Tan 510 Degrees: Exact Value, Unit Circle And Examples

What Does Tan 510 Degrees Mean?

In a right triangle, the tangent of an acute angle is the opposite side divided by the adjacent side. That definition only covers angles below $90^\circ$, so angles like $510^\circ$ need the unit circle.

On the unit circle (radius $1$, centred at the origin), an angle is measured anticlockwise from the positive $x$-axis, and its radius meets the circle at $(\cos\theta, \sin\theta)$. The tangent is $\frac{y}{x}$, the slope of that radius.

Four terms carry the rest of the work:

• A quadrant is one of the four regions the axes cut the plane into, numbered I to IV anticlockwise. Tangent is positive in I and III and negative in II and IV.

• Coterminal angles end at the same position; they differ by whole turns of $360^\circ$, so $510^\circ$ and $150^\circ$ are coterminal.

• The reference angle is the acute angle between the radius and the $x$-axis. For $150^\circ$ it is $180^\circ - 150^\circ = 30^\circ$.

• ASTC ("All, Sine, Tangent, Cosine") is a memory aid for which functions are positive in Quadrants I, II, III and IV.

One more term matters when you run the calculation backwards.

The principal value of an inverse function is the single output it is defined to return; for inverse tangent that output lies strictly between $-90^\circ$ and $90^\circ$. So $\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)$ gives $-30^\circ$, not $510^\circ$, even though both angles have this tangent.

Standard-Angle Tangent Reference Table

Every angle that shares $150^\circ$'s position on the circle, or sits $180^\circ$ away from it, has the same tangent. The table lists the standard angles from $0^\circ$ to $180^\circ$ with both forms.

The angles $-30^\circ$, $150^\circ$, $330^\circ$, $510^\circ$, $690^\circ$ and $870^\circ$ all give $-\frac{\sqrt{3}}{3}$. They are $180^\circ$ apart, which is the period of tangent.

The radian twin of $150^\circ$ has its own page: tan 5π/6 carries the same value from the radian side.

Where Does Tan 510 Degrees Show Up, And Why Does It Matter?

Tan 510 degrees is the slope of any line that points in the $150^\circ$ direction. A line through the origin at $150^\circ$ falls about $0.577$ units for every unit it moves right, so its slope is $-\frac{\sqrt{3}}{3}$.

Angles past $360^\circ$ appear whenever something keeps turning: a steering wheel, a robot joint or a turntable that rotates $510^\circ$ ends up facing the same way as one that rotates $150^\circ$. Engineers reduce the angle first and then read the direction, which is exactly the move this page teaches.

How Do You Find The Exact Value Of Tan 510 Degrees?

Every route below lands on $-\frac{\sqrt{3}}{3}$. The first two are the ones most textbooks show; the others are useful checks.

Method 1: Coterminal Angle, Then Reference Angle

Remove one full turn:

$$510^\circ - 360^\circ = 150^\circ$$

$$\tan 510^\circ = \tan 150^\circ$$

$150^\circ$ is in Quadrant II with reference angle $30^\circ$. Tangent is negative there, so:

$$\tan 150^\circ = -\tan 30^\circ = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$

The value of tan 30 degrees comes from the 30-60-90 triangle, where the side opposite $30^\circ$ is $1$ and the adjacent side is $\sqrt{3}$.

Method 2: The Unit Circle Coordinates

The point at $150^\circ$ (and so at $510^\circ$) is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.

$$\tan 510^\circ = \frac{y}{x} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}}$$

Multiply top and bottom by $\sqrt{3}$ to rationalise: $-\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$.

Method 3: Tangent's 180 Degree Period

The tangent function repeats every $180^\circ$, not every $360^\circ$, because the points at $\theta$ and $\theta + 180^\circ$ are opposite each other on the circle and both coordinates flip sign together. That makes $\frac{y}{x}$ unchanged.

$$510^\circ - 3 \times 180^\circ = -30^\circ$$

$$\tan 510^\circ = \tan(-30^\circ) = -\tan 30^\circ = -\frac{\sqrt{3}}{3}$$

The last step uses the fact that tangent is an odd function: $\tan(-\theta) = -\tan\theta$. More on repeating functions sits on the periodic function page.

