This is a reposting of a previous question whose answer was accepted, but after having introduced a crucial difference in the hypothesis which would now require a different answer.
Considering this trivial (at $a=\pi$) definite integrals inequality:
$$
\mbox{with} \;a \in (0, \pi]\; \mbox{such that} \;\; \int_{0} ^{a} \cos(x) dx \; = \; 0 \; \; \; \;\Longrightarrow \; \; \; \; \int_{0} ^{a} \sin(x) dx \; \neq \; 0
$$
Inserting now a multiplying function $f(x)$ (a not identically zero continuous real valued function of real variable), and after having tried to visualize the evolution of the total signed area for various $f(x)$ examples, it would appear to me that also the following inequality might hold:
$$
\mbox{with} \;a \in (0, \pi]\; \mbox{such that} \;\; \int_{0} ^{a} \cos(x)\;f(x) \; dx \; \; = \; 0 \; \; \;\Longrightarrow \; \; \; \; \int_{0} ^{a} \sin(x) \;f(x) \; dx \; \neq \; 0
$$
is this true ?
P.S.: I have further restricted the domain for $a$, as I believe it is in this way easier to visualize various examples of $f(x)$ in terms of total signed area when plugged into that integral, while still keeping myself interested in this sort of inequalities. Moreover, if the above inequality is false for some particular counterexample, it would be interesting to explore which additional conditions on $f(x)$ might instead make it true.
The following edit/addition is to illustrate an approach which so far I was unable to develop into a definitive answer.
Let us consider a smooth $f(x)$ as the red curve here below:
if we consider it as the first period of the following periodic function:
then, $f(x)$ can be represented by a Fourier series (note that the period of $f(x)$ is $\pi$)
$$
a_0 \; +\; \sum _{k=1}^{\infty}{a_k} \cos(2kx) \, + \, \sum _{k=1}^{\infty}{b_k} \sin(2kx)
$$
whose elementary sines and cosines periods fit an integer number of times in $[0, \pi]$. The hope I had was that it might be possible to identify some helpful symmetry condition (concerning the sum of the corresponding signed areas) which could then lead to prove that if one of the two integrals above is zero the other cannot. But perhaps that might be easier if we restrict our inequality to just $a=\pi/2$ and $a=\pi$ (meaning: for $f(x)$ such that one of the two integrals vanishes at $a=\pi/2$, or at $a=\pi$, then the other integral must be $\neq 0$).
Best Answer
We can see the space of functions as infinite dimensional with a scalar product
$\langle f|g\rangle=\int_0^\pi f(x)g(x)dx$.
Then to find the desired counterexample a starting function wich has to be not a multiple of the given elements of the set $g_i\in\{\sin,\cos\}$ is taken, let say $f$.
The orthogonalisation procedure runs as :
$f_1(x)=f(x)-\frac{\langle f|g_1\rangle}{\langle g_1|g_1\rangle}g_1(x)$
Which implies $f_1$ has a 0 integral with $g_1$.
Then the procedure is applied on it with $g_2$
The above solution is with
$f(x)=\sin(2x)$, then
$\langle f|sin\rangle=0$ and
$\langle f|cos\rangle=4/3$. The norm squared of cos in 0 to Pi is
$\int_0^\pi\cos(x)^2dx=\frac{\pi}{2}$, hence
$\sin(2x)-\frac{8}{3\pi}\cos(x)$ is a solution.
$x^2$ or others starting functions could work too, so there are infinitely many possibilities.