Sin pi/12: Exact Value (√6−√2)/4 and Decimal

What Is The Value Of Sin pi/12?

Sin pi/12 equals $\dfrac{\sqrt{6}-\sqrt{2}}{4}$, and as a decimal that is $0.2588$ to four places. Written in radians the angle is $\frac{\pi}{12}$; written in degrees it is $15^\circ$. The two are the same angle, because a straight angle of $180^\circ$ equals $\pi$ radians, so dividing both by $12$ gives $15^\circ = \frac{\pi}{12}$.

$$\sin\frac{\pi}{12} = \sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588$$

The value is positive, and it is small. That matches intuition: $15^\circ$ is a shallow angle, close to $0^\circ$, and $\sin 0^\circ = 0$, so the sine should sit just above zero. Its partner value, $\cos\frac{\pi}{12} = \frac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659$, is close to $1$ for the same reason. You can confirm the sine on a calculator set to the correct mode, but the surd form is the exact answer, and the sections below show where it comes from.

How Do You Find Sin pi/12 Using The Difference Identity?

The cleanest route is to notice that $15^\circ$ is the gap between two angles you already know: $15^\circ = 45^\circ - 30^\circ$. That lets you use the sine angle difference identity, one of the sum and difference identities.

The identity itself:

$$\sin(A - B) = \sin A \cos B - \cos A \sin B$$

Set $A = 45^\circ$ and $B = 30^\circ$, then substitute the standard values $\sin 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, and $\sin 30^\circ = \frac{1}{2}$. These come straight from the values of sin at π/4 and π/6 on the standard table.

Substitute first:

$$\sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$$

$$\sin 15^\circ = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$$

Multiply each pair:

$$\sin 15^\circ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$$

Combine over the common denominator:

$$\sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}$$

Final answer: $\sin\frac{\pi}{12} = \dfrac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588$.

A quick note on the split. You could also write $15^\circ = 60^\circ - 45^\circ$ and use $\cos 60^\circ$, $\sin 60^\circ$, and the $45^\circ$ values instead. The arithmetic looks different on the page, yet it lands on the identical surd. Any correct decomposition of $15^\circ$ into table angles gives the same answer, which is a useful check when you are unsure.

Can You Find Sin pi/12 With The Half-Angle Formula?

Yes, and it is worth doing, because a second method that lands on the same value is the strongest evidence the value is right. Notice that $15^\circ$ is exactly half of $30^\circ$, so the half-angle formula for sine applies:

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

Take $\theta = 30^\circ$, so $\frac{\theta}{2} = 15^\circ$, and use $\cos 30^\circ = \frac{\sqrt{3}}{2}$ from the cosine of π/6. The positive root is correct here, since $15^\circ$ lands in the first quadrant where sine is positive.

Substitute:

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

Clear the inner fraction by writing $1 = \frac{2}{2}$:

$$\sin 15^\circ = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2-\sqrt{3}}}{2}$$

This form, $\frac{\sqrt{2-\sqrt{3}}}{2}$, evaluates to $0.2588$, the same decimal as before. It is the same number wearing a nested-root disguise: the identity $\sqrt{2-\sqrt{3}} = \frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}}$ turns one into the other, so both simplify to $\frac{\sqrt{6}-\sqrt{2}}{4}$. Two independent methods, one value.

Where Does 15° Sit On The Unit Circle?

On the unit circle, Sin pi/12 is the height (the $y$-coordinate) of the point you reach after turning $15^\circ$ anticlockwise from the positive $x$-axis. The point sits in the first quadrant, low and far to the right, at coordinates $\left(\cos 15^\circ, \sin 15^\circ\right) = \left(\frac{\sqrt{6}+\sqrt{2}}{4}, \frac{\sqrt{6}-\sqrt{2}}{4}\right) \approx (0.9659, 0.2588)$.

Table: The unit-circle point for π/12, both exactly and as decimals.

