Scale a square rotated by an arbitrary angle so it covers the original square

Question

I have square images which I want to rotate by some random angle $a\in [0, 2\pi[$ radians. After rotating, I want to scale them, so they cover the entire area of the original image. In a last step, I will crop them to the original size. Which function describes the scale factor $x$ required to resize the rotated image? I think it should look something looking like $\frac{\sqrt{2}}{2}*|\sin{x}|$, but it is long ago, since I last had to actually solve a mathematical problem. Here, the red square indicates the boundaries of the original image for comparison.

Answer 1

Let's assume the original blue square to be of length 1. This would mean that the four right-angled triangles have a hypotenuse of length 1 with the smaller angle equal to a. Thus, the two other sides of these triangles are equal to $\sin(a)$ and $\cos(a).$ The total area of the four surrounding triangles would then be equal to $2\sin(a)\cos(a) = \sin(2a).$ Thus, the area of the scaled red square (step 3) would be equal to $1+2\sin(a).$

Equating the above expression to $x^2,$ you would get $x = \sqrt{1+2\sin(a)}$

Hope this helps!

Answer 2

As shown in the figure above the upright square is scaled and rotated such that the vertices of the original square lie on the sides of the rotated/scaled square. Applying the law of sines to $\triangle OAA' $,

$x=\dfrac{r'}{r} = \dfrac{ \sin(45^\circ + a ) }{ \sin 45^\circ } = \cos(a) + \sin(a) = \sqrt{1 + \sin(2a)} $