Book A Free Math Class

# Cos 15 Degrees - Exact Value (√6+√2)/4 Explained

[Trigonometry](/content/tag/trigonometry/index.html)

TL;DR

The value of cos 15 degrees is exactly $\frac{\sqrt{6}+\sqrt{2}}{4}$, about 0.9659. This article derives it by writing 15° as 45° − 30°, locates the angle on the unit circle, gives a standard-angle table, and works through examples and the mistakes that catch students out.

## What Cos 15 Degrees Means

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

The right-triangle definition — adjacent over hypotenuse — also applies because $15°$ is acute. But $15°$ is not an angle whose ratio you can read off a $30$-$60$-$90$ or $45$-$45$-$90$ triangle, so the value must be built from angles you already know rather than looked up.

## How to Find the Value of Cos 15 Degrees

The direct route writes $15°$ as the difference of two standard angles. Readers usually arrive with one question.

**Difference formula or half-angle formula?** Both reach the same value. The difference formula reads $15°$ as $45° - 30°$ and is the more direct; the half-angle formula reads $15°$ as half of $30°$ and suits angles that are naturally half of a known one. This article uses the difference formula.

### Method 1: The 45° − 30° difference formula

The cosine difference identity is

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

Note the sign: cosine of a difference uses a **plus**, where sine of a difference used a minus. Set $A = 45°$, $B = 30°$, and substitute $\cos 45° = \frac{\sqrt{2}}{2}$, $\cos 30° = \frac{\sqrt{3}}{2}$, $\sin 45° = \frac{\sqrt{2}}{2}$, $\sin 30° = \frac{1}{2}$:

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

$$\cos 15° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2}$$

$$\cos 15° = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6}+\sqrt{2}}{4}.$$

**Final answer:** $\cos 15° = \frac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659.$

### Method 2: From the unit circle (the check)

Mark $15°$ and measure the horizontal distance from the origin to the foot of the point — that is the cosine. It reads about $0.9659$, which matches the surd once you substitute $\sqrt{6} \approx 2.449$ and $\sqrt{2} \approx 1.414$:

$$\frac{2.449 + 1.414}{4} = \frac{3.863}{4} \approx 0.9659.$

## Examples of Cos 15 Degrees

### Example 1

**Evaluate $4\cos 15°$.**

$$4\cos 15° = 4\cdot\frac{\sqrt{6}+\sqrt{2}}{4} = \sqrt{6}+\sqrt{2} \approx 3.863.$$

### Example 2

**Find $\cos 15°$ using the difference form $\cos(45° - 30°)$, but watch the operation sign.**

The tempting move is to copy the _sine_ rule and subtract the products. Try it: $\cos 45°\cos 30° - \sin 45°\sin 30° = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588$. That cannot be cos 15° — it is far too small for an angle so close to $0°$, where cosine should be near $1$. The cosine _difference_ formula uses a plus:

$$\cos 15° = \cos 45°\cos 30° + \sin 45°\sin 30° = \frac{\sqrt{6}+\sqrt{2}}{4} \approx 0.9659.$$

The $0.2588$ that the wrong sign produced is actually $\sin 15°$, a useful tell that the sign was flipped.

### Example 3

**Show that $\cos 15° - \sin 15° = \frac{\sqrt{2}}{2}$, given $\sin 15° = \frac{\sqrt{6}-\sqrt{2}}{4}$.**

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

The $\sqrt{6}$ terms cancel — the mirror of the [sin 15 degrees](/content/math/trigonometry/sin-15-degrees/index.html) sum, where the $\sqrt{2}$ terms cancelled instead.

### Example 4

**Verify $\cos^2 15° + \sin^2 15° = 1$.**

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

The [Pythagorean identity](/content/math/trigonometry/pythagorean-identities/index.html) holds: the cross terms cancel.

### Example 5

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

$15° = 15 \times \frac{\pi}{180} = \frac{\pi}{12}$ radians, so $\cos\left(\frac{\pi}{12}\right) = \frac{\sqrt{6}+\sqrt{2}}{4}$. Converting between [radians and degrees](/content/math/geometry/radians-to-degrees/index.html) does not change the value.

## Common Mistakes With Cos 15 Degrees

### Mistake 1: Using the minus sign from the sine formula

**Where it slips in:** The cosine difference formula sits one line below the sine one, and the signs are opposite.

**Don't do this:** Writing $\cos(45° - 30°) = \cos 45°\cos 30° - \sin 45°\sin 30°$.

**The correct way:** Cosine of a difference uses a **plus**: $\cos(A-B) = \cos A\cos B + \sin A\sin B$. The reliable habit is a magnitude check — cos 15° is for an angle near $0°$, so the answer must be close to $1$; a result near $0.26$ means the sign was wrong.

### Mistake 2: Splitting the cosine of a difference

**Where it slips in:** Reaching for $\cos 45° - \cos 30°$ as if cosine were linear.

**Don't do this:** $\cos 15° = \cos 45° - \cos 30° = \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \approx -0.159$.

**The correct way:** Cosine does not distribute over subtraction, and a _negative_ answer for a Quadrant I angle is impossible. Apply the full identity. The learner meeting compound angles for the first time often subtracts the function values directly — the negative result is the giveaway.

### Mistake 3: Leaving the answer as a decimal

**Where it slips in:** A calculator returns $0.9659$, but the question asks for the exact value.

**Don't do this:** Writing $\cos 15° = 0.9659$ when "exact" is required.

**The correct way:** Keep $\frac{\sqrt{6}+\sqrt{2}}{4}$. The decimal is rounded; the surd is exact and is what the question is checking.

## Key Takeaways

- **Cos 15 degrees** equals the exact surd $\frac{\sqrt{6}+\sqrt{2}}{4}$, about $0.9659$.

- It is built by writing $15°$ as $45° - 30°$ and applying the cosine difference formula — note the plus sign.

- The angle sits in Quadrant I on the unit circle, so the value is positive and close to one.

- The most common error is borrowing the minus sign from the sine difference formula.

## Practice These Before Moving On

1. Derive $\sin 15°$ from the $45° - 30°$ split and confirm $\cos^2 15° + \sin^2 15° = 1$.

2. Use $\cos 2\theta = 2\cos^2\theta - 1$ with $\theta = 15°$ to recover $\cos 30° = \frac{\sqrt{3}}{2}$.

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