I normally love Eddie's videos, but this one gets it quite wrong. He fails to point out that θ (in radians) is equal to the arc-length divided by the radius (i.e. *not* an arbitrary way of measuring angle). And for small angles, this arc-length becomes very close to the opposite-side-length, h. Therefore, arc-length/r ≈ h/r ≈ h/(r-x) In other words, θ ≈ sinθ ≈ tanθ.
Right. Via his same argument, theta * 10 is also approaching 0, and then sin(theta) would be approximately equal to theta * 10, which it obviously isn't.
@@joshuaronisjr This isn't his argument, this is an accepted mathematical proof thats used regularly. He was just demonstrating it. You're seeming to neglect that theta is considered to be infinitely small.
@@Chris-hn4lp When he says that sin(theta) = theta for small theta, he does so by saying that they both become infinitely small. Take the example of z=4x and y=x, where x is variable. As x comes infinitely close to 0 (and indeed, all the time - but that misses the point) we could say that z and y both approach 0. But we wouldn't say that z=y as x goes to 0. No, we would say that z=4y. The same thing with theta - just because theta and sin(theta) both go to 0, that doesn't mean that they can be considered equal for infinitely small theta (I'm not saying that they don't approximate one another - I'm just saying his argument isn't enough). As another example take z=e^(2x) and y=e^(x). As x goes to 0, z is approximately 2 times y, even though they both go to 0.
For those who are interested, you can use the same setup to prove that theta is approx sin theta by considering that the area of the triangle approaches the area of the sector as theta approaches 0. So you have 1/2r^2theta = 1/2(r-x)h. We can cancel out the half and as theta -> 0, x->0, so we can set x = 0. Therefore, r^2theta = rh. So rtheta = h. But we know, h = rsintheta, therefore, rtheta = rsintheta which implies theta = sintheta for small angles of theta.
Being a math teacher must be brutal. It's hard enough when the students actually want to learn, but when the first question they always ask is "when am I ever going to use this", you're fighting a losing battle. I almost think mandatory math education might be a bad idea. If classes were small and only had motivated and interested students, that could only be a good thing, imo
Pie is the same for all circles because that number is the relation between the circles diameter and the circumference of the same circle. In a perfect circle there will always be pi numbers of diameters around the circumference no matter how big or small it is.
You haven't stated the proof completely yet. If sin x, tan x approaches 0 as x tends to 0 implies sin x, tan x could be approximated to the same thing for small x, then since x/100, 2x, x^100000000000000 all approaches 0 as x tends to 0, can't you also say that they are basically the same thing at x near 0? You have to also proof that, the rate of which the difference between sin x, tan x and x shrinks is much higher than that of which sin x shrinks, in other words, the limit of (sin x - x)/x is 0 as x tends to 0
If the limits as x approaches 0 of f(x) and g(x) are the same (let's call this limit a), then f(x) ~= g(x) as x approaches a. For example, take f(x)=x and g(x)=x^2. When x approaches 0, f(x)~=0 and g(x)~=0 , so we can say that f(x)~=g(x) when you get close enough to 0
@@antonyzhilin if f(x) / g(x) -> 1, multiply by g(x) on both sides and you have f(x) -> g(x), so when you approach a certain value of x, the two functions are equal
@@Jack_Frost Your definition is weaker. For example, take f = 2x, g = x. Then lim f(x) = lim g(x) = 0, but lim (f(x) / g(x)) =/= 1. So to prove sin(x) ~ tan(x) ~ x in one sense is significantly easier than the other
@@Jack_Frost You're quite wrong here. In fact, f(x)=x *does not* approximate g(x)=x^2 for small values of x. Try it out: f(0.01) = 0.01 but g(0.01) = 0.0001 - different by two orders of magnitude! Eddie Woo's derivation of the small-angle approximations is also wrong. It is insufficient to merely show that all three functions approach zero - you need to show that they do so with 1:1 proportionality. After all, the angle measured in degrees also approaches zero, but this is not a good approximation for sin θ.
