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you have a left-handed coordinate system, where $x$ increases right, and $y$ down. Given axis-aligned bounding boxes' upper left corner coordinates, width, and height, you wish to find the axis-aligned bounding boxes after rotation by some angle around their center.
It looks like positive angles indicate counterclockwise rotation. In a left-handed coordinate system, this means rotation by $\varphi$ around point $(x_0 , y_0)$ is
$$\begin{cases}
x^\prime = x_0 + (x - x_0) \cos(\varphi) - (y - y_0) \sin(\varphi) \\
y^\prime = y_0 + (x - x_0) \sin(\varphi) + (y - y_0) \cos(\varphi) \end{cases}$$
where $(x, y)$ are the coordinates before the rotation, and $(x^\prime , y^\prime)$ after the rotation.
Let's say $( x_{min} , y_{min} )$ are the upper left coordinates, $w$ is the width, and $h$ is the height of the axis-aligned bounding box. Then,
$x_{min}$ is the left edge of the axis-aligned bounding box,
$y_{min}$ is the top edge of the axis-aligned bounding box,
$x_{max} = x_{min} + w$ is the right edge of the axis-aligned bounding box, and
$y_{max} = y_{min} + h$ is the bottom edge of the axis-aligned bounding box.
The four corners of the axis-aligned bounding box are therefore $( x_{min} , y_{min} )$, $( x_{max} , y_{min} )$, $( x_{min} , y_{max} )$, and $( x_{max} , y_{max} )$. When it is clear that we are talking about axis-aligned bounding boxes, this is often written in simplified form as $( x_{min} , y_{min} ) - ( x_{max} , y_{max} )$; i.e. as the diagonal line segment (in increasing coordinates).
If the center of rotation is at the center of the axis-aligned bounding box, then
$$\begin{cases}
x_0 = x_{min} + \frac{w}{2} = \frac{x_{min} + x_{max}}{2} \\
y_0 = y_{min} + \frac{h}{2} = \frac{y_{min} + y_{max}}{2} \end{cases}$$
Let's say we have an axis-aligned bounding box, and we know $x_{min}$, $y_{min}$, and have calculated $x_{max}$, $y_{max}$, $x_0$, and $y_0$, and that we want to rotate this by $\varphi$ (positive if clockwise, negative if counterclockwise, in this left-handed coordinate system).
There are two ways of calculating the axis-aligned bounding box after the rotation. First way is to calculate the four corner point coordinates:
$$\begin{aligned}
x_1 &= x_0 + (x_{min} - x_0) \cos \varphi - (y_{min} - y_0) \sin\varphi \\
y_1 &= y_0 + (x_{min} - x_0) \sin \varphi + (y_{min} - y_0) \cos\varphi \\
x_2 &= x_0 + (x_{max} - x_0) \cos \varphi - (y_{min} - y_0) \sin\varphi \\
y_2 &= y_0 + (x_{max} - x_0) \sin \varphi + (y_{min} - y_0) \cos\varphi \\
x_3 &= x_0 + (x_{min} - x_0) \cos \varphi - (y_{max} - y_0) \sin\varphi \\
y_3 &= y_0 + (x_{min} - x_0) \sin \varphi + (y_{max} - y_0) \cos\varphi \\
x_4 &= x_0 + (x_{max} - x_0) \cos \varphi - (y_{max} - y_0) \sin\varphi \\
y_4 &= y_0 + (x_{max} - x_0) \sin \varphi + (y_{max} - y_0) \cos\varphi
\end{aligned}$$
so that after the rotation, we pick the smallest and largest coordinates:
$$\begin{aligned}
x_{min} &= \min( x_1 , x_2 , x_3 , x_4 ) \\
y_{min} &= \min( y_1 , y_2 , y_3 , y_4 ) \\
x_{max} &= \max( x_1 , x_2 , x_3 , x_4 ) \\
y_{max} &= \max( y_1 , y_2 , y_3 , y_4 ) \\
w &= x_{max} - x_{min} \\
h &= y_{max} - y_{min} \end{aligned}$$
The other way is to note that since the original shape is a rectangle, and we rotate around its center, two of the corners are just mirrored around the center; and we can simply find the largest coordinates in magnitude with respect to the rotation center:
$$\begin{aligned}
x_{\Delta} &= \left \lvert \frac{w}{2} \cos \varphi \right \rvert + \left\lvert \frac{h}{2} \sin \varphi \right \rvert \\
y_{\Delta} &= \left \lvert \frac{w}{2} \sin \varphi \right \rvert + \left\lvert \frac{h}{2} \cos \varphi \right \rvert \\
\end{aligned}$$
where $\lvert \; \rvert$ refer to taking the absolute value. Then, you can update the axis-aligned bounding box properties using
$$\begin{aligned}
x_{max} &= x_0 + x_{\Delta} = x_{min} + \frac{w}{2} + x_\Delta \\
y_{max} &= y_0 + y_{\Delta} = y_{min} + \frac{h}{2} + y_\Delta \\
x_{min} &= x_0 - x_{\Delta} = x_{min} + \frac{w}{2} - x_\Delta \\
y_{min} &= y_0 - y_{\Delta} = y_{min} + \frac{h}{2} - y_\Delta \\
w &= 2 x_\Delta \\
h &= 2 y_\Delta
\end{aligned}$$
where the operations need to be done in the order shown (but you can omit $x_{max}$ and $y_{max}$, so that all right sides use the values prior to rotation (except for $x_\Delta$ and $y_\Delta$, of course).
