1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trig help on unit square

  1. Jul 8, 2013 #1
    0. The app does not have the template built in so ill do my best, sorry i dont have a computer right now.

    1. This is a personal project, and there is a lot to it but I'm cutting out down to where i need help.

    Imagine a 2x2 grid of squares (2d for now). The center vector is at (0,0), and the length of one side of a square is 1m (or just think of one, distance doesnt matter right now). So essentially the grid is 2m by 2m. Now a arrow/line is abstract so it can help us meassure the distance from the center to a point on the outer rim of the grid.. in otherwords think of this like the unit circle (trig) but in a square... This is actually the direction where an object will be projected but thats unrelated to my question.

    2. So what i wanted was to obtain that length of arrow/line and actually the cordinates on the grid.. i can figure that out easily because im given the angle of the arrow.. and yes so if the arrow is at the cordinate (0,1) that is a 90 degree angle. I can use a triangle for reference but since its a square i cant seem to figure out how to write a function that can pull this off from 0 degrees to 360.. every 45 degrees i have to switch which function it will use.. and I been using the tool of sin(A)/a=sin(B)/b.. as well as soh-cah-toa because i use the right angle to help me find the distance of the arrow.

    3. So my question is how can i make this into one function rather then 8 different ones that switch after x amount of degrees. This is also a computer program so id like to keep it optimal.. my math course is entering calc 1 this fall so pre-calc.. I wonder if I'm not seeing something or is there other tools avaliable to help me figure this out?

  2. jcsd
  3. Jul 8, 2013 #2
    This should return the length for any angle:


    where x is your angle in degrees, % is the modulus function, floor returns the next lowest integer, and sec is the secant function.

    I would have a hard time explaining this without drawing a picture, but here goes... The distance to the edge of the unit square can always be expressed as 1/cos(q)=sec(q), where q is an angle between 0 and 45. This value should rise from 0-45 as x goes from 0-45, fall from 45-0 as x goes from 45-90, and so on. Can you see this? A picture would help. Anyway, the expression (1-2*floor[(x%90)/45]) is equal to 1 when x is from 0-45, 90-135, etc. and equal to -1 when x is from 45-90, 135-180, etc. Then {(1-2*floor[(x%90)/45])*x}%90 ends up creating the required graph for q.
    Last edited: Jul 8, 2013
  4. Jul 9, 2013 #3
    Wow good idea thanks a lot!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted