I'm stuck at attempting to prove the following statement:
$$\forall \theta \in \mathbb R, \exists (a, b, c) \neq (0, 0, 0) \in \mathbb R^3$$ such that $$\forall x \in \mathbb R, a\sin(x)+b\sin(x+\frac{\pi}{2})+c\sin(x+\theta)=0$$
(I already showed that $(\forall x, a\sin(x)+b\sin(x+\frac{\pi}{2}) = 0) \implies (a, b) = (0, 0)$)
I would highly appreciate any hint!
(my ultimate purpose is to conclude from this that the dimension of $Vect(\{f_\theta \in \mathbb R^{\mathbb R} | f_\theta(x) = \sin(x+\theta)\})$ is 2)
Well, if $\theta$ is fixed, then $f_\theta$ spans only one dimension.
If $\theta$ ranges over $\Bbb R$, it will indeed have dimension $2$, exactly because of this exercise.
Hints: $\sin(x+\frac\pi2)=\cos x$, we can fix $c=1$, and use the trigonometric addition theorem.