Let $\mathbb T=\{z\in \mathbb C: z=e^{i x}$ for some real $x \}$, then $L^2 (\mathbb T)$ is a Hilbert space with the inner product $\langle f,g\rangle=\frac 1 {2\pi}\int _0^{2\pi}f (x)\overline{g(x)} dx$. The functions $e_n=e^{inx}$, $n $ integer, form an orthonormal basis for $L^2 (\mathbb T)$, and $\langle f,e_n \rangle = \hat f (n)$.
From here in the course we introduced the real form of Fourier transform, using $e_n =\cos (nx) +i \sin (nx)$ in the linear combination $\sum\langle f,e_n \rangle e_n$, and since the inner products between the $\cos nx $ and the $\sin mx$ are zero, the $ \cos (nx),\sin (nx)$ form an orthonormal basis.
So now I come to the part that is not clear: if I do $\langle f,\cos (nx) \rangle + \langle f,\cos (-nx) \rangle =$ $=2\langle f,\cos (nx) \rangle$ (since the cosine is even) I obtain the same result as doing the calculations described in the paragraph above; however the sine is odd so $\langle f,\sin (nx) \rangle + \langle f,\sin (-nx) \rangle =0$, that clearly is different from the coefficient of $\sin (nx)$ that one finds proceeding as above. I just can't find the mistake, thank you in advance.
Consider $f:\mathbb{T}\to\mathbb{C}$ with $f\in L^2(\Bbb{T})$.
For the basis $(e^{inx})_{n\in\Bbb{Z}}$, $$ \hat{f}(n)=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{inx}\,dx. $$
If one considers the real basis $\{\sin (x), \sin (2x), \cdots, \}\cup\{1,\cos(x),\cos(2x),\cdots\}$, then $$ \langle f(x), \cos(nx)\rangle=\langle f(x),\frac{e^{inx}+e^{-inx}}{2}\rangle =\frac{1}{2}\big( \hat{f}(n)+\hat{f}(-n) \big) $$ and $$ \langle f(x), \sin(nx)\rangle=\langle f(x),\frac{e^{inx}-e^{-inx}}{2i}\rangle =\frac{1}{2i}\big( \hat{f}(n)+\hat{f}(-n) \big). $$