The right-triangle picture agrees. Drop a vertical line from the circle point to the $x$-axis, and you build a right triangle with hypotenuse $1$ (the radius). The side opposite the $15^\circ$ angle has length $\sin 15^\circ$, so $\text{sine} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sin 15^\circ}{1}$, the same number as the height on the circle. For the tangent line and the reciprocal ratios at this point, see the unit circle with tangent, and if the radian label is new, start with what a radian is.

Table: Sin pi/12 next to the neighbouring special angles, so you can see where it fits.

The neighbours have their own reference pages: the sine of π/6, the sine of π/3, and the full trigonometric table for every standard angle at once.

Why Is Sin pi/12 Equal To (√6 − √2)/4?

The short reason: $15^\circ$ is not a standard table angle, but it is built from angles that are. That single fact explains both why the answer is a surd and why you can find it exactly.

• It is not on the basic table. The angles $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, and $90^\circ$ have memorised sine values. $15^\circ$ is not among them, so there is no ready-made entry to read off.

• But it is reachable from the table. Since $15^\circ = 45^\circ - 30^\circ$ and also $15^\circ = \frac{30^\circ}{2}$, the difference identity and the half-angle formula both express it through angles that are on the table. That is why an exact value exists at all.

• Combining surds keeps surds. The table values for $45^\circ$ and $30^\circ$ already carry $\sqrt{2}$ and $\sqrt{3}$. Multiplying and subtracting them produces $\sqrt{6}$ and $\sqrt{2}$, and there is no way to cancel the roots. So the exact answer has to be a surd.

There is a neat pairing hiding here. Because $15^\circ$ and $75^\circ$ add to $90^\circ$, they are complementary, and the cofunction identities say $\sin 15^\circ = \cos 75^\circ$ and $\cos 15^\circ = \sin 75^\circ$. The same value therefore shows up under two different names, which is worth remembering when a problem hands you $75^\circ$ instead.

Who Discovered The Values Behind Sin pi/12?

Nobody woke up one morning and "discovered" $\sin 15^\circ$. The value is a consequence of the sine function itself, and building tables of that function took mathematicians across several civilisations more than a thousand years.

Two later mathematicians carried the sine table forward:

• Aryabhata (476–550 CE, India) tabulated the sine function directly, in a form he called jya, in his astronomical work the Aryabhatiya around 500 CE. The word "sine" traces back through Arabic to his term.

• Madhava of Sangamagrama (c. 1340 – c. 1425, India) found infinite series for sine and cosine roughly two centuries before they reappeared in Europe, giving a way to compute values like $\sin 15^\circ$ to as many decimal places as patience allowed.

Where Is Sin pi/12 Used In The Real World?

A $15^\circ$ angle and the difference identity that unlocks it are not classroom curiosities. They show up wherever precise angles between known directions matter.

• Surveying and navigation: surveyors constantly find one bearing as the difference of two measured bearings, and a $15^\circ$ separation is common. The exact sine keeps rounding error out of a chain of distance calculations.

• Waves and signals: a $15^\circ$ phase shift between two alternating signals is written using $\sin 15^\circ$, and combining offset waves is exactly the sum-and-difference identity at work.

• Antenna and radar arrays: steering a phased array to a target relies on the sine of the small angle between elements, where $15^\circ$-scale angles are routine.

• Computer graphics: rotating an object by $15^\circ$ multiplies its coordinates by $\sin 15^\circ$ and $\cos 15^\circ$, so the exact value feeds straight into the rotation.

• Astronomy: the field that started it all still measures small angular separations between objects, the same job Hipparchus faced.

One shallow angle, and the same square-root value quietly turns up in maps, sound, radio, screens, and the night sky.

What Are The Most Common Mistakes With Sin pi/12?

These four errors account for most wrong answers on $\sin 15^\circ$, confirmed against calculator-mode guides and step-by-step walkthroughs for this exact angle.

Splitting the sine across the subtraction.

Where it slips in:

A student writes $\sin(45^\circ - 30^\circ) = \sin 45^\circ - \sin 30^\circ$, treating sine as if it distributes over subtraction.

