We have that the function f defined on [-$\pi$,$\pi$] is continuous, and we want to prove the following statement about its Fourier coefficients $c_n$
$$c_n = c_{-n}\;\; \text{for all}\; n\; \Rightarrow \text{f even}. $$
I am using the convention
$$c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t)e^{-int}dt .$$
I have come up with two ways to show it but I am not sure if there are any flaws with method 2.
Method 1: $$c_n = c_{-n} \iff \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t)e^{-int}dt = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t)e^{int}dt $$
The variable change $t\to -t$ in the second integral gives the equality $$ \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t)e^{-int}dt = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(-t)e^{-int}dt \\ \iff \frac{1}{2\pi} \int_{-\pi}^{\pi} (f(t)-f(-t))e^{-int}dx = 0 $$ Now, the integrand in the last integral is a continuous function and for such functions the integral is zero iff the integral is identically zero over that interval. Hence
$$ (f(t)-f(-t))e^{-int} = 0 \iff f(t) = f(-t), t\in[-\pi,\pi] \iff \text{f even}. $$
Method 2:
$$ c_n = c_{-n} \iff \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t)e^{-int}dt = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t)e^{int}dt \\ \iff \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t)(e^{int}-e^{-int})dx = 0 \iff \int_{-\pi}^{\pi} f(t)sin(nt) dt = 0 \\ \Rightarrow f(t)sin(nt)\;\; \text{odd}\; \Rightarrow f(t)\; \text{even}.$$
I am unsure of whether I can conclude that f(t)sin(nt) is an odd function. I guess there is the possibility that the integral over the symmetric interval vanishes for other reasons. I think the implication only goes in the direction of "if g is odd then $\int_{-a}^a g(t)dt = 0$".
I also wanted to ask if the statements $$ c_n = -c_{-n} \; \Rightarrow \; \text{f odd}. $$ and $$\overline{c_n} = c_{-n} \; \Rightarrow \; \text{f real-valued}$$ are proven in the same manner as the above.