You stated that sin theta approaches 0 as theta approaches 0. But that does not necessary mean sin theta approximately equals theta for small theta. You need more convincing proof for that. Again go back to the calculus for that part. Using the Taylor series of sin, sin theta is approximately equals theta for small theta.
You don't need calculus, just the fact that θ (in radians) is the arc-length around the circle divided by the radius. For small angles, the length 'h' in Eddie's diagram becomes comparable to the arc-length. So h/r = sin θ approximates arc-length/r = θ
Thomas Kim yes, I was meaning to agree with you that Mr Woo‘s explanation here is incomplete. My point is that you can explain it properly without venturing anywhere near calculus (which these students haven’t covered yet) - let alone Taylor series, which is well beyond that.
@@thomaskim5394 Also by using calculus you defeat the point of small angle approx like this. This is needed to find the derivatives of sine and cosine which find the Taylor series for themselves.
@@llcstudios1846 Read my other commitments about the Calculus textbook. Many Calculus textbooks have a great and complete proof without using Calculus. The proof in the video is not complete.
*ENGINEERING INTENSIFIES*
Eddie you are the Teacher of teachers !
Thank you
he fucked up tho, cause 0 is sin0 which is tan0 throw approximations
genuinely so amazing, I hope the best for this teacher
I normally love Eddie's videos, but this one gets it quite wrong.
He fails to point out that θ (in radians) is equal to the arc-length divided by the radius (i.e. *not* an arbitrary way of measuring angle). And for small angles, this arc-length becomes very close to the opposite-side-length, h.
Therefore, arc-length/r ≈ h/r ≈ h/(r-x)
In other words, θ ≈ sinθ ≈ tanθ.
Right. Via his same argument, theta * 10 is also approaching 0, and then sin(theta) would be approximately equal to theta * 10, which it obviously isn't.
@@joshuaronisjr This isn't his argument, this is an accepted mathematical proof thats used regularly. He was just demonstrating it. You're seeming to neglect that theta is considered to be infinitely small.
@@Chris-hn4lp When he says that sin(theta) = theta for small theta, he does so by saying that they both become infinitely small. Take the example of z=4x and y=x, where x is variable. As x comes infinitely close to 0 (and indeed, all the time - but that misses the point) we could say that z and y both approach 0. But we wouldn't say that z=y as x goes to 0. No, we would say that z=4y. The same thing with theta - just because theta and sin(theta) both go to 0, that doesn't mean that they can be considered equal for infinitely small theta (I'm not saying that they don't approximate one another - I'm just saying his argument isn't enough). As another example take z=e^(2x) and y=e^(x). As x goes to 0, z is approximately 2 times y, even though they both go to 0.
May I have the link to part two of the teaching?
For those who are interested, you can use the same setup to prove that theta is approx sin theta by considering that the area of the triangle approaches the area of the sector as theta approaches 0. So you have 1/2r^2theta = 1/2(r-x)h. We can cancel out the half and as theta -> 0, x->0, so we can set x = 0. Therefore, r^2theta = rh. So rtheta = h. But we know, h = rsintheta, therefore, rtheta = rsintheta which implies theta = sintheta for small angles of theta.
LLC Studios or as theta tends to 0 h ~ arc length so h=r theta therefore r theta =rsin theta cancelling the r gives theta=sintheta
I’m watching this for fun.
I thought you were going to use maclaurin series for this one but this is good too
Can you please help me in taking the coordinates in matrix form??
Being a math teacher must be brutal. It's hard enough when the students actually want to learn, but when the first question they always ask is "when am I ever going to use this", you're fighting a losing battle. I almost think mandatory math education might be a bad idea. If classes were small and only had motivated and interested students, that could only be a good thing, imo
Sir, you are great...
I want to meet you personally in India.
Perfect explanation . Thanks
Crazy. Excellent teacher. Beyong words.
Watching this in 4k is soo good
Awesome sir
I am looking 30yo sir I have learned all these at my school...But now I understand maths..U made me see maths
Another part of the fundamental theorem of engineering.
That sin(theta)=theta.
