Given the distance of a point from a rotation axis in coordinates (for example, point A is 10 x units and -20 y units from axis) when the rotation angle is 0, how do I calculate the position of the point for any rotation angle (in radians)?

I've tried "x_pos = axis_x + x_distance * cos(angle); y_pos = axis_y + y_distance * -sin(angle);" but this only seems to work in some cases

I do it this way: (probably not the best method, but it works)

I use the vector length of pivot.x,pivot.y -> blit_point.x,blit_point_y and the angle of the vector (8.0 and 4.0 dec degree in that case)

let's say "18".

head_pos_ is the coordinate of the pivot of the head bitmap.
head_angle is the angle of rotation.

eyes_x and y are the "blitting point's" coord

float eyes_x = head_pos_x + int(18.0 * cosinus(head_angle -90.0))
+ int(8.0 * cosinus(head_angle));
float eyes_y = head_pos_y + int(18.0 * sinus(head_angle -90.0))
+ int(4.0 * sinus(head_angle));

Assuming 0 degrees is 'right',
x = cos(rotation) * distance + offsetamount_x
y = sin(rotation) * distance + offsetamount_y

+Rotation would be counter-clockwise.

I had this code right for a long time, but didn't really understand how radians worked. Remember to increment in decimals and add in conditions such as

rotation+=0.0036;

if(rotation>M_PI*2) rotation=0;
if(rotation<0) rotation=M_PI*2;

Without limiters your line or whatever you're drawing/calculating goes CRAZY =P

And, Epsi, what's that -90.0 for? To get it into the 'right' plane? I mean, so that 0 isn't pointing right, but instead, point up?

yup, or at least if I remember correctly. What I did was a little dirty hack (the angle is not even the same on x/y axis), so I'll be happy if another solution shows up

For rotation of a point around the origin over an angle , use

To rotate around an arbitrary point, translate to that point first, then rotate, then translate back. It wasn't entirely clear to me what you wanted to do, hope this helps.

That's quite wrong: there's nothing wrong with angles outside this range.

Mine should be correct:

For 0 degrees is 'up'
x = sin(rotation) * distance + offsetamount_x
y = cos(rotation) * distance + offsetamount_y

+Rotation would be clockwise --> or just make the 'distance' negative in the x calculation for counter-clockwise.

For 0 degrees is 'down'
x = sin(rotation) * distance + offsetamount_x
y = cos(rotation) * -distance + offsetamount_y

Default: clockwise

For 0 degrees is 'left'
x = cos(rotation) * -distance + offsetamount_x
y = sin(rotation) * distance + offsetamount_y

Default: clockwise

Thank you for all your responses, I'll try them later today. But there is one problem with most of your solutions, that I have to calculate the distance, and for that I have to use sqrt() which is slow. Evert's solution doesn't use the distance but I've never done matrix multiplications

You mean use sqrt() for acquiring the magnitude?
sqrt(x+x*y+y)?

Evert's matrix is just 1337. It's ok to use following:

x' = x*cos(v) - y*sin(v)
y' = x*sin(v) + y*cos(v)

I usually reinvent this wheel by dividing the point (x, y) to two "vectors" (x, 0) and (0, y).
Then I rotate these separately.
First becomes (x*cos(v), x*sin(v)) and the second becomes (-y*sin(v), y*cos(v)).
Then I add them and I get the formulae above.
This all is for a coordinate system where y grows upwards and v grows anti-clockwise, zero being x-axis direction.

You fail it. sqrt(x*x + y*y).

Epsi's code is the "right" way to multiply by Evert's matrix, but it mixes up sine & cosine by ignoring the fact that sin(a - 90.0 degrees) == cos(a) and cos(a - 90.0) == -sin(a) because the sine & cosine of an angle are 90 degrees out of phase.

Evert's method is the best. You should always use vectors and not angles. Ever wondered why phisics is mainly based in vectors? Because they work very well. You can also include the translations in Evert's matrix, making it a 3x3 matrix. I would write how, but I don't know how to use the mock up well...

Note: homogeneous coordinates.

The w should be 1

Well I tried the matrix multiplication but it didn't work. Maybe it's because the coordinate and/or angle system is different? Mine is like this:
http://mega.ist.utl.pt/~vbfr/coord.png

Aye, you're inverting the y axis, due to screen mapping, just switch the y by -y.

I tried that and it still didn't work x and y go to strange places...

EDIT: The actual code I have is like this:

x = sin(angle) * xPos + cos(angle) * -yPos + OFFSET_X;
y = -cos(angle) * xPos + sin(angle) * -yPos + OFFSET_Y;

You need this:
x = (sin(angle) * (xPos-offset_X)) + (cos(angle) * (-y-offset_Y)) + offset_x;
Where offset_x is centerOfRotation(x);

Still doesn't work... I feel that there is something very wrong because the position of the point changes much more than it should (it goes all over the screen when it should be near the center).

Anyway, OFFSET_X should be the x distance of the point to the axis right? And what you said is to change xPos (the x position of the axis) to (xPos - OFFSET_X) and yPos to (yPos - OFFSET_Y) right?

Maybe the problem has to do with the way I'm doing things. What I'm trying to do is set the position of a gun on a ship. The center of the ship is the rotation axis, and when the screen is rendered, everything is rotated except the ship, that is always motionless. I can't just set the gun coordinates because it's rotated, so I have to compensate for that rotation so the gun is always at the same position on the ship.

The offsext_x I meant was the center of the ship, the rotation point.

That depends on whether you are transforming a point or a direction.

So in that case, what's xPos? The position of the point I want to rotate, or the distance of that point to the axis (in x)?

Let's say that you want to rotate the square attached using the blue point as the rotation center.
You have to:
for each point:
T(x,y) = (x',y');
x' = ((sin(angle) * (x-blueCenterX)) + (cos(angle) * (y-blueCenterY))) + blueCenterX;

[edited}

I'm guessing it's the position of the other object...

Thank you, it works perfectly now

I just had to change "x-blueCenterX" to "-(y-blueCenterX)" and "y-blueCenterY" to "x-blueCenterX".

I'm really glad that you acknowledged everyone who tried to help!

[edit]
Sarcasm taken back!

This doesn't make sense to me, but glad to help.

I wasn't sure if I should've acknowledged everyone that actually helped, or everyone that tried to help. You're right, I should acknowledge everyone that tried, because I do appreciate any help even if it wasn't part of the solution I used. Now I can't change it Sorry for that, it won't happen again.