Unit Circle Fractions

[Chasney913]

I'm new here, so just let me know if I'm doing something wrong.

Anyway, in my math class this year, we're going to be doing unit circle calculations, such as determining the co-ordinates of a point in relation to the angle, which is for angle x, (cos x, sin x).

I was wondering if there was an easy way to convert sin x and cos x into fractions easily. For example, sin 45 is approx. 0.707 and that is sqrt(2)/2. Is there an easy way to convert the sin and cos of any angle into an equivalent fraction?

If there isn't, it would be great to know, so I don't have to worry about it.
If there is, I'd like to know, so I can make a program, however, this is not about programming, because I'd like to be able to know how to do it on paper.

Thanks in advance.

P.S. Take a look at the illustration on the wikipedia page for "Unit Circle" if you have no idea what I'm talking about.

[Weregoose]

Hi, and welcome to United-TI!

The only angles that can be turned into neat fractions like √(a)/b are those that you've probably heard so much about: 0°, 30°, 45°, 60°, 90°. For these, you can remember that their sines correspond to √(0)/2, √(1)/2, √(2)/2, √(3)/2, and √(4)/2, or you can simply square the long-winded decimals to find fractions to go underneath the square root signs (for example, sin(60°)² = 3/4, so sin(60°) = √(3/4)) and then simplify the radicals (√(3/4) = √(3)/√(4) = √(3)/2) to get your results.

Other angles have to be computed using difficult expressions that can sometimes involve infinitely nested radicals. So, it really depends on how complicated you want your program to be. sin(48°) as an input would make an ideal program show some equivalent of (√(7-√(5)+√(30-6√(5))))/4, and for such answers we'd be looking into Multiple-Angle Formulas and the like. I'm willing to offer support, but note that I'd also be learning something in the process. :)

P.S., You can contact an administrator to have your username changed.

[Chasney913]

OK, thanks.

I doubt we'll be doing anything that involves angles other than 0, 30, 45, 60 and 90, so I don't need to write a program to figure them out, as I can just memorize them.

I've been reading this forum for quite some time, and I thought it would be a good place to post about something like this.

I have another question, relating to physics.

I'm writing a program to simulate the trajectory of a projectile, and I have it doing the following:

- Determine three points so I can draw an arc. I'm using:
- (Starting height,0) ; (0,Landing distance) ; (X calculated from arbitrary time, Y calculated from arbitrary time)
- Calculate arc.
- Display.

I'm looking for two things. One, I need to know how to find the x at the max y, so the curve will be more precise. Right now, it's off by about one or two meters, and I would like it to be more accurate. I need to isolate x in the following, but I don't know how.

y = h + tan a - (g(x^2))/(2((v cos a)^2))

Sorry about the redundant parentheses, I'm trying to make sure it is clear where the exponents apply.

Second, I would like to apply air resistance, so that the simulation is more realistic. Resistance is dependent on mass, so I can simply request that as an input, but I do need to know how to calculate the above points, taking drag into account.

I hope I'm being clear enough, and again thanks for your help on the unit circle problem.

[simplethinker]

Welcome to UTI!

I've tried to do the same thing myself, and after hours and hours of work I think I know what you're trying to get.
First, the air resistance is more complicated because it is dependent on shape and distribution of mass. The equations for finding it include multiple vector fields and are incredibly complicated. In most cases you don't even need to factor the resistance in because the biggest influence it has is close to the initial launching, so most velocity measurement tools will record afterwards, thus the largest component of air resistance is already factored in.

I assume you started with the parametric projectile equations of:
x=v*t*cos(a)
y=h+v*t*sin(a)-.5g*t^2

and solved/substituted for 't' to obtain (you missed the 'x' multiplying tan(a) ):
y=h+x*tan(a)-(g*x^2)/(2*v^2*cos(a)^2)

To find the peak, differentiate with respect to x and then solve for zero. The final equation to find the x value of the peak is:
x=(v^2)*sin(2a)/(2*g)

A couple more notes about this: In most situations you can ignore the initial height 'h', and when solving for one of the values assume that y=0 to make it easier on yourself, and then add the 'h' back into the final result.

