Saw this question in a test, but got stuck, here is my attempt:
\begin{align} \frac{2}{\pi} \int_{0}^{2\pi} f(e^{i\theta})\cos^2\bigg(\frac{\theta}{2}\bigg)\,d\theta &= \frac{2}{\pi} \int_{0}^{2\pi} f(e^{i\theta})\frac{1+\cos(\theta)}{2}\,d\theta \\ &= \frac{1}{\pi} \int_{0}^{2\pi} f(e^{i\theta}) + \frac{1}{\pi} \int_{0}^{2\pi} f(e^{i\theta})\cos(\theta)\,d\theta \\ &= 2f(0) + \frac{1}{\pi} \int_{0}^{2\pi} f(e^{i\theta})\cos(\theta)\,d\theta . \end{align}
Now it only remains to show $$ \frac{1}{\pi} \int_{0}^{2\pi} f(e^{i\theta})\cos(\theta)\,d\theta = f'(0), $$ but this where I am stuck. From Cauchy's formula: \begin{align} f'(0) &= \frac{1}{2\pi i} \int_{0}^{2\pi} \frac{f(e^{i\theta})}{e^{2i\theta}}ie^{i\theta}\,d\theta \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} \frac{f(e^{i\theta})}{e^{i\theta}}\,d\theta \\ &= \frac{1}{2\pi} \int_{0}^{2\pi} f(e^{i\theta})e^{-i\theta}\,d\theta \\ &=\frac{1}{2\pi} \int_{0}^{2\pi} f(e^{i\theta})\cos(\theta)\,d\theta - \frac{i}{2\pi} \int_{0}^{2\pi} f(e^{i\theta})\sin(\theta)\,d\theta, \end{align} Which is clearly not $$ \frac{1}{\pi} \int_{0}^{2\pi} f(e^{i\theta})\cos(\theta)\,d\theta$ unless $\int_{0}^{2\pi} f(e^{i\theta})\sin(\theta)\,d\theta = 0. $$ I haven't a clue as $$ \int_{0}^{2\pi} f(e^{i\theta})\sin(\theta)\,d\theta = 0 $$ or whether this actually makes sense. Any help would be appreciated.