The problem asks for the simplification of the expression $\sin 60^\circ \cos 15^\circ - \cos 60^\circ \sin 15^\circ$.

1. Problem Description

The problem asks for the simplification of the expression $\sin 60^\circ \cos 15^\circ - \cos 60^\circ \sin 15^\circ$.

2. Solution Steps

The given expression is of the form $\sin A \cos B - \cos A \sin B$.

Recall the trigonometric identity:

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

In this case, $A = 60^\circ$ and $B = 15^\circ$. Therefore, the expression can be simplified as follows:

$\sin 60^\circ \cos 15^\circ - \cos 60^\circ \sin 15^\circ = \sin(60^\circ - 15^\circ) = \sin(45^\circ)$

We know that $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$. However, this result is not in the given options. Let's look at the options carefully. The options are $\frac{1}{\sqrt{2}}$, $\frac{\sqrt{3}+1}{2\sqrt{2}}$, $\frac{\sqrt{3}}{2}$, and $\frac{1}{2}$.

Since the correct simplification should have been $\sin(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. I notice the original problem is slightly wrong. It should be $\sin 60^\circ \cos 15^\circ - \cos 60^\circ \sin 15^\circ$. It looks like there is a typo. The problem statement is

$\sin 60^\circ \cos 15^\circ - \cos 60^\circ \sin 15^\circ = \sin (60^\circ - 15^\circ) = \sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.

This answer is among the multiple choices.

However, the question could be asking for something different.

Let's try the following:

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

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

Consider $A = 60^\circ$ and $B = 15^\circ$, then we have:

$\sin(60^\circ) = \frac{\sqrt{3}}{2}$

$\cos(60^\circ) = \frac{1}{2}$

We also have $15^\circ = 45^\circ - 30^\circ$, so we can find $\sin(15^\circ)$ and $\cos(15^\circ)$ using the angle subtraction formulas.

$\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$

$\cos(15^\circ) = \cos(45^\circ - 30^\circ) = \cos(45^\circ)\cos(30^\circ) + \sin(45^\circ)\sin(30^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$

So, $\sin 60^\circ \cos 15^\circ - \cos 60^\circ \sin 15^\circ = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} - \frac{1}{2} \cdot \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{18} + \sqrt{6}}{8} - \frac{\sqrt{6} - \sqrt{2}}{8} = \frac{3\sqrt{2} + \sqrt{6} - \sqrt{6} + \sqrt{2}}{8} = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

This result is among the multiple choices. Now, consider the case where it is $\sin (60-15)$ instead.

$\sin(60^\circ - 15^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{2}}$ is one of the options.

3. Final Answer

$\frac{1}{\sqrt{2}}$