2. $$ \cos \left(15^{\circ}\right) $$

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To find $$ \cos(15^\circ) $$, we can use the cosine subtraction formula. We know that:

$$
15^\circ = 45^\circ - 30^\circ
$$

Using the cosine subtraction formula:

$$
\cos(a - b) = \cos a \cos b + \sin a \sin b
$$

we can substitute $$ a = 45^\circ $$ and $$ b = 30^\circ $$:

$$
\cos(15^\circ) = \cos(45^\circ) \cos(30^\circ) + \sin(45^\circ) \sin(30^\circ)
$$

Now, we need the values of $$ \cos(45^\circ) $$, $$ \cos(30^\circ) $$, $$ \sin(45^\circ) $$, and $$ \sin(30^\circ) $$:

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

Substituting these values into the formula:

$$
\cos(15^\circ) = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)
$$

Calculating each term:

$$
\cos(15^\circ) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}
$$

Combining the terms:

$$
\cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}
$$

Thus, the value of $$ \cos(15^\circ) $$ is:

$$
\boxed{\frac{\sqrt{6} + \sqrt{2}}{4}}
$$
$$ \cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} $$

2. \( \cos \left(15^{\circ}\right) \)

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