# TI89 Failing to solve trig system of equation

1. Feb 5, 2014

### bwkerbow

I am in degrees
I am remembering to include the restriction (|0<x<90)
This is the only equation I've had trouble doing on my ti89

In order to solve a problem I am plugging in the following equations:

520cos(x) = 490cos(y)
490sin(y) + 520sin(x) = 678

Doing this by hand with some Pythagorean trig identities gets me the correct answers of x = 44.0188 degrees and y = 40.2583 degrees and yet for some reason my calculator is not returning these.

Here is exactly what I am plugging in:

solve(520*cos(x) = 490*cos(y) and 490*sin(y) + 520*sin(x) = 678,x)|0<x<90

but rather than getting a value for x and y I am getting (after about a 7 second wait, pretty long)
x = 57.2958 (0.017453 * arcsin(.942308 * (sin(y)-1.38367)) + 6.28319*(@n11 + .5)) and...(insert more similar stuff here)

anyway the point is it isnt returning the correct answers that I was got (and checked) by hand and I'm not sure why.

Thank you so much in advance for your help, I'm trying to take full advantage of this calculator im trying to learn.

2. Feb 6, 2014

### tycoon515

First I would rewrite your trigonometric functions to gain a better sense of their domain and range.
520cos(x) = 490cos(y) is equivalent to cos(y) = 52cos(x)/49. Since cos(y) is always between -1 and 1, it follows that we can only let 52cos(x)/49 be between -1 and 1, so we want cos(x) between -49/52 and 49/52. Geometrically speaking, this means we don't quite let cos(x) get to the bounds of its range (-1 and 1), so we don't quite let x get to 0 or 180. The domain is defined between supplementary ( equidistant from 90° ) angles, arccos(49/52) ≈ 19.6° and arccos(-49/52) ≈ 160.4°. We can obtain all positive intervals by shifting this interval up by whole number multiples of 360°. Since cos(-x) = cos(x), this means our domain is symmetric over the y-axis. To get the other half of the domain, consider all the negative versions of these intervals. Also cos(x) is always between -1 and 1, so we want 49cos(y)/52 to be between -1 and 1 as well, which it is, so the range is all real numbers. As for 490sin(y) + 520sin(x) = 678, you could rewrite it in the following way: sin(y) = (339 - 260sin(x))/245. Again, sin(y) is always between -1 and 1; however, (339 - 260sin(x))/245 is always between 79/245 and 599/245, so we need to let it be between 79/245 and 1. This means we only let sin(x) get as low as 47/130, but we let it get all the way to 1. The domain is defined between arccos(47/130) ≈ 21.2° and arccos(1) = 90°. We can obtain all intervals by shifting this interval by integer multiples of 360°. Also sin(x) is always between -1 and 1; however, (339 - 245sin(y))/260 is always between 47/130 and 146/65, so we need to let it be between 47/130 and 1. This means we only let sin(y) get as low as 79/245, but we let it get all the way to 1. The range is defined between arcsin(79/245) ≈ 18.81° and arcsin(1) = 90°. It is important to note that both of your functions take angles as input and return angles as output. Also, if you look at the graphs of the two functions, you will see that each of the pieces bounded by every domain interval looks the same. Solving algebraically, square both sides of the first to get 270400cos2(x) = 240100cos2(y). For the second, subtract 520sin(x) from both sides and then square both sides, yielding 240100sin2(y) = 270400sin2(x) - 705120sin(x) + 459684. Then combine both of the x-sides and both of the y-sides to make one big equation: 270400sin2(x) + 270400cos2(x) - 705120sin(x) + 459684 = 240100sin2(y) + 240100cos2(y). Factor out a 270400 from the sin2(x) and cos2(x) terms and 240100 from the sin2(y) and cos2(y) terms, and these simplify to 1. Then you have 270400 - 705120sin(x) + 459684 = 240100. Simplify to get sin(x) = 489984/705120, thus x = arcsin(5104/7345) ≈ 44.0° plus or minus any whole number multiple of 360°. If you graph both functions simultaneously and use the intersect feature, you will see that these functions do indeed coincide at ≈ 44.0° plus or minus any whole number multiple of 360°. That's how you should check where two functions are equal, provided you can express them explicitly ( y in terms of x ). For these, you should input y1=arccos(52cos(x)/49) and y2=arcsin((339-260sin(x))/245).

Last edited: Feb 6, 2014
3. Feb 6, 2014

### bwkerbow

Oh yes that makes perfect sense. I don't like it, but I understand. Thank you so much tycoon515