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? Well, over here, I get a one and on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities.
The reciprocal of one is still going to be one but now, since I'm taking the reciprocal of this here, it's gonna be greater than or equal to the absolute value of the sine of theta over the absolute value of theta, and that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants.
You can think about this theta approaching zero from that direction or from that direction there, so that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta is gonna have the same sign.
It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value, so I can erase those.
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 9 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? I wouldn't say the differences in language and notation were the challenging part.
Show 1 more comment. Anshul Laikar 6 6 bronze badges. The picture isn't immediately convincing. Add a comment. Santosh Linkha Santosh Linkha John Joy John Joy 7, 1 1 gold badge 21 21 silver badges 30 30 bronze badges.
Using an area based proof, to me, seems like putting the cart before the horse. See axiom 2 archive. Thank you for pointing me to that! Community Bot 1. FUZxxl 8, 5 5 gold badges 26 26 silver badges 47 47 bronze badges. Now we have to prove that this sine behaves like the sine we learned in high school.
This proof is compelling but it is not really a proof. Alex Alex Yuval Filmus Yuval Filmus Thank you for your answer though. Supreeth Narasimhaswamy Supreeth Narasimhaswamy 1, 9 9 silver badges 22 22 bronze badges. Mark Viola Mark Viola k 12 12 gold badges silver badges bronze badges. Also, where you used the sqeeuze theorem, you could equally have used the Fundamental Theorem of Calculus and the Inverse Function Theorem.
It can be shown that the inverse function is the well-known sine function. And while one can apply the FOC instead of the squeeze theorem, there is no advantage in doing so. So, why on earth are you leaving these comments? The squeeze arguments are just unpacking the proof of FOC on a special case. The FOC itself is proved using a very similar squeeze argument to the one people are using. The areas of the sectors are simply the integrals that are being differentiated. In tkr's argument, he's instead differentiating an arclength integral, which gives the same result anyway.
Show 2 more comments. Nice job! I think Archimedes used a weaker form of your Lemma to get the circumference of a circle. Timur Zhoraev Timur Zhoraev 4 4 silver badges 7 7 bronze badges. The answer given above might be helpful for them. Michael Hardy 1. Davide Giraudo k 64 64 gold badges silver badges bronze badges.
Madhu Madhu 1, 3 3 gold badges 16 16 silver badges 33 33 bronze badges.
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