1. A box of mass m is sitting on an incline of 45 degrees and it requires an applied force F up the incline to get the box to begin to move. What is the maximum coefficient of static friction?

I know that the parallel component of the gravitational force is mgsin45 and the perpendicular component is mgcos45. The maximum static frictional force would be (normal force mgcos45)*the static coefficient. The question asked about when the box is just about to move so the upward and downward forces are equal and static friction is at a max. But the explanation states F=static coefficient*mgcos45+static coefficient*sin45 and from there you can solve for the coefficient. Why is the static coefficient multiplied to sin45 is it bc both gravitational and static forces are acting in the same direction opposing motion?

answer is {2F^(1/2)/(mg)}-1

How did they get this answer?