Your explanations are amazing 🤩 I always have always felt like I was missing something when trying to learn mathematical concepts. The “Why” factor... Today was the day i found the usefulness of proofs. Any one can learn how to change a light bulb but knowing why to change it is another story and without this key piece of the puzzle we are all left in the dark. Thank you for the amazing content.
There's easier way to prove this; The derivative of sin(x) = lim ((sin(z)-sin(x))/(z-x)), z->x Using the equation : sin(a)-sin(b) = 2 cos((a+b)/2) sin((a-b)/2) we get : lim ((2 cos((z+x)/2) sin((z-x)/2))/(z-x)), z->x the limit of 2 cos((z+x)/2) as z->x is 2 cos (x) the limit of sin((z-x)/2)/(z-x) as z->x can be solved by substitution; y = (z-x), when (z->x) (y->0) so we get : lim sin(y/2)/y as y->0 then we substitute g = y/2 so y = 2g so we get : lim sin(g)/2g as g->0 = 1/2 lim sin(g)/g as g->0 = 1/2 So finally we get (1/2)*2cos(x) = cos(x) hope this was clear
@@valle2353 Also, it's pretty intuitive visually anyway. The smaller an arc is, the better it will resemble a straight line. Naturally, as the limit approaches 0, it eventually becomes a perfect line. Hence lim->0 sinx/x=1
I like just using exponents. We know the d/dx of exp(x), so we can just write sin(x) as (exp(ix) - exp(-ix) )/(2i) and it's trivial from there. But I know of course the proofs done here are meant to avoid complex numbers. Sometimes complex numbers make things simple though :D
There is a geometric way to solve this limit and the other limit stated in the video. It is based on the squeeze theorem (as said in the video) and therefore does not require any notion of derivative
Would have been nice to see the entire proof. Especially the part that is useful in the proof. I know how to manipulate the expressions to get it to look like that but how do i show the lim =1 and 0?
This page has UA-cam links to the proofs you mentioned: www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/modal/a/proving-the-derivatives-of-sinx-and-cosx
x is a different variable from Δx. Since limiting Δx to 0 doesn't change the value of x, sin x doesn't need to be in the limit. This is just to separate it to create a limit that we know the answer to.
Wait a minute. You cannot just arbitrarily multiply something by -1. That isn't allowed. Are you muliplying the sin and the (cosx-1) each by -1? Because that would be ok, -1 X -1 = 1. Why did you elide that step?
He is multiplying (cos dx - 1) by -1 to get (1 - cos dx), and to compensate for this, he multiplies again by -1 to get that minus sign in front of the parenthesis. So all in all, he multiplies that term by -1 twice :)
it is incorrect to claim that the limit of sin(Δx ) / Δx as Δx = cos(x) as Δx approaches 0. Here is why: If sin(Δx)/ Δx = 1 just because you have made some mathematical manipulation with trig-formulas to arrive at the inequality that 1>/ = sin(Δx)/ Δx >/= cos(Δx), and as Δx---> 0, 1= sin(Δx) = cos(0)=1, and basing yourself on the "squeeze" theorem, you jump to the conclusion that sin(Δx)/Δx must also = 1, it'll mean that sin(Δx) = Δx. But, you will never eve be able to prove it in any way. Let's see why you guys have made a fundamental mistake here. To avoid making it confusing, let Δx = θ . According to the trig-graph of the unit circle, R = 1, y is the side of the right triangle and parallel to the Y axis of the unit circle, and the arc- length subtended by the 2 radii which create the angle θ is S= θR ---> θ = S/R, sin(θ) = y/R Limit of sin(θ)/θ = Lim {( y/R )/ S/R )= y/S , as θ approaches 0. There is no drawing function here for me to draw the graph. If you draw the graph yourself, you'll see that y is perpendicular to the X axis, and always smaller than the arc-length S because y is also the perpendicular of another smaller triangle of which the base is on the positive X axis , and the positive X axis = radius R of the circle, and the hypotenuse is the segment connecting the 2 ends of the arc S. So, the hypotenuse is greater than the perpendicular y, while it is smaller than the arc-length S. it means that the perpendicular y in sin(θ) = y/R is smaller than the arc-length S. Since y is always < S, y/S is always < 1---> As θ approaches 0, Lim (sin(θ)/ θ) = y/S is always smaller than 1. Substituting Limit of sin(θ)/θ i= < 1( smaller than 1) as θ approaches 0, which is the same as substituting Limit of sin(Δx) / Δx) =
No one has ever claimed that sin(x)/x is between 1 and cos(x). In reality, sin(x)/x is squeezed between cosec(x) and cos(x), meaning that as x goes to 0, sin(x)/x is squeezed between 1 and 1 and therefore has to be equal to 1. You can find that inequality geometrically. The squeeze theorem is rigorous if applied correctly. You should seek epsilon-delta definitions to understand how limits truly work. I wish you a happy Real Analysis class
You mention that in other videos you will complete the proof. In which of your videos do you show the complete proof. You might provide a reference number or name for a videos that are used in subsequent subjects.
