How to translate the center of an equirectangular projection?

An equirectangular projection treats the x coordinate as the angle theta around the vertical axis going from 0 to 360 degrees. These angles match the longitudinal angles of a globe. The y coordinate is treated as the angle phi that represents the latitudes of a globe. They typically go from +90° at the zenith (or north pole) to -90° at the nadir (or south pole).

So to translate the center of projection of an image by some amount (x0, y0), you can simply add x0 to the x value of every coordinate (wrapping around to the other side at the edges). The y coordinate is similar - add y0 to every coordinate. However, instead of wrapping around, you need to reflect the value. So if your new coordinate is less than 0, simply take the absolute value. If it's greater than the height, take the difference between the new value and the height of the image and subtract that from the height of the image. If it was less than 0 or greater than height then you need to also add half the width to the x coordinate.

So putting it in pseudocode:

newY = oldY + translateY;
if (newY < 0)
{
    newY = fabs(newY);
    translateX = fmod((translateX + (width / 2.0)), width);
}
else if (newY >= height)
{
    newY = height - (newY - height);
    translateX = fmod((translateX + (width / 2.0)), width);
}
newX = oldX + translateX;
if (newX < 0)
{
    newX = width + newX;
}
else if (newX >= width)
{
    newX = newX - width;
}