EvilZone

This is so funny! Thank you! I tried it with some random images and it worked nicely, is there any specific reason behind the choice of the thresholds and the symbols you used or did you just try and try until you got it right?


I agree that taking the argument from command line instead of forcing you to call your image "hcdev.jpg" would be definitely better.

I believe it's a really personal thing, but I'll still give you my 2 cents.


I think that trying to calm down when under big stress is often not gonna work out if the situation is really heavy. Instead, accept your current mood and channel all those energies onto something else. You know, hit a punchball until you can't feel your knuckles, go running for 2 hours, listen to "angry" music and sing along. Find a way to pour out.


If you are unable to do that, try masturbating. It really does wonders for me at times, it completely erases both good and bad thoughts off my mind for a while. Just don't rush it, take time for yourself!

When 0<x<pi/2 then prove this,
sinx.tanx > 2(1-cosx)

First, an important question: with sinx.tanx you mean sen(x) * tan(x), i.e. the dot represents the multiplication? Or the dot the Americans use to divide the decimals?

In the first case, it's a simple inequality:

sin(x)*tan(x) > 2(1-cos(x))
(sin(x)*sin(x)) / cos(x) + 2cos(x) - 2 > 0
(sin(x)*sin(x) + 2cos(x)*cos(x) -2cos(x)) / cos(x) > 0
cos(x) > 0 in (0, pi/2) (the interval is open) so we can ignore it:
sin(x)^2 + cos(x)^2 + cos(x)^2 -2cos(x) > 0
1 + cos(x)^2 -2cos(x) > 0
(cos(x) - 1)^2 > 0
It's a square, thus always positive except when the argument is 0:
cos(x) - 1 = 0
cos(x) = 1 -> x = pi/2 + 2kpi (k in Z) but again the interval is open and this never happens. QED

Again, IF this is what you meant with the expression you wrote. This is very basic algebra, almost no trigonometry involved besides the fundamental identity.

I'm sorry I have no way to format it and it looks horrible but I hope you get the idea.

Anyway, input some numbers and you get that it never ever goes beyond the value of sinxtanx, so the statement is true and Anti-A's are false. Proven.

I absolutely don't mean to be rude, but this is never the way you prove something. This goes against the very idea of a proof; you CAN prove that something is wrong if you find at least one example in which the given statement is false (counterexample), but you can never say that something is true just because inserting some random numbers you never end up in a contradiction.
To cite one big example, almost everyone is sure that one of the most important open Math problems is true, but there is still no proof and nobody can be sure. They did A LOT of empiric experiments, I believe they tried billions of numbers and it always worked so far, but it means nothing if not that there's a good reason to place your bet on the "true".
Also OP's question is to prove that something is true in (0, pi); it doesn't matter wether the statement is true or false outside it (in your case, no need to prove that it's false outside it).
Also (I know I'm getting annoying) talking about quadrants isn't the best way to look at the problem imho: it's a simple inequality in one variable, so just imagine a straight line (representing the real numbers) and all you care about is one segment without the extremes.

Please don't take all I said as a critique but rather as an effort to bring some knowledge around. I believe everyone should be entitled to correct someone else's mistakes to help him/her pursue his goals. For this reason, if I did any mistake myself (and I often do a lot of them) I beg you to point them out