dryga

because sine and cosine can be considered to be the same function, since they are related by [math]\sin x = \cos (x + \frac{\pi}{2})[/math]. Tangent, OTOH, is defined by [math] \tan x = \frac{\sin x}{\cos x} [/math], so it looks very different from both sine and cosine. Plots of sine, cosine and tangent can be found here (note the similarity between sine and cosine): http://mathworld.wolfram.com/Sine.html http://mathworld.wolfram.com/Cosine.html http://mathworld.wolfram.com/Tangent.html

The concepts of modulo and congruences is textbook discrete mathematics. I got to take a (rather shallow) course on it in high school, and I took it as an elective in my first year of college (I think I took it at the same time as Calc II, actually). It was a very fun course, a lot of first-years took it since it didn't require a lot of knowledge before taking it. And group theory/rings/fields etc is still the most aesthetically pleasing math I've seen I don't know if that answers anything. Anyway, I hadn't heard of the AIME tests or anything before, but fwiw I looked at a few sample questions and it seems to be almost only discrete math stuff. If you have the opportunity I highly recommend taking a course on it

But 1 is equal to .999 repeating.

Okay, so maybe I should start checking how old the threads I reply to are.

I think Cauchy series multiplication would get messy very fast. The easiest way to show it, I think, is to first show that [math] \cos x = \frac{e^{ix} + e^{-ix}}{2}[/math] and [math] \sin x = \frac{e^{ix} - e^{-ix}}{2i}[/math] (you just use the Taylor series for [math]e^x[/math]). Once you have the functions written that way, it's easy to show that [math]\sin^2 x + \cos^2 x = 1[/math], and it'll follow that the absolute values for both sine and cosine are smaller than or equal to one.

I haven't seen anyone explain why this is wrong yet, even though everyone seems to agree that the limit is 2. What you're doing is factoring out an x from the square root, so you get this: [math] \lim_{x\to -\infty} [x + |x| \sqrt{1+\frac{4}{x}+\frac{1}{x^2}}] [/math] So far so good. But then you're saying that what's left under the square root sign is going towards one as x goes to negative infinity, so that the limit above should go towards [math] \lim_{x\to -\infty} [x - x] = 0 [/math]. This would be true if x were finite -- but since x in this case is approaching infinity, multiplying it with something that's just approaching one doesn't necessarily leave it unchanged. It could converge to anything, or diverge, for all we know.

the easiest way to prove it would be to just differentiate f f'(x) = 3 + 1/x f' is positive wherever f is defined, so f is one-to-one