Tan 15 Degrees: Exact Value And How To Find It

What Is The Value Of Tan 15 Degrees?

Tan 15 Degrees equals $2 - \sqrt{3}$ in exact form, and about $0.2679$ as a decimal to four places. Written with the angle in radians, the same fact reads $\tan\frac{\pi}{12} = 2 - \sqrt{3}$, because $15^\circ = \frac{\pi}{12}$.

The value is a clean surd, not a rounded approximation. That is unusual for an angle this small, and it happens because $15^\circ$ is built from angles you already know.

$$\tan 15^\circ = \tan\frac{\pi}{12} = 2 - \sqrt{3} \approx 0.2679$$

The tangent function of any angle is the ratio of the opposite side to the adjacent side in a right triangle, or the ratio $\frac{\sin}{\cos}$. For $15^\circ$ both routes give the same surd, as the sections below show.

How Do You Find The Value Of Tan 15 Degrees?

Write $15^\circ$ as the difference of two familiar angles, $45^\circ - 30^\circ$, then apply the tangent difference formula. This is the fastest exact route and the one every ranking source uses.

The tangent difference identity is:

$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

Set $A = 45^\circ$ and $B = 30^\circ$. You need two standard values: $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Substitute them in:

$$\tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$$

$$\tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}}$$

Multiply the top and bottom by $\sqrt{3}$ to clear the inner fractions:

$$\tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$$

This is correct but not yet simplified. Rationalise by multiplying the top and bottom by $\sqrt{3} - 1$:

$$\tan 15^\circ = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2}$$

$$\tan 15^\circ = 2 - \sqrt{3} \approx 0.2679$$

Final answer: $\tan 15^\circ = 2 - \sqrt{3} \approx 0.2679$.

You can reach the identical surd with the sum and difference identities for sine and cosine, computing $\sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$, then dividing. The division simplifies back to $2 - \sqrt{3}$, so the answer is route-independent.

The Half-Angle Route (A Second Check)

Because $15^\circ$ is also $\frac{30^\circ}{2}$, the half-angle formula gives a quick independent confirmation:

$$\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$$

With $\theta = 30^\circ$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$:

$$\tan 15^\circ = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}$$

Two different formulas, one answer. That agreement is the strongest sign the surd is right.

Where Does 15 Degrees Sit On The Unit Circle?

On the unit circle, $15^\circ$ lands in Quadrant I, just above the positive $x$-axis, where both coordinates are positive. The tangent of the angle is the $y$-coordinate divided by the $x$-coordinate of the point where the terminal side meets the circle.

For $15^\circ$ that point is $(\cos 15^\circ, \sin 15^\circ) \approx (0.9659,\ 0.2588)$. Dividing gives:

$$\tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{0.2588}{0.9659} \approx 0.2679$$

This is the double anchor worth holding on to. From the right triangle, $\tan 15^\circ$ is opposite over adjacent. From the unit circle, it is the $y$-coordinate over the $x$-coordinate. Same number, two pictures.

What Are The Tangent Values Of Nearby Angles?

The special angles from $0^\circ$ to $90^\circ$ make a useful reference, with $15^\circ$ and $75^\circ$ filling the gaps between the textbook values. Notice the symmetry: $\tan 15^\circ = 2 - \sqrt{3}$ and $\tan 75^\circ = 2 + \sqrt{3}$ are reciprocals of each other.

Table: Tangent of the special first-quadrant angles in degrees and radians.

For the full grid of sine, cosine, and tangent at every standard angle, see the trigonometric table. If the radian column feels unfamiliar, what is a radian explains why $15^\circ$ becomes $\frac{\pi}{12}$.

Why Is Tan 15 Degrees Equal To 2 − √3?

The surd is not a coincidence of arithmetic. It falls out of three facts working together, each of which you can point to.

• 15° is constructible from known angles. It is $45^\circ - 30^\circ$, and it is also half of $30^\circ$. Both routes use only $30^\circ$, $45^\circ$, and $60^\circ$ values, whose tangents are already exact surds ($\frac{1}{\sqrt{3}}$, $1$, $\sqrt{3}$).

• Quadrant I keeps the sign positive. The terminal side of $15^\circ$ lies where $x > 0$ and $y > 0$, so $\tan 15^\circ = \frac{y}{x}$ is positive. Any negative answer signals a slip.

• A small angle means a small tangent. Near $0^\circ$ the opposite side is tiny next to the adjacent side, so the ratio is well below $1$. The value $0.2679$ fits: it is small, positive, and less than $\tan 30^\circ = 0.5774$.

Put together, the geometry forces a positive surd smaller than one, and the difference formula pins it down to exactly $2 - \sqrt{3}$. The number and the picture agree.

Who Discovered How To Find Angles Like Tan 15 Degrees?

Long before anyone wrote $\tan 15^\circ$, astronomers needed the ratios for angles that were not neat fractions of a right angle. They built tables, and the trick they invented for filling the gaps is the ancestor of the difference formula on this page.

Two earlier figures laid the ground for angle tables:

• Hipparchus of Nicaea (c. 190–120 BCE, Greece) compiled one of the first tables of chords, the earliest known systematic record of how a ratio changes with its angle.