"Let's see if I can draw a relatively straight line"
*draws perfectly straight
because he's using an app that can automatically draw straight lines
@@vknl99 no he is just a legend.
@Braylon Brennan It's paid you spammer
@@vknl99 He just locked the cursor for y axis
@@ruzzcraze1862 if he were a legend, he must have drawn a "relatively" straight line
I love this guys voice and the way he accentuates on words. He just makes anything sound so interesting
The Brazil version is goku voice
No you're telling lie.
@@motiwaledeepak6530 what...??
@@motiwaledeepak6530 ...h.. He.. H.... Hi. Ghbhhhb.
idk but it's kinda hot 🚶♀️‼️
Your explanations are amazing 🤩
I always have always felt like I was missing something when trying to learn mathematical concepts.
The “Why” factor...
Today was the day i found the usefulness of proofs.
Any one can learn how to change a light bulb but knowing why to change it is another story and without this key piece of the puzzle we are all left in the dark.
Thank you for the amazing content.
There's easier way to prove this;
The derivative of sin(x) = lim ((sin(z)-sin(x))/(z-x)), z->x
Using the equation : sin(a)-sin(b) = 2 cos((a+b)/2) sin((a-b)/2) we get :
lim ((2 cos((z+x)/2) sin((z-x)/2))/(z-x)), z->x
the limit of 2 cos((z+x)/2) as z->x is 2 cos (x)
the limit of sin((z-x)/2)/(z-x) as z->x can be solved by substitution; y = (z-x), when (z->x) (y->0) so we get :
lim sin(y/2)/y as y->0 then we substitute g = y/2 so y = 2g so we get :
lim sin(g)/2g as g->0 = 1/2 lim sin(g)/g as g->0 = 1/2
So finally we get (1/2)*2cos(x) = cos(x)
hope this was clear
actually i never saw this kind of proof. Why can you use z instead of dx
Dx=x-xo or in this case x-z@@nagys36snn
Please can you link the video of the cos x Lim ∆x tends to 0
I mean the one you were talking about cos x and sin x is 1 and 0 respectively
I'm confused, the title of this video is "proof of derivative of sin(x)" and then at 5:01 he says he's not going to do the proof in this video.
No he says tat he won't be proving sinx/x =1....he did derive the proof
Sinx/x can be derived from Euler theorem and talor series
you need the derivative of sinx to proof eulers theorem so you cant use it to proof sindx/dx = 1 since you need the derivative "first".
@@valle2353 it can be found via the squeeze theorem. Sal has already done a video on this
@@valle2353 Also, it's pretty intuitive visually anyway. The smaller an arc is, the better it will resemble a straight line. Naturally, as the limit approaches 0, it eventually becomes a perfect line. Hence lim->0 sinx/x=1
Great proof, Sal! Thanks! I enjoy learning with Khan Academy; it’s a pastime not a sad-time.
I like just using exponents. We know the d/dx of exp(x), so we can just write sin(x) as (exp(ix) - exp(-ix) )/(2i) and it's trivial from there.
But I know of course the proofs done here are meant to avoid complex numbers. Sometimes complex numbers make things simple though :D
I once heard that the derivative of a function is the measure of the rate of change of that function. Pls could you elaborate what that means??
Thank you
can we use l'hopital's rule for the last part
Would like to ask as to whered you get the "cosx+sin∆x+sinxcos∆x" from sin(x+∆x)?
Myrrh Cast it's just a formula you have to know. Search up sin angle addition or something. Doesn't come up tooo often, but very useful
Also it's cosx · sindx, not +
he got if from the trigo. iden. sum & diff. of 2 angles
sin(A+B) = sinAcosB + cosAsinB
@@a1exanderparra thaankksss.
. I rarely see it.
@@a1exanderparra formula you have to understand
Okay, but what if I don't know what a limit is and why is the lim od sin x that thing you wrote there???
You cannot do this because lim sinx/x =1 based on taylor expansion around zero which relies on knowing the derivative in the first place!