In pseudocode, the rotation operation can be written as
Function rotate_aabb(xmin, ymin, width, height, angle):
Let c = cos(angle)
Let s = sin(angle)
Let halfw = 0.5*width
Let halfh = 0.5*height
Let xdelta = abs(c*halfw) + abs(s*halfh)
Let ydelta = abs(s*halfw) + abs(c*halfh)
Let xmin = xmin + halfw - xdelta
Let ymin = ymin + halfw - ydelta
Let width = 2*xdelta
Let height = 2*ydelta
Return (xmin, ymin, width, height)
End Function
which can be implemented using one sin, one cos, six multiplications, eight additions or subtractions, and four absolute values.
When rotating e.g. around the upper left corner, we have fewer symmetries:
Function rotate_aabb(xmin, ymin, width, height, angle):
Let c = cos(angle)
Let s = sin(angle)
Let x1 = c*width
Let y1 = s*width
Let x2 = -s*height
Let y2 = c*height
Let x3 = x1 + x2
Let y3 = y1 + y2
Let txmin = min(0, x1, x2, x3)
Let txmax = max(0, x1, x2, x3)
Let tymin = min(0, y1, y2, y3)
Let tymax = max(0, y1, y2, y3)
Let width = txmax - txmin
Let height = tymax - tymin
Let xmin = xmin + txmin
Let ymin = ymin + tymin
Return (xmin, ymin, width, height)
End Function
and we need one sin and one cos, four multiplications, six additions or subtractions, and twice an operation to find the maximum and minimum among four values, one of which is zero.
In C (C99 or later), you can implement min(0, x1, x2, x3) as fmin(fmin(0.0, x1), fmin(x2, x3)) and max(0, y1, y2, y3) as fmax(fmax(0.0, y1), fmax(y2, y3)), although
txmin = 0.0;
if (txmin > x1) txmin = x1;
if (txmin > x2) txmin = x2;
if (txmin > x3) txmin = x3;
tymax = 0.0;
if (tymax < y1) tymax = y1;
if (tymax < y2) tymax = y2;
if (tymax < y3) tymax = y3;
and similarly for txmax and tymin, should generate optimal code on architectures with hardware min/max instructions, if optimizations are enabled. For example, on x86-64 with GCC 5.4.0 with -O2, the above generates a sequence of minsd and maxsd instructions (and no jumps, so no slowdown due to conditional expressions at all).
The notable downside of calculating the rotated bounding box based on the previous bounding box, is that unless the rotation is a multiple of 90°, the bounding box will become larger after every rotation.
Best Answer
Most programming languages provide an $\operatorname{atan2}$ function that deals with quadrants correctly. If you have a point $(x,y) \ne (0,0)$, then $\operatorname{atan2}(y,x)$ gives the counter-clockwise angle from the positive $x$-axis to $(x,y)$, in the range $(-\pi,\pi]$. Since you want the clockwise angle from the negative $x$-axis, it is enough to observe that when $\operatorname{atan2}(y,x) = 0$ the angle you want is $\pi$, and when it is $\pi/2$ you want $\pi/2$, so in general what you want is $\pi-\operatorname{atan2}(y,x)$, which lies in the range $[0,2\pi)$.