Proof: Derivative of Sin is Cos (Version 2)

[@physicsgod]

For true mathematician or physicist it’s like heaven.❤ *one of the best video i have ever watched on calculus on KZbin love from india* 😊❤

[@jameshale9093]

At 7:03, I think the cos(h) should be at the bottom line and the bottom yellow line should not extent your the perimeter of the circle. Correct? So, you’d see that line approach 1. As is, the cos(h) line is already one, since this is a unit circle and it extends from the center to the perimeter.

[@stevegerard966]

Thank you for taking the time to show the 'how' and 'why' rather than just providing information about a concept. Now it makes sense.

[@htg4666]

Thank you for sharing this beautiful video. There is a small mistake in minute 07:00 , you defined the radius of circle as "cos h" , but the "cos h" is part of yellow line which is located inside the right triangle and compose the base of triangle.

[@divine6104]

Thank you. I was so confused about that.

[@dietrichschoen7340]

Thank you very much. Your lecture is fantastic! Finally I understud why sin(x)/x=1 is. After hours trying to understand this! Again: Thank you very much.

[@jacknisen]

I just wanted to see the formal proof of the cosine limit. You hand waved that.

[@Kumurajiva]

Math can be beautiful

[@avijitdey992]

That was wrong. Cosx is the adjacent side in a unit circle. Not the hypotenuse. But in unit circle as angle h->0 we have cosx -> 1 as the radius is 1

[@adamlea6339]

I don't get the explanation of the limit (cos(h)-1)/h. I understand visually that cos(h) -> 1 as h -> 0 but I don't see how it immediately follows that (cos(h)-1)/h -> 0 as h -> 0 since this tends to 0/0 which is undefined.

[@martinepstein9826]

Yeah, the title is misleading. He doesn't prove anything, he just says "these things look like they're closer together".

[@samyaspapa]

Yeah, his proof is wrong. Basically, you need to construct the "outside triangle" where the hypotenuse is a secant line of the circle. The height is sin(h) and the base is (1-cos(h)). Using trig identities: 1-cos(h) = 2sin(h/2)sin(h/2). Then you can solve for the secant line length which is 2sin(h/2). Then you use the limit that the perimeter of an N-gon approaches the circumference of a circle as N goes to infinity. This means that (2pi/h) times the secant line length equals 2pi. Which after some substitution will lead you to sin(h)/h -> 1. Then with the previous identity for 1-cos(h) you end up with (1-cos(h))/h equalling 0*1which is zero.

[@Kumurajiva]

Fabulous visual

[@deekeyclasses6793]

Just you make it very simple 😮

[@PJ-wg7vh]

Very well explained.

[@geraldramos3961]

Great video! How did you do the animations?

[@Adler09]

Great video, this is how math should be interpreted

[@hqs9585]

5:50 big jum, did you use l'Hopital rule i.e. lim h->0 of sih/h is limit h->0 of cosh/1 (differentiate top and button).

[@ss76845]

@7:30 in a circle ... at any angle, cos h will always be equals to value 1. because the radius of circles is same no matter what is their angle. Please correct me

[@seroujghazarian6343]

The hypothenuse is NOT cos h because it's the radius.

[@carultch]

The hypotenuse is equal to 1 by definition, because it is a unit circle.

[@haniperetz4123]

at 9:02 if H becomes zero does it not mean that sin h divided by h is also zero. since sin(0) = 0 ???

[@spudhead169]

Circular reasoning.

[@dougr.2398]

Why radians are used seems to be skated over… just stated as factual without explaining its necessity

[@dougr.2398]

@Eucalypticus the ratio of circumference to diameter is a pure number, pi, and has no dimensions as it is feet divided by feet or meters over meters. Radians or degrees enter slightly differently. Degrees come from a Babylonian measure that is a multiple of sixty, and is to some “degree” (pun) nearly the number of days in the year, plus five festival days. Pi radians is 180° only because there are 2 π of them το make a unit circle

[@dougr.2398]

@Eucalypticus I agree that radians are a more “natural” measure. I never disputed that. In fact, imposing 360° on a circle was arbitrary and capricious, as I already indirectly indicated.

