Find the value of 2 sin 45 ∘ cos 15

Explanation

To find the value of \( 2 \sin 45^\circ \cos 15^\circ \), we can follow these steps:

### Step 1: Identify the values of \( \sin 45^\circ \) and \( \cos 15^\circ \)

We know from trigonometric values that:
\[ \sin 45^\circ = \frac{\sqrt{2}}{2} \]
To find \( \cos 15^\circ \), we can use the cosine angle subtraction formula:
\[ \cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \]
Using known values:
\[ \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2} \]
Substituting these values in:
\[ \cos 15^\circ = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \]

### Step 2: Substitute the values into the expression

Now we substitute \( \sin 45^\circ \) and \( \cos 15^\circ \) into the expression:
\[ 2 \sin 45^\circ \cos 15^\circ = 2 \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \]

### Step 3: Simplify the expression

The \( 2 \) and \( \frac{1}{2} \) cancel out:
\[ = \frac{\sqrt{2}(\sqrt{6} + \sqrt{2})}{4} \]
Distributing \( \sqrt{2} \):
\[ = \frac{\sqrt{12} + \sqrt{4}}{4} = \frac{2\sqrt{3} + 2}{4} \]
This simplifies to:
\[ = \frac{2(\sqrt{3} + 1)}{4} = \frac{\sqrt{3} + 1}{2} \]

### Final Result

Thus, the value of \( 2 \sin 45^\circ \cos 15^\circ \) is:
\[ \frac{\sqrt{3} + 1}{2} \]