Don't do this:

Sine is not linear. $\sin(A - B)$ is never $\sin A - \sin B$. That shortcut gives $\frac{\sqrt{2}}{2} - \frac{1}{2} \approx 0.207$, which is wrong.

The correct way:

Use the full identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$. Every term pairs a sine with a cosine, and skipping the cross terms is what breaks the result.

Using the plus formula instead of the minus formula.

Where it slips in:

A student reaches for $\sin(A + B)$ after correctly splitting $15^\circ = 45^\circ - 30^\circ$, flipping the middle sign to a plus.

Don't do this:

Do not mix the two identities. The plus formula computes $\sin 75^\circ$, not $\sin 15^\circ$, and lands on $\frac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659$ instead.

The correct way:

Match the sign to the angle. A subtraction of angles takes the subtraction identity, so the middle sign stays negative: $\sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$.

Leaving the denominator as 2 instead of 4.

Where it slips in:

A student multiplies $\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}$ but writes the result as $\frac{\sqrt{6}}{2}$, forgetting that the two denominators of $2$ multiply to $4$.

Don't do this:

Do not carry only one factor of $2$. Both fractions contribute a denominator, so $\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{4}$, not $\frac{\sqrt{6}}{2}$.

The correct way:

Multiply numerators and denominators separately, then combine: $\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6}-\sqrt{2}}{4}$.

Leaving the calculator in the wrong angle mode.

Where it slips in:

A student types $\sin(\pi/12)$ with the calculator set to degrees, or types $\sin(15)$ with it set to radians, and copies down a value that does not match.

Don't do this:

Do not trust the display before checking the mode. In degree mode, $\sin(\pi/12)$ reads the input as $0.2618^\circ$ and returns about $0.0046$, nothing like the real answer.

The correct way:

Match the mode to the units. Set radians for $\frac{\pi}{12}$, set degrees for $15^\circ$, and confirm you land near $0.2588$ either way.

Practice Problems On Sin pi/12

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

1. Convert $\frac{\pi}{12}$ radians to degrees.
   (Answer: $\frac{\pi}{12} \times \frac{180^\circ}{\pi} = 15^\circ$.)

2. State $\sin\frac{\pi}{12}$ as a decimal to four places.
   (Answer: $0.2588$.)

3. Use $\cos(A - B) = \cos A \cos B + \sin A \sin B$ to find $\cos 15^\circ$.
   (Answer: $\frac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659$.)

4. Use the cofunction identity to write $\sin\frac{\pi}{12}$ as a cosine.
   (Answer: $\sin\frac{\pi}{12} = \cos\frac{5\pi}{12} = \cos 75^\circ$.)

5. Find $\tan\frac{\pi}{12}$ by dividing $\sin\frac{\pi}{12}$ by $\cos\frac{\pi}{12}$.
   (Answer: $\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} = 2-\sqrt{3} \approx 0.2679$.)

6. Without a calculator, decide which is larger: $\sin\frac{\pi}{12}$ or $\sin\frac{\pi}{6}$.
   (Answer: $\sin\frac{\pi}{6} = 0.5$ is larger, since sine grows as the angle grows from $0^\circ$ to $90^\circ$.)

Where Should You Go Next After Sin pi/12?

You now have the value, two derivations, and the unit-circle picture. A few natural doors open from here.

1. Sum and difference identities. The tool that unlocked $15^\circ$ also unlocks $75^\circ$, $105^\circ$, and any angle you can build from the table.

2. Trigonometric ratios of specific angles. Lock in the $30^\circ$, $45^\circ$, and $60^\circ$ values that every derivation on this page leaned on.

3. Trigonometric ratios in radians. Get comfortable moving between $\frac{\pi}{12}$ and $15^\circ$ so neither unit slows you down, starting from sin, cos, and tan.

If your child is meeting these identities for the first time, a live Bhanzu trainer teaches the difference formula from the unit circle up, so the surd form feels earned rather than memorised, in the Bhanzu trigonometry program.