Yeah, I was struggling to understand why in vibrations we were equating sin theta to theta and this video made it perfectly clear.
Great visualization of it. I also want to ask if you could explain later on why pie is the same for big and small circles.
Pie is the same for all circles because that number is the relation between the circles diameter and the circumference of the same circle. In a perfect circle there will always be pi numbers of diameters around the circumference no matter how big or small it is.
Why does it look like you're doing the nae nae in the thumbnail 😂😂
ahahahahhhahahahah omggg
Osm explanation sir
Nice lecture
You haven't stated the proof completely yet. If sin x, tan x approaches 0 as x tends to 0 implies sin x, tan x could be approximated to the same thing for small x, then since x/100, 2x, x^100000000000000 all approaches 0 as x tends to 0, can't you also say that they are basically the same thing at x near 0?
You have to also proof that, the rate of which the difference between sin x, tan x and x shrinks is much higher than that of which sin x shrinks, in other words, the limit of (sin x - x)/x is 0 as x tends to 0
Another Video for another math tutorial.
❤💙💚💜💛
Wow.
NGUYÊN TV chúc bạn thành công
I like how your videos abruptly end, like if a limit exist. 🌿
Lol
If the limits are the same, the functions are the same, wat?
If there was ~= 0 at the end, I'd understand, but in-between relationship isn't implied
If the limits as x approaches 0 of f(x) and g(x) are the same (let's call this limit a), then f(x) ~= g(x) as x approaches a.
For example, take f(x)=x and g(x)=x^2. When x approaches 0, f(x)~=0 and g(x)~=0 , so we can say that f(x)~=g(x) when you get close enough to 0
Jack Frost I see. I'm just used to equivalence meaning “same order”, i.e. f(x) / g(x) -> 1
@@antonyzhilin if f(x) / g(x) -> 1, multiply by g(x) on both sides and you have f(x) -> g(x), so when you approach a certain value of x, the two functions are equal
@@Jack_Frost Your definition is weaker. For example, take f = 2x, g = x. Then lim f(x) = lim g(x) = 0, but lim (f(x) / g(x)) =/= 1. So to prove sin(x) ~ tan(x) ~ x in one sense is significantly easier than the other
@@Jack_Frost You're quite wrong here. In fact, f(x)=x *does not* approximate g(x)=x^2 for small values of x.
Try it out: f(0.01) = 0.01 but g(0.01) = 0.0001 - different by two orders of magnitude!
Eddie Woo's derivation of the small-angle approximations is also wrong. It is insufficient to merely show that all three functions approach zero - you need to show that they do so with 1:1 proportionality. After all, the angle measured in degrees also approaches zero, but this is not a good approximation for sin θ.
You stated that sin theta approaches 0 as theta approaches 0. But that does not necessary mean sin theta approximately equals theta for small theta. You need more convincing proof for that. Again go back to the calculus for that part. Using the Taylor series of sin, sin theta is approximately equals theta for small theta.
You don't need calculus, just the fact that θ (in radians) is the arc-length around the circle divided by the radius. For small angles, the length 'h' in Eddie's diagram becomes comparable to the arc-length. So h/r = sin θ approximates arc-length/r = θ
@@MattColler That is the point. Mr. Woo does not explain even though many calculus textbooks explain only using geometry.
Thomas Kim yes, I was meaning to agree with you that Mr Woo‘s explanation here is incomplete.
My point is that you can explain it properly without venturing anywhere near calculus (which these students haven’t covered yet) - let alone Taylor series, which is well beyond that.
@@thomaskim5394 Also by using calculus you defeat the point of small angle approx like this. This is needed to find the derivatives of sine and cosine which find the Taylor series for themselves.
@@llcstudios1846 Read my other commitments about the Calculus textbook. Many Calculus textbooks have a great and complete proof without using Calculus. The proof in the video is not complete.
SinA = A = tanA = A, cosA = 1
Can someone rectify
Any Physics students here?
Here! :D
Sin x / x = 1 as x -> 0
But i didn't knew for tan, it's quite obvious know ^^'
My favorite is pi = 3
First
First