Also, when typing long formulas, it's easiest to use the '*' instead of parentheses for multiplying, but use parentheses for powers

If you need any more help with this kind of stuff PM me, I've put an immense amount of time into calculating projectile motion and don't mind helping.


EDIT: I just read your other post in the TI-Basic forum. The wikipedia article for projectile motion mentions time as a parameter, not as time itself. Parametric equations are a way to represent a function of two variables that are related by another variable, the parameter. An easier way to look at parametric functions is to think of them as coordinates defined by functions instead of numbers.
Let's say that there are two functions, X(t) and Y(t). To find the graph of the function, you can calculate the coordinates as ( X(t) , Y(t) ) for each 't'.
Here's an example:
X(t)=t-1
Y(t)=t^2
on your calculator, hit 'mode' and go down to to selections "func par pol seq" and select 'par', then change the window settings to Tmin=-9 , Tmax=9 , Tstep=1 , Xmin=-9.4 , Xmax=9.4 , Ymin=-62 , Ymax=62
and then graph it, now switch back to 'func' in the mode menu and graph the function y=(x+1)^2 with the same window settings (except there are no 't' settings), and you get the same graph.
With this example, the parametric equations actually look more complicated, but like with the projectile equations (and many others), they make it easier to define a curve, and they can tell you when a curve is at a point, rather than just the point itself.
another EDIT: The X(t) , Y(t) definition of projectile motion uses time because in most real-life applications, you want to know where it is at a certain time. The path of satelites can be represented as a standard equation, but since there are so many in earth's orbit, it looks like they would all crash into eachother. When their paths are defined parametrically, you can see that they all cross many of the same points, but at different times, which means we can still watch TV and use cell phones

[Chasney913]

Thanks.
I did start with the parametric equations, and I did substitute to find the other equation. (And by that, I mean I looked it up on Wikipedia.)

About air resistance not mattering, if you go to Wikipedia (can you tell where I do my learning?) and go to the "trajectory of a projectile" page, and scroll down to the air resistance part, it seems to indicate that drag does have a large effect. (Mind you, it is Wikipedia.)

I would like to have the time be realistic, so if the function is supposed to slow down and speed up, and take a certain amount of time, it should do that.

[simplethinker]

What type of projectile are you using? Most of my work has been in the flight of arrows, bullets, and rockets, which are designed to be aerodynamic (I'm into archery, and I was trying to figure out the angle needed to get an arrow a set distance, and then when I started reading up on it I got sidetracked and interested in methods of stabilization). When I did go through the trouble of calculating out the drag, it did not have as large an effect as you would think. But if you were doing something like a ball, which is not as aerodynamically effiecient as arrows, then you can disregard what I said earlier and calculate out the drag. The exact calculations are complex, so I think the easiest way to figure them out would be to search online for a website with the drag constants for the object you're using. I can't remember where I found them, but if I remember I'll PM you.

Using time as the parameter is only to make it easier and clearer to define. The actual speed doesn't change. As you can see from the equations, the only thing that changes is the rate of change in horizontal position. for X(t)=v*t*cos(a) , if you would graph it in function mode pretending that 't' is 'x', you would get a line (type in y=v*x*cos(a) on your calculator), which means it's increasing at a constant rate. The horizontal rate does change, but the actual speed across the screen never changes. When graphing a parametric function, the parameter time has absolutely nothing to do with actual time.

I hope I'm not too confusing

EDIT: I was just looking at the wikipedia entry. I think you clicked on the link at the bottom of the page and went to 'trajectory of a projectile' and then looked at the bottom of that page that had 'trajectory of a projectile with air resistance'. In it, it's talking about resistance to a particle, where the shape of the object is irrelevent, and it says does not apply for fast moving particles. Since particles are really small, the only factor you need to worry about is mass and velocity, but with larger objects the shape and distribution of mass is the most important.