Method 4: A Subtraction Identity

Write $510^\circ = 540^\circ - 30^\circ$ and use $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ with $\tan 540^\circ = 0$:

$$\tan 510^\circ = \frac{0 - \tan 30^\circ}{1 + 0 \times \tan 30^\circ} = -\tan 30^\circ = -\frac{\sqrt{3}}{3}$$

Is $-\frac{1}{\sqrt{3}}$ The Same As $-\frac{\sqrt{3}}{3}$?

Yes. They are the same number, $-0.5773502692$, written two ways. Most textbooks and answer keys give the rationalised form $-\frac{\sqrt{3}}{3}$, which has no square root in the denominator.

How Do You Write Tan 510 Degrees In Terms Of Other Trig Functions?

Since $\sin 510^\circ = \frac{1}{2}$ and $\cos 510^\circ = -\frac{\sqrt{3}}{2}$, every other function gives the same tangent. Each line below evaluates to $-\frac{\sqrt{3}}{3}$.

Examples Of Tan 510 Degrees

Example 1

Find the exact value of $\tan 510^\circ$ using a coterminal angle.

$$510^\circ - 360^\circ = 150^\circ$$

$$\tan 150^\circ = -\tan 30^\circ = -\frac{\sqrt{3}}{3} \approx -0.5774$$

Example 2

A student reduces $\tan 510^\circ$ and gets $+\frac{\sqrt{3}}{3}$. Find the error and the correct value.

The student's working: $510^\circ - 360^\circ = 150^\circ$, then $150^\circ = 180^\circ - 30^\circ$, so $\tan 150^\circ = \tan 30^\circ = \frac{\sqrt{3}}{3}$.

That breaks on a quick check. The point at $150^\circ$ is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, with a negative $x$ and a positive $y$, so $\frac{y}{x}$ must be negative.

The reference angle gives the size, and the quadrant gives the sign separately. In Quadrant II tangent is negative:

$$\tan 510^\circ = -\tan 30^\circ = -\frac{\sqrt{3}}{3}$$

Example 3

Evaluate $\sin 150^\circ + \cos 510^\circ + \tan 4110^\circ - \tan 210^\circ$.

Reduce each angle first:

• $\cos 510^\circ = \cos 150^\circ = -\frac{\sqrt{3}}{2}$

• $4110^\circ = 22 \times 180^\circ + 150^\circ$, so $\tan 4110^\circ = \tan 150^\circ = -\frac{\sqrt{3}}{3}$

• $210^\circ = 180^\circ + 30^\circ$, so $\tan 210^\circ = \tan 30^\circ = \frac{\sqrt{3}}{3}$

Now combine, using sixths:

$$\frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3} = \frac{3}{6} - \frac{3\sqrt{3}}{6} - \frac{4\sqrt{3}}{6}$$

$$= \frac{3 - 7\sqrt{3}}{6} \approx -1.5207$$

Example 4

Find the value of $\cos 570^\circ \times \sin 510^\circ$.

$$\cos 570^\circ = \cos 210^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$$

$$\sin 510^\circ = \sin 150^\circ = \sin 30^\circ = \frac{1}{2}$$

$$\cos 570^\circ \times \sin 510^\circ = -\frac{\sqrt{3}}{2} \times \frac{1}{2} = -\frac{\sqrt{3}}{4} \approx -0.4330$$

Notice that $\sin 510^\circ$ stays positive while $\tan 510^\circ$ is negative; the quadrant decides each sign on its own.

Example 5

Find the exact value of $\tan 555^\circ$ by writing it as $\tan(510^\circ + 45^\circ)$.

Use $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, from the sum and difference identities, with $\tan 510^\circ = -\frac{1}{\sqrt{3}}$ and $\tan 45^\circ = 1$:

$$\tan 555^\circ = \frac{-\frac{1}{\sqrt{3}} + 1}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$

$$= \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} \approx 0.2679$$

Check: $555^\circ - 3 \times 180^\circ = 15^\circ$, and $\tan 15^\circ = 2 - \sqrt{3}$.