[@carultch]

@@dougr.2398 Because it would make the video excessively longer, when he already has radians covered in another video.

[@BlackbodyEconomics]

Very cool - I've never seen it done like this. Thanks man :)

[@erenjager4220]

I don't understand how the slope = cos(theta)? from minute 1:19

[@tarannum7884]

literally me neither but maybe it's like slope is basically theta...right...and we know that to calculate it we do like tan(theta) = sin(theta)/theta...we can write tan(theta) as sin(theta)/cos(theta) and then do math to find that theta = cos(theta) but it doesn't really make sense

[@cipciop77]

I don t agree with the explanatation of the limit.

[@tariqmehmoodraza9961]

hi, awesome video. could you please suggest what apps or software can be used to make such videos. what app are u specifically using for such an interactive geometrical stuff? regards

[@gentlemandude1]

How can the hypotenuse be "cosh" by definition (6:57)? Shouldn't it be equal to sqrt([cosh]^2+[sinh]^2) by Pythagorean theorem? Am I missing something?

[@shivamsharanlall672]

Actually he did a bit mistake. Just replace the 1 at the base of the triangle with the cos h. Here the length of the base of the triangle is cos h but the radius of circle along the base is 1. (7:15) As the triangle becomes smaller and smaller, cos h approaches the complete length of radius of circle along the base which is 1. (7:40)

[@gentlemandude1]

Thanks, that makes sense. I thought I was crazy. I wish the video's creator would fix that mistake. It's very confusing for people who are trying to learn the concept.

[@shivamsharanlall672]

@@gentlemandude1 in which class do you read? And where are you from?

[@petermm8119]

@@gentlemandude1 Me, too! I thought no, cos h cannot be 1, which is the hypotenuse, and the hypotenuse is not the same length as the adjacent side! Oof. Thought I was losing it!

[@dougr.2398]

Egregious error!! Any radius of the circle = 1 as it is the unit circle, which should be stated at the outset. If you look at the actual cos (h) length, it does approach one.

[@travelindiawithme8267]

Amazing video Sir👍

[@lowlightevangelist9431]

Bravo, bravo!

[@peterbauer8461]

can you explain more fully when you say later in the video concerning the triangle that "the hypotenuse is cos h by definition"

[@sameerathreya9253]

peter bauer, it is wrong...The part of the base subtended by sin (theta) and the origin is cos (theta) (Conventional). But, the line parallel to it originating from intersection point of the elevated radius ("hypotenuse") and the arc to Y-axis is the actual Cosine.

[@sameerathreya9253]

Or maybe cos "h" means something different.

[@청동이-u6p]

continuing speaking out Ah~ Ah~ , Thank youCal1fun!

[@peterlohnes1]

This is an excellent visualization and helps to understand the results, but doesn't really prove it (which is a very convoluted process than involves proving sin(h)/h is 1 as h approaches zero and (1-cos(h))/h is zero as h approaches zero. It is a very difficult proof and I believe Khan academy shows it well...not easy though

[@haideralam1]

Thanks🎉

[@wenzelbotha8077]

You're good.

[@Kumurajiva]

So neat

[@Ahmed_alduolaimi]

thank you it helped my alot

[@abdouliebah6792]

i dont understand

[@37no37]

thanks, where is the links.

[@jackkalver4644]

This is not proof! Fortunately, I have come up with a proof that is much easier to follow than the whole process of finding the limits and angle-sum identity and then using (sin(x+h)-sin x)/h to derive cos x. It uses the unit circle and the derivative of a parametric function. Someday I may make a video version of it.

[@안정만-d2g]

최고

[@weqe2278]

This is atrocious.