Write the expression as a function of $\theta$, with no angle measure involved. \[ \cos \left(45^{\circ}-\theta\right) \] \[ \cos \left(45^{\circ}-\theta\right)= \] (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Step 1 ：We know that \(\cos (a - b) = \cos a \cos b + \sin a \sin b\).

Step 2 ：Substitute \(a = 45^\circ\) and \(b = \theta\) into the formula, we get \(\cos (45^\circ - \theta) = \cos 45^\circ \cos \theta + \sin 45^\circ \sin \theta\).

Step 3 ：From the solution of QuestionA, we know that \(\cos 45^\circ = \frac{\sqrt{2}}{2}\) and \(\sin 45^\circ = \frac{\sqrt{2}}{2}\).

Step 4 ：Substitute \(\cos 45^\circ = \frac{\sqrt{2}}{2}\) and \(\sin 45^\circ = \frac{\sqrt{2}}{2}\) into the formula, we get \(\cos (45^\circ - \theta) = \frac{\sqrt{2}}{2} \cos \theta + \frac{\sqrt{2}}{2} \sin \theta\).

Step 5 ：Factor out \(\frac{\sqrt{2}}{2}\), we get \(\cos (45^\circ - \theta) = \frac{\sqrt{2}}{2} (\cos \theta + \sin \theta)\).

Step 6 ：\(\boxed{\cos (45^\circ - \theta) = \frac{\sqrt{2}}{2} (\cos \theta + \sin \theta)}\) is the final answer.