Example 6

An arm of length 2 m rotates $510^\circ$ anticlockwise from the positive $x$-axis. Find the coordinates of its tip and the slope of the arm.

$$x = 2\cos 510^\circ = 2 \times \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \approx -1.732$$

$$y = 2\sin 510^\circ = 2 \times \frac{1}{2} = 1$$

$$\text{slope} = \frac{y}{x} = \frac{1}{-\sqrt{3}} = -\frac{\sqrt{3}}{3} = \tan 510^\circ$$

The step that gets skipped here is the one that matters: reducing $510^\circ$ before reading the coordinates, rather than trying to picture a turn and a half directly.

Where Do Students Trip Up On Tan 510 Degrees?

Mistake 1: A Calculator In Radian Mode

Where it slips in: Typing $\tan(510)$ on a calculator or search bar that works in radians.

Don't do this: Copying $1.7928$. That is the tangent of $510$ radians, an angle of about $29{,}221^\circ$ that behaves like $60.85^\circ$ for tangent.

The correct way: Switch to degree mode, or enter $\tan\left(\frac{17\pi}{6}\right)$ in radian mode. A positive answer for an angle that reduces to Quadrant II is the warning sign; the rusher who copies the first number on the screen never sees it.

Mistake 2: Dropping The Quadrant II Sign

Where it slips in: Right after finding the reference angle $30^\circ$.

Don't do this: Writing $\tan 510^\circ = \tan 30^\circ = \frac{\sqrt{3}}{3}$.

The correct way: Treat the sign as a separate step: reduce the angle, find the quadrant, then attach the sign from ASTC. The first instinct after spotting the familiar $30^\circ$ is to write its textbook value and stop, and that habit is exactly what flips the answer here.

Mistake 3: Measuring The Reference Angle From The Y-Axis

Where it slips in: Reading $150^\circ$ as "$60^\circ$ past the vertical".

Don't do this: Using $60^\circ$ as the reference angle and writing $\tan 510^\circ = -\sqrt{3}$.

The correct way: The reference angle is always measured to the $x$-axis: $180^\circ - 150^\circ = 30^\circ$. Then $\tan 510^\circ = -\tan 30^\circ = -\frac{\sqrt{3}}{3}$, which is much smaller in size than $-\sqrt{3} \approx -1.7321$.

Where Can You Get Help With Trigonometry Values Like Tan 510 Degrees?

Reducing large angles, choosing the quadrant sign and rationalising the answer are the three steps that most need a second pair of eyes. A trigonometry tutor, a high school math tutor, live math classes online or regular math tutoring can turn that routine into a habit.

Key Takeaways

• Tan 510 degrees equals $-\frac{\sqrt{3}}{3}$ (or $-\frac{1}{\sqrt{3}}$), about $-0.5774$.

• In radians the angle is $\frac{17\pi}{6}$; after one full turn it is $150^\circ$, in Quadrant II.

• The reference angle is $30^\circ$ and Quadrant II makes tangent negative, so $\tan 510^\circ = -\tan 30^\circ$.

• Tangent repeats every $180^\circ$, so $\tan 510^\circ = \tan 330^\circ = \tan(-30^\circ)$.

• A positive calculator answer of $1.7928$ means radian mode.

Practice These Before Moving On

1. Find $\tan 870^\circ$.

2. Find $\cot 510^\circ$ and $\sec 510^\circ$ in exact form.

3. Show that $1 + \tan^2 510^\circ = \sec^2 510^\circ$ using exact values.

4. Find $\tan 1020^\circ$.

5. A line through the origin makes an angle of $510^\circ$ with the positive $x$-axis. What is its slope, and which way does it fall?

Answers:

1. $870^\circ - 2 \times 360^\circ = 150^\circ$, so $\tan 870^\circ = -\frac{\sqrt{3}}{3}$.

2. $\cot 510^\circ = -\sqrt{3}$ and $\sec 510^\circ = \frac{1}{\cos 510^\circ} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$.

3. $1 + \left(-\frac{\sqrt{3}}{3}\right)^2 = 1 + \frac{1}{3} = \frac{4}{3}$, and $\left(-\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}$.

4. $1020^\circ - 5 \times 180^\circ = 120^\circ$, so $\tan 1020^\circ = -\sqrt{3}$.

5. Slope $-\frac{\sqrt{3}}{3} \approx -0.577$; the line falls about $0.577$ units for each unit to the right.

If question 4 goes wrong, check whether you removed $180^\circ$ steps (allowed for tangent) and then took the sign of $120^\circ$ from Quadrant II.

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