How would you solve this equation system with trigonometric functions?

Thread replies: 6
Thread images: 1

Anonymous
2016-04-04 02:17:07 Post No. 7977245
[Report] Image search: [Google]

File: sistema_trigonometrico.png (18 KB, 1280x340) Image search: [Google]

Anonymous 2016-04-04 02:17:07 Post No. 7977245 [Report]

How would you solve this equation system with trigonometric functions? (If it even have a solution).

$x*(cos(\theta) - 1) + y*(sin(\theta)) = 0 $
$x*(-sin(\theta)) + y*(cos(\theta) - 1) = 0 $

I'm getting the eigenvectors for a rotation matrix. Since I'm such a dumbass with trigonometric functions, it was for me a full challenge to get to the eigenvalues, but I did it with the help of 4chan/sci.

Now I'm stuck again finding the eigenvectors for the corresponding eigenvalues. The equation in the image correspond to an eigenvalue 1 (I get to the equation following some videos in YT on how to find the eigenvectors).

If it's not trouble and someone could at least give me some tip, clue, a trigonometric identity, tell me if I'm missing something obvious or tell me some of that magical things mathematicians do with trigonometric functions.

Thanks in advance.

>>

Anonymous 2016-04-04 02:20:05 Post No.7977249
[Report]

Anonymous 2016-04-04 02:20:05 Post No.7977249 [Report]

>>7977245
well, the rotation preserves both x and y
so the angle must be a multiple of 2*pi

>>

Anonymous 2016-04-04 02:21:02 Post No.7977252
[Report]

Anonymous 2016-04-04 02:21:02 Post No.7977252 [Report]

I don't know hot to call Tex code here >.<
$x*(cos(\theta) - 1) + y*(sin(\theta)) = 0 $
$x*(-sin(\theta)) + y*(cos(\theta) - 1) = 0 $

>>

Anonymous 2016-04-04 02:25:28 Post No.7977260
[Report]

Anonymous 2016-04-04 02:25:28 Post No.7977260 [Report]

>>7977249
let me elaborate. a 2x2 matrix (a linear transformation from R^2 to R^2) which has the form
[ a, b]
[-b, a]
is actually isomorphic to a complex linear transformation from C to C, which is defined by multiplying by (a+bi), or (a-bi) depending on how you want to define i.

so this matrix you're looking at, where T(x, y) = (x, y) is actually the equation

(cost - isent)(x+yi) = x+y
which is exactly
z*e^(it) = z

so e^(it)=1, giving t = k*2pi

>>

Anonymous 2016-04-04 02:26:29 Post No.7977262
[Report]

Anonymous 2016-04-04 02:26:29 Post No.7977262 [Report]

>>7977260
equation should be
(cost - isent)(x+yi) = x+yi

>>

Anonymous 2016-04-04 03:08:17 Post No.7977358
[Report]

Anonymous 2016-04-04 03:08:17 Post No.7977358 [Report]

>>7977260
nicely done anon.