cosine

Who defines what a monster is? What will keep this definition of monster from changing from culture to culture, generation to generation, when people no longer share your views? Edit: I shouldn't have posted this, I'm sure this has probably been asked somewhere in these 181 posts...

Well [math]rcos(\theta)[/math] and [math]rsin(\theta)[/math] are the trig functions for angle theta and radius r. Does that answer your question?

I'm an undergraduate right now, but in a program that is prepping me for my intent to go for my Ph. D. So I'm ultimately working on my Ph. D. But in the short term I'm working on my Bachelor's... so...?

Yeah that distinction helps too.

Yep. I hopefully will have more to contribute as I take higher mathematics courses... by the way you got me to do it with the link in your sig. =)

Haha same here. I just started an account... no idea what to say that isn't already there...

Haha, Ecoli showed me about the Four 4's challenge going on in the General Mathematics forum. After making a fool of myself in my first post, I stumbled around the rest of SFN, and now we're here!

Hm my professor explained to me how to make a good parametrication, since we proved that The transformation on the cline (circle in mobius geometry) is real for a given point z if it is on the cline determined by the three points. So setting the transformation equal to a real number will provide an parametrication.

your welcome!

Wow, I feel stupid. After the cosine formula manipulation, the problem becomes: [math]cos(x) = -sin(x)[/math] So, [math]cos(x)\frac{1}{cos(x)} = -sin(x)\frac{1}{cos(x)}[/math] [math]tan(x) = -1[/math] this happens at [math]x = \tfrac{3\pi}{4} + k\pi [/math] For all integer [math]k[/math] There, that should work. P.S. So the first few positive x that will satisfy the equation are [math]\tfrac{3\pi}{4}, \tfrac{7\pi}{4}, \tfrac{11\pi}{4}, \tfrac{15\pi}{4}, \tfrac{19\pi}{4},...[/math]

Hey, There is a formula: [math]cos(A+B) = cos(A)cos(B)-sin(A)sin(B)[/math] so, [math]cos(x+\tfrac{\pi}{2}) = cos(x)cos(\tfrac{\pi}{2})-sin(x)sin(\tfrac{\pi}{2})[/math] [math]cos(x+\tfrac{\pi}{2}) = - sin(x)[/math] so your problem becomes [math]cos(x) = sin(x)[/math] Lemma:[math]cos(x) =/= 0[/math] because if [math]cos(x) = 0[/math], then [math] sin(x) =/= 0[/math] So, [math]cos(x)\frac{1}{cos(x)} = sin(x)\frac{1}{cos(x)}[/math] [math]tan(x) = 1[/math] this happens at [math]x = {x| x = \tfrac{\pi}{4} + k\pi} [/math] For all integer [math]k[/math] Does that help?

Yeah my professor gave me a parametrication based on the theorem that the the transformation of z is real iff it is on the cline (also called a m-circle, or just a circle in Mobius geometry) of z_1, z_2, and z_3.

The fall of the Soviet Union was actually driven into the ground by their years of military work in the Middle East. You know, invading Afgahnistan and all over a number of years will deplete someone's resources... Ironic? No, just sad.