Consider the function $$ f(\theta)=\frac{1}{2}(a+i b) e^{i \theta}+\frac{1}{2}(a-i b) e^{-i \theta} $$ where \(a\) and \(b\) are real numbers. Show that \(f\) can be written in the form $$ f(\theta)=C_{1} \cos \theta+C_{2} \sin \theta $$ and determine the values of \(C_{1}\) and \(C_{2}\).

Euler's formula states that \(e^{ix} = \cos x + i \sin x\). By applying this formula, we can rewrite the given function \(f(\theta)\) as $$ f(\theta) = \frac{1}{2}(a+ib)(\cos \theta + i \sin \theta) + \frac{1}{2}(a-ib)(\cos (-\theta) + i \sin (-\theta)). $$ Since \(-\theta\) just results in opposites for sin while the cos is the same, we can go ahead and replace \(-\theta\) with \(\theta\) in our expression to simplify it.

Replace the trigonometric functions with their equivalent counterparts and simplify the expression: $$ f(\theta) = \frac{1}{2}(a+ib)(\cos \theta + i \sin \theta) + \frac{1}{2}(a-ib)(\cos \theta - i \sin \theta). $$ Now, distribute the constants to both trigonometric terms in each of the parentheses: $$ f(\theta) = \frac{1}{2}(a \cos \theta + ib \cos \theta + i a \sin \theta - b \sin \theta) + \frac{1}{2}(a \cos \theta - ib \cos \theta - i a \sin \theta - b \sin \theta). $$ Next, combine like terms: $$ f(\theta) = (a \cos \theta - b \sin \theta) + i (b \cos \theta + a \sin \theta). $$ Since a real function does not have imaginary parts, we only need the real part of the function, which gives: $$ f(\theta) = a \cos \theta - b \sin \theta. $$

Comparing the simplified function to the required form, we can see that $$ f(\theta) = C_{1} \cos \theta + C_{2} \sin \theta = a \cos \theta - b \sin \theta. $$ Thus, we can conclude that \(C_1 = a\) and \(C_2 = -b\). Therefore, the function can indeed be written in the required form with \(C_1 = a\) and \(C_2 = -b\).