $\theta$, it is $\tan\theta = \frac{\sin\theta}{\cos\theta}$, and on a circle it reads as the $y$-coordinate divided by the $x$-coordinate of the point where the angle's radius lands. So $\tan\frac{7\pi}{6}$ asks a precise question about a specific spot on the circle.

The angle $\frac{7\pi}{6}$ measures $210^\circ$ once you convert from radians using $180^\circ = \pi$. That rotation stops in the third quadrant, the region past $\pi$ and before $\frac{3\pi}{2}$, where both coordinates are negative. Its reference angle, the acute gap to the horizontal axis, is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$.

Where Tan 7pi/6 Shows Up

A $210^\circ$ heading points into the third quadrant of a compass, and the slope of a straight path along that bearing carries a rise-over-run of $\tan\frac{7\pi}{6}$. The same value appears when a rotating machine part is tracked past the half-turn mark, where phase angles run beyond $\pi$ and the tangent describes the ratio of vertical to horizontal displacement.

Because tangent repeats every $\pi$ radians, $\tan\frac{7\pi}{6}$ equals $\tan\frac{\pi}{6}$, the slope of a gentle $30^\circ$ incline. That periodic reuse is exactly what makes tangent useful in applications of trigonometry that track repeating motion.

Standard-Angle Tangent Reference Table

Seven-sixths of $\pi$ is a rotation past the halfway mark of the circle, so its tangent reuses a first-quadrant value with a sign fixed by the quadrant. Here are the tangents you meet most, from $0$ around into the third quadrant.

Notice the last row repeats the second. That is not a coincidence: $210^\circ$ and $30^\circ$ share the same reference angle, so their tangents match in size, and both come out positive.

How Do You Find the Exact Value of Tan 7pi/6?

There are three clean routes, and all land on $\frac{1}{\sqrt{3}}$.

Method 1: The reference-angle rule.

Find the reference angle, take the first-quadrant tangent, then apply the quadrant sign.

$$\frac{7\pi}{6} - \pi = \frac{\pi}{6}, \qquad \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$

In the third quadrant tangent is positive (the ASTC rule, where the "T" quadrant keeps tangent positive), so:

$$\tan\frac{7\pi}{6} = +\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$

Method 2: The unit circle.

Rotate the radius to $210^\circ$. The terminal point on the unit circle is $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$.

$$\tan\frac{7\pi}{6} = \frac{y}{x} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$$

The two minus signs cancel, which is why a third-quadrant tangent is positive.

Method 3: Sine over cosine.

$$\sin\frac{7\pi}{6} = -\frac{1}{2}, \qquad \cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$$

$$\tan\frac{7\pi}{6} = \frac{\sin\frac{7\pi}{6}}{\cos\frac{7\pi}{6}} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$$

Examples of Tan 7pi/6

Working with $\frac{7\pi}{6}$ is really working with $\frac{\pi}{6}$ plus a sign rule. The set below builds from a plain evaluation to a rationalised form and an equation. One recurring pattern worth naming: students who have just learned that sine and cosine are negative in the third quadrant often expect the tangent to be negative too, and stamp a minus sign on an answer that should be positive.

Example 1

Evaluate $6\tan\frac{7\pi}{6}$.

$$6\tan\frac{7\pi}{6} = 6 \times \frac{1}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \approx 3.464$$

Example 2

Find $\tan\frac{7\pi}{6}$ by deciding the sign first.

Wrong attempt. A student notes that $\frac{7\pi}{6}$ is past $\pi$, so sine and cosine are both negative, and concludes the tangent is negative: $-\frac{1}{\sqrt{3}}$.

That breaks under a quick check. Tangent is $\frac{\sin}{\cos}$, and here that is $\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$. A negative divided by a negative is positive, so the minus sign cannot survive.

Correct. In the third quadrant, tangent is positive. The reference angle gives the size, $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$, and the sign is $+$, so $\tan\frac{7\pi}{6} = \frac{1}{\sqrt{3}}$.

Example 3

Write $\tan\frac{7\pi}{6}$ in rationalised form.

$$\tan\frac{7\pi}{6} = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

Both $\frac{1}{\sqrt{3}}$ and $\frac{\sqrt{3}}{3}$ are correct; the rationalised $\frac{\sqrt{3}}{3}$ is the standard exam form.

