Prove this!

[Anonxtreme]

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

[Kulverstukas]

What kind of question is that? best to/of what?

[MSteph]

I got this!

I must admit that I jumped right into the equation and tried to find myself in trigonometric identities, while that was definitely not the way to go.

I did this in a completely rational way, so bear with me. I'm certain there are other ways to solve this, but I felt comfortable following this train of logic... Enjoy pals:


First of all, if you could please direct your attention to the basic, high school Unit Circle:

http://www.regentsprep.org/Regents/math/algtrig/ATT5/600px-Unit_circle_angles_svg.jpg

And then you know that "All Students Take Calculus" way to remember in what specific quadrants are cosine, sine, and tangent negative or positive, because I'm sure your high school teacher kept repeating it the first day of Pre-Calc/Trig class. Here's a reference picture for convenience:

http://www.regentsprep.org/Regents/math/algtrig/ATT3/signchart.gif

Now, the conditions for this proof are that 0<x<pi/2.

This implies that we are trying to prove the statement using only Quadrant I.

I did by proving all Anti-A statements false (while the statement we're proving is A, of course), so I went to the rest of the Quadrants themselves.

In QII, sinxtanx = - n (negative n), because sinx would be positive and tanx would be negative, so +sinx-tanx = -n. However, the right side, which is 2(1-cos), remains positive, because in QII, cosx is negative, so 1 - (-cosx), is equal to something that is greater than 1.

Therefore, -n > +n, is a false statement.

Upon examining the other Quadrants, we get the same statement, -n > +n, which cannot be true.

Therefore, all Anti-A's have been proven false. The only thing that remains is to prove A to be true.

Well, if we are under the condition that 0 < x < pi/2, then we get +n > +n upon the same line of examination (where n just means any random number). So far, so good. Let's go on to the details.

In order that the right side of the statement increases in value, then cosine must be decreasing (because of the 1-cos part of it). As cosine decreasing, it is moving closer to the y-axis, where its value is equal to 0, and where the value of sine is equal to 1. Therefore, the value of sine is also increasing. By extension, the value of tangent (which is just sine/cosine), is also increasing, but by a greater rate than sine, because the cosine value is growing smaller.

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.



TL;DR? Really? I wouldn't have read it either.

[parad0x]

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

Come on bro! It isn't that hard.
Its one of the question of basics of trigonometry. You can't solve even this?


Here's my advice:
 Take a paper and pencil and a book and try this.

[ZTP0]

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

[Traitor4000]

Now that may have been easy now go prove p=np 

[rasenove]

Now that may have been easy now go prove p=np 

n = 1; problem solved. NEXT!