Considering this trivial definite integrals inequality: $$ \mbox{with} \;a \in (0, 2\pi)\; \mbox{such that} \;\; \int_{0} ^{a} \cos(x) dx \; = \; 0 \; \; \; \;\Longrightarrow \; \; \; \; \int_{0} ^{a} \sin(x) dx \; \neq \; 0 $$ 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: $$ \mbox{with} \;a\; \mbox{such that} \;\; \int_{0} ^{a} \cos(x)\;f(x) \; dx \; \; = \; 0 \; \; \;\Longrightarrow \; \; \; \; \int_{0} ^{a} \sin(x) \;f(x) \; dx \; \neq \; 0 $$ is this true ?
NOTE: the last edit is posted as a new question, because I have accepted @5xum answer and said last edit has now changed the question to an extent which would require a different answer.
It is not true, for example, you could have $\forall x: f(x)=0$, in which case both integrals would be $0$.
For a less trivial example, take any function $g(x)$, and define $$f(x) = g(x) - A\cos(x) - B\sin(x)$$ where $$A=\frac{1}{\pi} \int_0^{2\pi} \cos(x) g(x) dx$$ and $$B= \frac{1}{\pi} \int_0^{2\pi} \sin(x)g(x)dx.$$
Then, you have
$$\begin{align}\int_0^{2\pi} f(x)\cos(x)dx &=\int_0^{2\pi} g(x)\cos(x)dx - A\int_{0}^{2\pi}\cos^2(x)dx - B\int_{0}^{2\pi}\cos(x)\sin(x)dx \\&=\int_0^{2\pi} g(x)\cos(x)dx - A\cdot \pi - B\cdot 0\\ &=\int_0^{2\pi} g(x)\cos(x)dx - \pi\cdot \frac{1}{\pi} \int_0^{2\pi} \cos(x) g(x) dx = 0\end{align}$$
and you would also have
$$\begin{align}\int_0^{2\pi} f(x)\sin(x)dx &=\int_0^{2\pi} g(x)\sin(x)dx - A\int_{0}^{2\pi}\sin(x)\cos(x)dx - B\int_{0}^{2\pi}\sin^2(x)dx \\&=\int_0^{2\pi} g(x)\sin(x)dx - A\cdot 0 - B\cdot \pi\\ &=\int_0^{2\pi} g(x)\sin(x)dx - \pi\cdot \frac{1}{\pi} \int_0^{2\pi} \sin(x) g(x) dx = 0\end{align}$$
and this presents a counterexample to your question.