Example 4

Confirm the periodic link $\tan\frac{7\pi}{6} = \tan\frac{\pi}{6}$.

Tangent repeats every $\pi$, so adding $\pi$ to an angle leaves the tangent unchanged.

$$\tan\left(\frac{\pi}{6} + \pi\right) = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$

Since $\frac{\pi}{6} + \pi = \frac{7\pi}{6}$, the two are equal.

Example 5

Solve $\tan\theta = \frac{1}{\sqrt{3}}$ for $\theta$ in $[0, 2\pi)$.

Tangent hits $\frac{1}{\sqrt{3}}$ at its reference angle $\frac{\pi}{6}$, then again half a turn later.

$$\theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6}$$

Both solutions are valid, which is why $\frac{7\pi}{6}$ turns up as a standard answer on the tangent.

Where Do Students Trip Up on Tan 7pi/6?

Mistake 1: Making the third-quadrant tangent negative

Where it slips in: Right after learning that sine and cosine are negative in the third quadrant, when the sign rule for tangent gets over-applied.

Don't do this: Writing $\tan\frac{7\pi}{6} = -\frac{1}{\sqrt{3}}$ because "everything is negative down there."

The correct way: Tangent is a ratio of two negatives in the third quadrant, so it is positive. The learner who anchors on ASTC instead of copying the sine sign gets this right every time; the one who reasons from a half-remembered rule is the one who flips it.

Mistake 2: Using the wrong reference angle

Where it slips in: Subtracting from the wrong axis, using $\frac{3\pi}{2} - \frac{7\pi}{6}$ or $2\pi - \frac{7\pi}{6}$ instead of measuring from $\pi$.

Don't do this: Taking the reference angle as $\frac{\pi}{3}$ and reporting $\tan\frac{7\pi}{6} = \sqrt{3}$.

The correct way: In the third quadrant, the reference angle is $\theta - \pi$. Here that is $\frac{7\pi}{6} - \pi = \frac{\pi}{6}$, giving $\frac{1}{\sqrt{3}}$, not $\sqrt{3}$.

Mistake 3: Confusing the value with tan 5pi/6

Where it slips in: Both angles have reference angle $\frac{\pi}{6}$, so the sizes match and only the sign separates them.

Don't do this: Reporting $\tan\frac{7\pi}{6}$ and $\tan\frac{5\pi}{6}$ as the same number.

The correct way: $\frac{5\pi}{6}$ is in the second quadrant where tangent is negative, so tan 5pi/6 $= -\frac{1}{\sqrt{3}}$, while $\tan\frac{7\pi}{6} = +\frac{1}{\sqrt{3}}$. Same size, opposite sign.

Key Takeaways

• Tan 7π/6 equals $\frac{1}{\sqrt{3}}$, or $\frac{\sqrt{3}}{3}$, about $0.5774$, a positive value because the angle is in the third quadrant.

• The reference angle is $\frac{\pi}{6}$, so the size matches $\tan\frac{\pi}{6}$, and periodicity gives $\tan\frac{7\pi}{6} = \tan\frac{\pi}{6}$.

• The unit-circle point is $\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$, and $\frac{y}{x}$ returns the same $\frac{1}{\sqrt{3}}$.

• The classic slip is reporting a negative answer; two negatives divide to a positive.

To work through more angles like this with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor, or math classes online.

Practice These Before Moving On

1. Evaluate $\tan\frac{7\pi}{6} + \tan\frac{\pi}{6}$.

2. A bearing runs at $210^\circ$. Using $\tan\frac{7\pi}{6}$, find the rise for every $3$ units of horizontal run.

3. Solve $\sqrt{3},\tan\theta = 1$ for all $\theta$ in $[0, 2\pi)$.

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Read More

• Sin, cos, and tan explained — the three core ratios and how they connect.

• Tangent formula and derivation — where $\tan\theta = \frac{\sin\theta}{\cos\theta}$ comes from.

• Trigonometric ratios of specific angles — the full standard-angle set.

• Tan pi/6 exact value — the reference-angle twin that fixes the size of tan 7pi/6.

• Coterminal angles — why angles a full turn apart share a tangent.