• Claudius Ptolemy (c. 100–170 CE, Roman Egypt) refined chord tables in the Almagest and used a subtraction result now called Ptolemy's theorem, an early cousin of the $\tan(A - B)$ formula that gives $\tan 15^\circ$ today.

Where Is Tan 15 Degrees Used In The Real World?

Shallow angles near $15^\circ$ turn up wherever a gentle slope or a small tilt matters, and the tangent is what converts that angle into a usable ratio.

• Accessibility and construction: a ramp's steepness is its rise over its run, which is the tangent of its angle. Gentle ramps and low roof pitches sit in the small-angle range where $\tan 15^\circ$ lives.

• Roads and rail: gradient signs and track banking are stated as slopes, and engineers move between the angle and the grade using the tangent.

• Optics and cameras: a lens tilt or a small angle of incidence is handled through its tangent when projecting where a ray lands.

• Computer graphics: rotating or shearing a sprite by a small angle uses the tangent of that angle inside the transformation.

• Navigation and surveying: measuring a height from a distance uses the tangent of the sight angle, and shallow sight lines fall in this range.

One shallow angle, one ratio, and it quietly sets the slope of ramps, roads, and rays of light.

What Are The Most Common Mistakes With Tan 15 Degrees?

These four errors account for most lost marks on this value. Each is verified against the "frequent errors" notes on ranking pages and the unit-circle help pages students search for.

Stopping at the unrationalised fraction.

Where it slips in:

A student reaches $\tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ and writes that as the final answer.

Don't do this:

Do not leave a surd in the denominator. The fraction is correct but not in simplest form, and most mark schemes want the clean surd.

The correct way:

Multiply the top and bottom by $\sqrt{3} - 1$ to get $\frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

Reading the angle as radians on a calculator.

Where it slips in:

A student types "tan 15" while the calculator is in radian mode and reads off $-0.8598$, the value of $\tan(15\text{ rad})$.

Don't do this:

Do not trust the display before checking the mode. $\tan 15^\circ$ is a small positive number near $0.27$, so a negative or large reading is a red flag.

The correct way:

Set the calculator to degree mode for $15^\circ$, or enter $\frac{\pi}{12}$ in radian mode. Both give $0.2679$.

Getting the quadrant sign wrong.

Where it slips in:

A student confuses $15^\circ$ with an angle like $165^\circ$ or $-15^\circ$ and attaches a negative sign to the answer.

Don't do this:

Do not assign a sign before locating the angle. Tangent is positive in Quadrant I and $15^\circ$ lands there.

The correct way:

Place the terminal side first. In Quadrant I both coordinates are positive, so $\tan 15^\circ = \frac{y}{x} > 0$, giving $+ (2 - \sqrt{3})$.

Confusing tan 15° with tan 75°.

Where it slips in:

A student recalls that $15^\circ$ and $75^\circ$ are linked and writes $\tan 15^\circ = \tan 75^\circ$.

Don't do this:

Do not equate the two tangents. The correct link is the cofunction identity: $\tan 15^\circ = \cot 75^\circ$, not $\tan 75^\circ$.

The correct way:

Use $\tan\theta = \cot(90^\circ - \theta)$. So $\tan 15^\circ = \cot 75^\circ = 2 - \sqrt{3}$, while $\tan 75^\circ = 2 + \sqrt{3}$ is its reciprocal.

Practice Problems On Tan 15 Degrees

Work each one, then check against the answer beside it.

1. Convert $15^\circ$ to radians.
   (Answer: $\frac{\pi}{12} \approx 0.2618$.)

2. Use $\tan(45^\circ - 30^\circ)$ to show $\tan 15^\circ = 2 - \sqrt{3}$.
   (Answer: substitute $\tan 45^\circ = 1$, $\tan 30^\circ = \frac{1}{\sqrt{3}}$, simplify to $\frac{\sqrt{3}-1}{\sqrt{3}+1}$, rationalise to $2 - \sqrt{3}$.)

3. Find $\cot 15^\circ$, the reciprocal of $\tan 15^\circ$.
   (Answer: $\frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3} \approx 3.7321$.)

4. Is $\tan 15^\circ$ positive or negative, and why?
   (Answer: positive, because $15^\circ$ lies in Quadrant I where $x$ and $y$ are both positive.)

5. Use the half-angle formula on $30^\circ$ to find $\tan 15^\circ$.
   (Answer: $\frac{1 - \cos 30^\circ}{\sin 30^\circ} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}$.)

6. Given $\tan 15^\circ = 2 - \sqrt{3}$, use a cofunction to find $\tan 75^\circ$.
   (Answer: $\tan 75^\circ = \cot 15^\circ = 2 + \sqrt{3} \approx 3.7321$.)

Where Should You Go Next After Tan 15 Degrees?

The exact value opens onto the wider machinery of angles. A few natural doors lead on from here.

1. Sum and difference identities. The tool that cracked $15^\circ$ also gives $75^\circ$, $105^\circ$, and any angle built from the standard set.

2. Sin, cos, and tan. Reconnect tangent to its two parent ratios and see how the three move together on the unit circle.

3. Unit circle with tangent. Watch how the tangent value grows as the angle sweeps from $0^\circ$ toward $90^\circ$.

If your child is building these foundations, a live Bhanzu trainer teaches angle values starting from why the difference formula works, in the Bhanzu trigonometry program.