Going for a proof by contradiction, lets assume both inequalities hold. It is clear that the statement can be re-written as $||f(x) - \sin(x)||_2 \leq \frac{2}{3}$ and $||f(x) - \cos(x)||_2 \leq \frac{1}{3}$. At this point, we can use Minkowski's inequality (given $f,g \in L^p(E)$, $||f+g||_p \leq ||f||_p + ||g||_p$), to get $$\sqrt{\int_0^\pi |(f(x) - \cos(x)) + (f(x) - \sin(x))|^2 dx} \leq ||f(x) - \cos(x)||_2+||f(x) - \sin(x)||_2 \leq 1$$ Trying to expand the left hand side seems way too messy and circular. I am not sure how to proceed from here. It somehow seems like I am overcomplicating this problem.
I was also looking into Holder's inequality, but once again, having to compute $||(f(x) - \sin(x))(f(x)-\cos(x))||_1$ seems very messy.
Any hint/help is appreciated!