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

## TL;DR  
The value of sin 15 degrees is exactly \( \frac{\sqrt{6}-\sqrt{2}}{4} \), about 0.2588. This article derives it by writing 15° as 45° − 30°, places the angle on the unit circle, gives a standard-angle table, and works through examples and the mistakes that cost marks.

The value of **sin 15 degrees** is \( \frac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588 \).

> **Quick Answer**:  
> **Result:** \( \sin 15°= \frac{\sqrt{6}-\sqrt{2}}{4} \)  
> **Decimal:** 0.2588 (to four places)  
> **In radians:** \( \sin(\frac{\pi}{12})= \frac{\sqrt{6}-\sqrt{2}}{4} \)  
> **Method shown:** difference formula \( \sin(45°-30°) \)  
> **Exact form:** \( \frac{\sqrt{6}-\sqrt{2}}{4} \)

## 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 \( (\cos 15°,\sin 15°) \), 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 it 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.  
### **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 \( B=30° \) and substitute the known standard values:  
\[ \sin 15°=\sin(45°−30°)=\sin 45°\cos 30°−\cos 45°\sin 30° \]  
\[ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} \]  
\[ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6}-\sqrt{2}}{4}. \]

**Final answer:** \( \sin 15°=\frac{\sqrt{6}-\sqrt{2}}{4} \approx 0.2588. \)

### **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.**  
\[ \sin 15°=\sin(60°−45°)=\sin 60°\cos 45°−\cos 60°\sin 45° \]  
\[ = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} - \frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{6}-\sqrt{2}}{4}. \]

### Example 3  
**Show that \( \sin 15°+\cos 15°=\frac{\sqrt{6}}{2} \).**  
\[ \frac{\sqrt{6}-\sqrt{2}}{4} + \frac{\sqrt{6}+\sqrt{2}}{4} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2}. \]

### Example 4  
**Verify \( \sin^2 15°+\cos^2 15°=1 \).**  
\[ \left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^2 + \left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \frac{16}{16} = 1. \]

### Example 5  
**Express 15° in radians and state the value.**  
\[ 15°=15×\frac{\pi}{180}=\frac{\pi}{12}. \]so \( \sin(\frac{\pi}{12})=\frac{\sqrt{6}-\sqrt{2}}{4} \).

## 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:** \( \sin 15°=\sin 45°−\sin 30°=\frac{\sqrt{2}}{2}−\frac{1}{2} \).  
**The correct way:** Sine does not distribute over subtraction.

### 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:** \( \sin(30°−45°)=-\sin 15° \).

### Mistake 3: Reporting only the decimal  
**Where it slips in:** Exam questions ask for the _exact_ value, but a calculator hands back 0.2588.

## 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.