To evaluate \(\cos^2\left(\frac{\pi}{12}\right)\), we can first find \(\cos\left(\frac{\pi}{12}\right)\).
We know that \(\frac{\pi}{12} = 15^\circ\). We can use the cosine angle subtraction formula:
\[ \cos\left(\frac{\pi}{12}\right) = \cos\left(15^\circ\right) = \cos(45^\circ - 30^\circ) \]
Using the cosine subtraction formula:
\[ \cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b) \]
So,
\[ \cos\left(15^\circ\right) = \cos(45^\circ)\cos(30^\circ) + \sin(45^\circ)\sin(30^\circ) \] \[ = \left(\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\right) \] \[ = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \] \[ = \frac{\sqrt{6} + \sqrt{2}}{4} \]
Now we need \(\cos^2\left(15^\circ\right)\):
\[ \cos^2\left(\frac{\pi}{12}\right) = \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2 \] \[ = \frac{(\sqrt{6} + \sqrt{2})^2}{16} \] \[ = \frac{6 + 2 + 2\sqrt{12}}{16} \] \[ = \frac{8 + 4\sqrt{3}}{16} \] \[ = \frac{2 + \sqrt{3}}{4} \]
Thus,
\[ \cos^2\left(\frac{\pi}{12}\right) = \frac{2 + \sqrt{3}}{4} \]
So, the correct answer is:
\(\frac{2 + \sqrt{3}}{4}\)