For $2\pi-$periodic and continuous $f$, does $\int_{-\pi}^{\pi} f(x) \cos (nx) = 0$ for $n = 0, 1, 2 \dots$ imply $f$ is odd? Similarly, does $\int_{-\pi}^{\pi} f(x) \sin (nx) = 0$ for $n = 0, 1, 2 \dots$ imply $f$ is even?
Motivation. If we are given $f$ is $2\pi$ periodic and we are given both conditions, we can show that $f \equiv 0$ by considering a sequence of trigonometric polynomials which converge uniformly to $f$. Can we get this same result by showing that the first individually shows that $f$ is odd and the second that $f$ is even so that we get $f \equiv 0$?
Let $$s_n(x) = \sum_{n = 1}^{\infty} b_n \sin (nx)$$ then $$s_n(x)\to f(x)\Rightarrow -s_n(x)=s_n(-x)\to f(-x) $$ also $$-s_n(x)=s_n(-x)\to -f(x)$$ uniqueness of limit implies $f(-x)=-f(x)$