Show that $\int_{-\pi}^{\pi} f(t) \sin (nt) dt \rightarrow 0 , \int_{-\pi}^{\pi} f(t) \cos (nt) dt \rightarrow 0$ For any integrable function $f:  S^1 \rightarrow \mathbb{C}$ , I need to show that $$ \int_{-\pi}^{\pi} f(t) \sin (nt) dt \rightarrow 0$$ $$ \int_{-\pi}^{\pi} f(t) \cos (nt) dt \rightarrow 0$$ as $n \rightarrow \infty$
Now, here is what I did - 
Using the Riemann-Lebesgue Lemma , I got $$\hat{f}(n) = \int_{-\pi}^{\pi} f(x) e^{- inx} dx \rightarrow 0 $$ as $n \rightarrow \infty$
Now breaking $e^{- inx}$ as $\cos(nx) - i\sin(nx)$ , I get
$$\int_{-\pi}^{\pi} f(x) (\cos(nx) - i\sin(nx)) dx \rightarrow 0 $$
If $f(x)$ is a real function then the proving is trivial. But I am not able to find a proof if $f$ is complex. 
I have seen lim$_{n\rightarrow \infty}\int _{-\pi}^\pi f(t)\cos nt\,dt$ but can someone suggest a different methodology from Fourier analysis point of view.

Um, I think you can just break $f$ into real and imaginary parts, and apply Riemann Lebesgue Lemma to each, just as you did for real valued $f$ here?
