Sin 15 Degrees - Exact Value (√6−√2)/4 Explained

The value of sin 15 degrees is $\dfrac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588$.

Quick Answer:

Result: $\sin 15° = \dfrac{\sqrt{6}-\sqrt{2}}{4}$

Decimal: $0.2588$ (to four places)

In radians: $\sin\left(\dfrac{\pi}{12}\right) = \dfrac{\sqrt{6}-\sqrt{2}}{4}$

Method shown: difference formula $\sin(45°-30°)$

Exact form: $\dfrac{\sqrt{6}-\sqrt{2}}{4}$ (equivalently $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$, exact, never terminates)

What Sin 15 Degrees Means

On the unit circle — a circle of radius $1$ centred at the origin — the sine of an angle is the $y$-coordinate of the point where the angle's radius meets the circle. Rotating $15°$ counterclockwise from the positive $x$-axis lands just above it in Quadrant I, at the point $\left(\cos 15°, \sin 15°\right)$, whose height above the $x$-axis is $\sin 15° = \frac{\sqrt{6}-\sqrt{2}}{4}$.

The right-triangle definition — opposite over hypotenuse — agrees here because $15°$ is acute, but $15°$ is not one of the angles whose ratio you can read straight off a $30$-$60$-$90$ or $45$-$45$-$90$ triangle. That is why the value has to be built from angles you do know, rather than looked up.

How to Find the Value of Sin 15 Degrees

The cleanest route writes $15°$ as the difference of two standard angles. A reader landing here usually asks one of two things first.

Should I use the difference formula or the half-angle formula?

Both give the same answer. The difference formula treats $15°$ as $45° - 30°$ and is the more direct of the two; the half-angle formula treats $15°$ as half of $30°$ and is handy when the target is naturally half of a known angle. This article uses the difference formula.

Method 1: The 45° − 30° difference formula

The sine difference identity is

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

Set $A = 45°$ and $B = 30°$, then substitute the known standard values $\sin 45° = \frac{\sqrt{2}}{2}$, $\cos 30° = \frac{\sqrt{3}}{2}$, $\cos 45° = \frac{\sqrt{2}}{2}$, $\sin 30° = \frac{1}{2}$:

$$\sin 15° = \sin(45° - 30°) = \sin 45°\cos 30° - \cos 45°\sin 30°$$

$$\sin 15° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2}$$

$$\sin 15° = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6}-\sqrt{2}}{4}.$$

Final answer: $\sin 15° = \dfrac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588.$

Method 2: From the unit circle (the check)

Mark $15°$ on the unit circle and drop a vertical from the point to the $x$-axis. The height of that segment is the sine. Measured, it is about $0.2588$, which matches the surd above once you substitute $\sqrt{6} \approx 2.449$ and $\sqrt{2} \approx 1.414$:

$$\frac{2.449 - 1.414}{4} = \frac{1.035}{4} \approx 0.2588.$$

Examples of Sin 15 Degrees

Example 1

Evaluate $4\sin 15°$.

$$4\sin 15° = 4\cdot\frac{\sqrt{6}-\sqrt{2}}{4} = \sqrt{6}-\sqrt{2} \approx 1.035.$$

Example 2

Find $\sin 15°$ by writing $15°$ as $60° - 45°$ instead.

The order of the angles is the part students rush. Subtracting in the wrong direction would give $\sin(45° - 60°) = -\sin 15°$, a negative number, which cannot be right because $15°$ is in Quadrant I where sine is positive. Keep the larger angle first:

$$\sin 15° = \sin(60° - 45°) = \sin 60°\cos 45° - \cos 60°\sin 45°$$

$$\sin 15° = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} - \frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}.$$

Same answer as the $45° - 30°$ split — the decomposition you choose does not change the value.

Example 3

Show that $\sin 15° + \cos 15° = \dfrac{\sqrt{6}}{2}$, given $\cos 15° = \dfrac{\sqrt{6}+\sqrt{2}}{4}$.

$$\frac{\sqrt{6}-\sqrt{2}}{4} + \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2}.$$

The $\sqrt{2}$ terms cancel, leaving a single surd — see cos 15 degrees for where that cosine comes from.

Example 4

Verify $\sin^2 15° + \cos^2 15° = 1$.

$$\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^2 + \left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \frac{(8-2\sqrt{12}) + (8+2\sqrt{12})}{16} = \frac{16}{16} = 1.$$

The Pythagorean identity holds: the cross terms cancel and the squares sum to one.

Example 5

Express $15°$ in radians and state the value.

$15° = 15 \times \frac{\pi}{180} = \frac{\pi}{12}$ radians, so $\sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{6}-\sqrt{2}}{4}$. Converting between radians and degrees leaves the value unchanged.

Common Mistakes With Sin 15 Degrees

Mistake 1: Splitting the sine of a difference

Where it slips in: The first instinct on $\sin(45° - 30°)$ is to subtract the sines.

Don't do this: Writing $\sin 15° = \sin 45° - \sin 30° = \frac{\sqrt{2}}{2} - \frac{1}{2} \approx 0.207$.

The correct way: Sine does not distribute over subtraction. The habit that fixes it is to reach for the identity $\sin(A-B) = \sin A\cos B - \cos A\sin B$ every time an angle is split — the value $0.2588$ from the identity exposes the $0.207$ shortcut as wrong on sight.

Mistake 2: Subtracting the angles in the wrong order

Where it slips in: Writing $15°$ as $30° - 45°$ and applying the formula literally.

Don't do this: $\sin(30° - 45°)$, which equals $-\sin 15°$ and flips the sign.

The correct way: Put the larger standard angle first, either $45° - 30°$ or $60° - 45°$, so the result stays positive, matching Quadrant I. The learner who has just met compound angles tends to subtract left-to-right out of reflex, and a quick quadrant check catches it.

Mistake 3: Reporting only the decimal

Where it slips in: Exam questions ask for the exact value, but a calculator hands back $0.2588$.

Don't do this: Writing $\sin 15° = 0.2588$ when the question says "exact."

The correct way: Keep the surd form $\frac{\sqrt{6}-\sqrt{2}}{4}$. The decimal is rounded; the surd is exact, and that distinction is exactly what the question is testing.

Key Takeaways

• Sin 15 degrees equals the exact surd $\frac{\sqrt{6}-\sqrt{2}}{4}$, about $0.2588$.

• It is built by writing $15°$ as $45° - 30°$ and applying the sine difference formula.

• The angle sits in Quadrant I on the unit circle, so the value is positive.

• The most common error is splitting $\sin(45°-30°)$ into a difference of sines — sine is not linear.

Practice These Before Moving On

1. Use the $45° - 30°$ split to derive $\cos 15°$, then check $\sin^2 15° + \cos^2 15° = 1$.

2. Evaluate $\sin 15° \cdot \cos 15°$ and confirm it equals $\frac{1}{2}\sin 30° = \frac{1}{4}$.

3. Convert $15°$ to radians and write the value as $\sin\left(\frac{\pi}{12}\right)$.

To work through more compound-angle values with a teacher, Bhanzu's trigonometry tutor and high school math tutor sessions cover the difference and half-angle formulas in depth, with math classes online for live practice.

Read More

• Sum and difference identities — the full family of compound-angle formulas.

• Cos 30 degrees — one of the standard angles sin 15° is built from.

• Trigonometric table — every standard-angle ratio in one chart.

• Trigonometric ratios — how sine, cosine, and tangent are defined.

• Trigonometric functions — the six functions and how they behave across the circle.