Cartesian coordinates and trig functions

Question

Sorry my math is very weak so it's difficult to research, this seems like it would be an obvious question that gets asked often.

I'm a budding software developer trying to write a watch face. The face is $240 \times 240$, for our purposes just consider $X(-120,+120)$ and $Y(-120,+120)$, easy enough to convert after.

Trying to derive $(X,Y)$ coordinates to draw hands based on hour/minute. Doing this case by case is easy enough. $\text{Minute} \times 6 = \theta$. This $\theta$ gets applied to $\sin / \cos$ depending on it's value.

The problem is writing it in a single function to derive. For example:

if(theta > 0 && theta <= 45) {
  X = sin(theta) * 120;
  Y = cos(theta) * 120;
}
else if (theta >= 45 && theta <= 90) {
  X = cos(theta) * 120;
  Y = sin(theta) * 120;
}

Then draw a line from [0,0] to $[X,Y]$.

Depending on the value the formula applied changes. This seems like it's surely a problem that somebody has had before. Any easy formula to account for this? Or it becomes 8 unique cases?

Answer 1

In clock the angle $\theta$ that you compute using $Minute\times6$ is the angle with respect you positive y axis in counter clockwise direction. For this theta you should write position of the minute hand which has a length $l$ as follows $$(x,y)= (l\sin\theta, l\cos\theta)$$ You don't have to worry which quadrant the minute hand is in, as that will be taken care of by the functions of $\sin$ and $\cos$. For example $$\theta=90 \implies(x,y)=(l,0)\\\theta=180 \implies(x,y)=(0,-l)\\\theta=270 \implies(x,y)=(-l,0)$$

Answer 2

Every two minutes the hour hand rotates by $\theta=1^0$, i.e., by half this time in mins and second hand by $60^0$

$$ (x,y) = 120 (\sin \theta, \cos \theta)$$

reckoning from $12\, O'$ clock position.