Basic Properties Of Fourier Series. 2.

Let's solve (a).

The Fourier series of the function $f$ is $$ \sum_{n=-\infty}^{\infty}\hat{f}(n)e^{in\theta} $$

By definition, this is $$ \sum_{n=0}^{\infty}\hat{f}(n)e^{in\theta} + \sum_{n=1}^{\infty}\hat{f}(-n)e^{i(-n)\theta} $$

Taking out the $n=0$ term in the first series, this is $$ \hat{f}(0)+\sum_{n=1}^{\infty}\hat{f}(n)e^{in\theta} + \sum_{n=1}^{\infty}\hat{f}(-n)e^{i(-n)\theta}\tag{1} $$

Applying Euler's formula we have $$ e^{in\theta}=\cos(n\theta)+i\sin(n\theta)\\ e^{i(-n)\theta}=\cos((-n)\theta)+i\sin((-n)\theta) $$

Since $\cos$ is an even function and $\sin$ is an odd function, this is $$ e^{in\theta}=\cos(n\theta)+i\sin(n\theta)\\ e^{i(-n)\theta}=\cos(n\theta)-i\sin(n\theta) $$ Substituting in $(1)$, we obtain $$ \hat{f}(0)+\sum_{n=1}^{\infty}\hat{f}(n)[\cos(n\theta)+i\sin(n\theta)] + \sum_{n=1}^{\infty}\hat{f}(-n)[\cos(n\theta)-i\sin(n\theta)] $$ But this equals $$ \hat{f}(0)+\sum_{n=1}^{\infty}\left\{\hat{f}(n)[\cos(n\theta)+i\sin(n\theta)] + \hat{f}(-n)[\cos(n\theta)-i\sin(n\theta)]\right\} $$ which is nothing more than $$ \hat{f}(0)+\sum_{n=1}^{\infty}\left\{[\hat{f}(n) + \hat{f}(-n)]\cos(n\theta) + i[\hat{f}(n) - \hat{f}(-n)]\sin(n\theta)\right\} $$