Can the formula for magnetic flux be re-written as $\Phi=AB\sin\alpha$ where $\alpha$ is the angle between the area and magnetic field? The formula for magnetic flux is generally written as (Formula 1) $$\Phi=AB\cos\theta$$ where $B$ is the magnetic field strength, $A$ is the 'area vector' and $\theta$ is the angle between the NORMAL of the area and the magnetic field.
Is it correct if this formula is re-written as (Formula 2) $$\Phi=AB\sin\alpha$$ where $\alpha$ is the angle between the area and the magnetic field? 
This ($\alpha$) is the angle that is generally given in questions and we always need to calculate ($90°-\alpha$) to get the angle $\theta$ needed for Formula 1. Since $\forall x, \cos x=\sin(90°-x)$, can the formaula for magnetic flux just be re-written in terms of this angle $\alpha$? 
Unless my calculations are wrong or there is some other special reason I don't know of, would Formula 2 be correct for all applications?
Edit: If it is correct then is there any reason why it isn't used on formula sheets or in textbooks?
 A: The answer is yes. However, the reason why people did not do it this way is because it is lengthy to calculate the angle $\alpha$. 
Think about it, when we have a surface, the mathematical description is always given as the normal vector to that surface. Meanwhile, if you want to find out $\alpha$, you need to first determine the projection of $\vec{B}$ on the surface, then find the angle between the that vector and $\vec{B}$. Hence, this process has to be repeated for different $\vec{B}$, which is really long compared to simple carry out the dot-product which yields the first formula.
A: But the definition of a surface is via a normal vector to that surface $d\vec{A}$ at every point on it. The only ambiguity is the 180 degree ambiguity of which way it points (always outward if it is a closed surface, or you can choose otherwise).
The total flux through that surface is 
$\Phi = \int \vec{B} \cdot d\vec{A}$, which reduces to $BA \cos{\theta}$ only in the case of a constant normal vector and uniform B-field over that area and where the $\cos \theta$ naturally arises from the definition of scalar product.
The version $\Phi = BA\sin \alpha$, where $\alpha = 90^{\circ} -\theta$, is clearly exactly equivalent in that simple case, but how would you express the flux in the more general case of a changing area vector or changing magnetic field vector? 
