I am not sure if it counts as proof, but I have seen this done by a High Schooler. Here you may see an elementary approach that starts from a very interesting result, see this problem. All you need is a bit of imagination. Don't you feel strange about why most of the proofs are done with a figure? For example,. It depends on your definition of the sine function.
I would suggest checking out the geometric proof in ProofWiki. This can easily be done using the picture below. Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest without appeal to geometry or differential calculus.
Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed. Since the circle sector is a subset of the quadrilateral, the perimeter of the circle sector is less than the perimeter of the quadrilateral.
In particular. The convexity of the disk follows from the triangle inequality: a disk is a closed ball for the euclidean distance.
Similarly we write three systems for the initial conditions:. The proof is completed when the Wallis product is used. Simple one is using sandwich theorem Which demonstrated earlier. You can use geogebra to see the visualization of this phenomena using geogebra. Here is a more fun one! This is a new post on an old saw because this is one of those things where that I can see how that, all too sadly, the way in which we've structured the current maths curriculum really doesn't make it possible to do these kinds of things the justice they deserve and I think, ultimately, that is a disservice to many learners.
The truth is, this limit cannot really be given an honest proof without an honest definition of the sine function, first. And that is not as easy as it seems. Even if we consider the simple notion from many trigonometric treatments that the sine is equal to the "length of the opposite side of the right triangle divided by the length of its hypotenuse", that doesn't truly solve the problem because there is actually a subtle missing element and that is that sine is not a function of a "right triangle" though you could define that if you wanted, and it'd be easy!
And actually parsing out what "angle measure" means, it turns out, is essentially equivalent to defining the sine function in the first place, so this approach is circular!
So how do we define sine, or angle measure? Unfortunately, any approach to this is such that it must involve calculus. Then we have the famous "impossible" problem of "trisection of the angle" which vexed even the ancient Greeks and for which people would keep trying to pound at until Pierre Wantzel finally proved it undoable over two thousand years after.
We are asking for a mathematical widget that can not only trisect, but 5-sect, sect, etc. Indeed, not only is the sine function not trivial, we could argue that even the exponential function is considerably easier to treat than sine, though I won't give such a treatment here.
Thus, how do we do it? Well, the key observation is that our "steady" angle measure is one which is, effectively, defined by the arc length of a segment of circle intercepted by the angle when drawn at the circle's center and projected outward. In particular, this should be "obvious" from the circularly-introduced geometric formula. Hence, we will begin with the arc question first and one will see that this answer will end up using a fair bit of Calculus II material to answer this Calculus I-level question about a supposedly pre-Calculus mathematical object.
Indeed, this is what the whole "radian measure" is: it's a measure of angles in terms of the arc length of the piece they cut from a unit circle i. Thus, what we have above is something called an arc length parameterization of the circle - and that tells us how we need to proceed. First, we need a separate definition of the arc length of a circle.
How do we get that? Well, we will obviously need a more elementary circle equation, first, than the one we just gave, and that means going to the simple algebraic definition,. And now this is where we then must introduce Calculus II-level concept - namely, integration for arc length. Since the "real", or base, function here is really the inverse function, i. By the power rule and chain rule,.
What I am saying is that, in fact, that is not truly honestly possible and reveals a weakness of the curriculum in that it doesn't actually follow the proper logical buildup of the mathematical edifice.
What really should be done is to leave trig for later , that is, skip trig and go for Calculus first. When I studied maths on my own, I did just that. In fact, I'd say, as many educators have suggested, that most people don't need either, but really need more statistics instead. Then for those who do pursue higher maths, if we've done algebra and statistics, we already have right there a lot of interesting material we can build on for calculus, including the exponential function.
That said, rote crunching is not something I suggest banning either but I suggest that ideas, concepts, and creativity should come first, then you get into those techniques because very often they are also still useful in analysis and being fluent at them can also make you able to solve problems more quickly, e.
This is a variant of robjohn's answer. One possible definition is here. This allows us to take the limit inside and we get. This is not a rigorous proof, but is instead an intuitive argument. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group.
Create a free Team What is Teams? Learn more. Ask Question. Asked 10 years ago. Active 11 days ago. Viewed k times. Show 11 more comments. Active Oldest Votes. Also, I didn't have a problem with using that arclength and area are proportional to the angle.
Show 24 more comments. Domates 4 4 silver badges 16 16 bronze badges. The definition of radians makes the picture above true. For whatever reason, this is the proof I like the most because it relates the tangent line at the point on the circle to the value we call "tangent". To each his own Show 4 more comments. That is how Euler viewed the matter. See his book on differential calculus. Michael Hardy Michael Hardy 1. Was it very difficult to read because of the notation and language of the time?
Well, the height of this line would be the y-coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that also worked for thetas that end up in the fourth quadrant, which will be useful, we can just insure that it's the absolute value of the sine of theta.
Now what about this blue line over here? Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one, so the tangent of theta is the opposite side.
The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value for sitting here in the first quadrant but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within the circle, so I can construct this triangle.
And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle. I'll rewrite it over here. I can just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color.
So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians, so this is theta over to two pis of the entire circle and we know the area of the circle. This is a unit circle, it has a radius one, so it'd be times the area of the circle, which would be pi times the radius square, the radius is one, so it's just gonna be times pi.
And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there 'cause we're talking about positive area.
And now let's think about this larger triangle in this blue color, and this is pretty straightforward. And so I can just write that down as the absolute value of the tangent of theta over two. Now, how would you compare the areas of this pink or this salmon-colored triangle which sits inside of this wedge and how do you compare that area of the wedge to the bigger triangle?
Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge and the area of the wedge is less than or equal to the area of the big, blue triangle. The wedge includes the salmon triangle plus this area right over here, and then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true and I'm just gonna do a little bit of algebraic manipulation.
Let me multiply everything by two so I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta which is less than or equal to the absolute value of tangent of theta, and let's see. Actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's gonna be the same thing as the absolute value of tangent of theta.
And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta and then I'm gonna divide this by an absolute value of the sine of theta.
And what do I get?
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