Please Register or Log in to view the hidden image! In the diagram, we have (x,y) = (cos t, sin t). If a third line is drawn tangent to the circle at (x,y) = (1,0), and the circle radius is extended through (cos t, sin t), where do both lines intersect?
The easy answer is because of the law of similar triangles: extending the radius outside the unit circle will preserve the ratio between the sides, hence y/x will be the same ratio; y/x = sin t/cos t = tan t, which gives y = x tan t which is where the y intersect will be for any similar triangle. But x = 1, so y = tan t.
It's a right triangle so that makes this much easier. You've got a right triangle with a base of 1 which is the adjacent to t, and we therefore know that tan(t) = opposite/1 = opposite. Thus tan(t) is the height (y coordinate) of the tangent line where the intersection occurs; the point of intersection is therefore (1, tan(t)).
Next question: how do we use the similar triangles to show that \( \lim_{t \to 0} \frac {sin\,t}{t} = 1 \)?
Connect the angle t to the length of the arc 't' on the unit circle from (1,0) to (cos t, sin t) (they have the same value), see if that helps.
You're gonna make me get out textbooks. And remember all that stuff they taught me in school when I was very much more handsome than I am now.
I ain't gonna do it with similar triangles. r = 1, so sin(t) = y. Thus, we must show that as t → 0, y → t. As t → 0, sin(t) → 0. Thus, as t → 0, y → t. QED.
No, you're supposed to show that \(\frac {sin\,t}{t} \to 1 \) as \( t \to 0 \). Not quite what you've done.
Hmmm, I'd say showing that y→t as t→0 shows that y/t → 1 as t→0, since y/y = t/t = 1 in the limit. I'm not clear on your objection. Do you disagree that if r = 1, sin(t) = y? You do see that sin(t)/t = y/t if r = 1, right? Are you familiar with the trigonometric functions, sin(t) = y/r, cos(t) = x/r, tan(t) = y/x? I didn't learn trig using triangles, I learned it using circles. It makes a lot more sense that way. I don't fuss with triangles in proofs unless they're right triangles. If I gotta do a non-right triangle I have the stuff for that, but I see no point in fooling around if I can use the functions.
. I'm not sure about getting full marks for that, though. I mean, you could show that cos t → 1 as t → 0. You could show that y = 1 → 1 as t → 0. Why the condescension? It's a simple enough problem, 1st year calculus, from a book called Freshman Calculus. Most students who leave the high school I went to know about trigonometry, if not why they had to learn it.
I'm not being condescending, I'm trying to figure out what your objection is. Sorry if that makes you feel all offended and stuff. Of course I could prove cos(t) → 1 as t → 0; and I'd do it exactly the same way, using those same circular techniques. I don't know what y = 1 →1 means. <- condescending As far as full marks, I proved it, are you arguing I didn't? Sorry I didn't use the same textbook you did. Get over it. <- condescending I figure if someone's gonna accuse me of being condescending-- especially without good evidence-- I'll get my money's worth.
You can walk this back if you want- you can apologize for calling me "condescending" when I was trying to be friendly and I'll apologize for being condescending twice in my next post. But don't do it again.
Look, you haven't "proved" that \( \frac {sin\,t}{t} \) approaches any value, you've shown that \( sin\,t \to 0 \). And, more rigorously you haven't shown that the limit being asked about is the same from above as from below (maybe that's worth extra points). And, you can use the fact that \( \lim_{t \to 0} (y = 1) \) exists. No condescension intended. Another point is, the limit corresponds to the slope of the sine function at zero; it's basically the method of secants, but that's the function, not the unit circle. The book does it like this: you can see from the geometry that \( y = sin\, t\) is smaller than \(t\) the arc length, and both are smaller than \(y = tan\, t\). So we have: \(sin\, t < t\) , which means we have \(\frac {sin\, t}{t} < 1\). And: \( t < tan\, t\) means \( \frac {cos\,t}{t} t < \frac {cos\,t}{t} tan\,t \equiv cos\,t < \frac {sin\, t}{t}\).
I showed that on a unit circle sin(t) = y, which means that sin(t)/t = y/t. Do you disagree? If you're going to be nasty we'll go one step at a time.
So, by the argument that uses the inequalities (obvious from the diagram), we have that \( \frac {sin\, t}{t} \) must lie between y = 1, and y = cos t. But since cos(o) = 1, this must be the value of the limit we're after. One more thing . . . it's been emotional.
No, one step at a time. I showed that on a unit circle sin(t) = y, which means that sin(t)/t = y/t. Do you disagree?
See, the way I figure it, you thought I was lying when I said I knew trig. And now you're looking like a fool for not believing me.
. Yes, but any "fool" can do that. What I can't see is how it gets you near a solution. The rest of your demonstration is: I don't follow that, what happened to \( \frac {sin\,t}{t} \)? Again, what happens to \( \frac {sin\,t}{t} \)? Can you see that it must be equal to the derivative of sin t at t = 0? This is true not just because cos(0) = 1. ed: I'll admit I can sort of see your argument which appears to be predicated on sin t ≈ t when t is small, so you have \( \frac {sin\,t}{t} ≈ \frac {t}{t} = 1 \) as t→0.