There is a geometric way to solve this limit and the other limit stated in the video. It is based on the squeeze theorem (as said in the video) and therefore does not require any notion of derivative
Awesome
good
What would this mean intuitively, does it mean that the tangent of a cosine function at different points varies as a negative sine function???
Yeah, it means that the slope of the tangent of a cosine function is the negative sine function
Thanks, great proof. My math book was super confusing when it came to this subject!
Aaron Georgeson দক্সগক্সক্সক্সক্স
যক্সক্সক্সক্স
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So if the derivative of sin(x)=cos(x) does that means if your trying to find the derivative of sin(x) at x=a , f'(a) by plugging in a for x in cos(x)
what
what crosshair do you use?
how am I only the second person to like this??
he paid for red dot sight
Finally here
Would have been nice to see the entire proof. Especially the part that is useful in the proof. I know how to manipulate the expressions to get it to look like that but how do i show the lim =1 and 0?
This page has UA-cam links to the proofs you mentioned:
www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/modal/a/proving-the-derivatives-of-sinx-and-cosx
Which program do you use to draw on?
How does he single out the cosx like that and put it in front of lim
x is a different variable from Δx. Since limiting Δx to 0 doesn't change the value of x, sin x doesn't need to be in the limit. This is just to separate it to create a limit that we know the answer to.
@@chrispyexe its like cos(x) is a constant (k) to the limit
Thanks a lot for this amazing proof
I have not understand anything
That's your fault
Its none of ur business
😂😂😂@@Zolnyx
Skill issue
Pause and ponder.
How does the squeeze theorem come into this? i.e. why does lim as dx --> 0 x sin dx/dx = 1?
nvm it's in previous video
What if he factor out sinx first then cosx?
Omg I’m first comment on a khan academy video! So happy! I love your videos btw
Thanks a lot
Playing with such crosshair hopefully make me good at math some day
Wait a minute. You cannot just arbitrarily multiply something by -1. That isn't allowed. Are you muliplying the sin and the (cosx-1) each by -1? Because that would be ok, -1 X -1 = 1. Why did you elide that step?
He is multiplying (cos dx - 1) by -1 to get (1 - cos dx), and to compensate for this, he multiplies again by -1 to get that minus sign in front of the parenthesis. So all in all, he multiplies that term by -1 twice :)
it is incorrect to claim that the limit of sin(Δx ) / Δx as Δx = cos(x) as Δx approaches 0. Here is why:
If sin(Δx)/ Δx = 1 just because you have made some mathematical manipulation with trig-formulas to arrive at the inequality that
1>/ = sin(Δx)/ Δx >/= cos(Δx), and as Δx---> 0, 1= sin(Δx) = cos(0)=1, and basing yourself on the "squeeze" theorem, you jump to the conclusion that sin(Δx)/Δx must also = 1, it'll mean that sin(Δx) = Δx. But, you will never eve be able to prove it in any way. Let's see why you guys have made a fundamental mistake here.
To avoid making it confusing, let Δx = θ . According to the trig-graph of the unit circle, R = 1, y is the side of the right triangle and parallel to the Y axis of the unit circle, and the arc- length subtended by the 2 radii which create the angle θ is S= θR ---> θ = S/R, sin(θ) = y/R
Limit of sin(θ)/θ = Lim {( y/R )/ S/R )= y/S , as θ approaches 0. There is no drawing function here for me to draw the graph. If you draw the graph yourself, you'll see that y is perpendicular to the X axis, and always smaller than the arc-length S because y is also the perpendicular of another smaller triangle of which the base is on the positive X axis , and the positive X axis = radius R of the circle, and the hypotenuse is the segment connecting the 2 ends of the arc S. So, the hypotenuse is greater than the perpendicular y, while it is smaller than the arc-length S. it means that the perpendicular y in sin(θ) = y/R is smaller than the arc-length S.
Since y is always < S, y/S is always < 1---> As θ approaches 0, Lim (sin(θ)/ θ) = y/S is always smaller than 1. Substituting
Limit of sin(θ)/θ i= < 1( smaller than 1) as θ approaches 0, which is the same as substituting Limit of sin(Δx) / Δx) =
No one has ever claimed that sin(x)/x is between 1 and cos(x). In reality, sin(x)/x is squeezed between cosec(x) and cos(x), meaning that as x goes to 0, sin(x)/x is squeezed between 1 and 1 and therefore has to be equal to 1. You can find that inequality geometrically. The squeeze theorem is rigorous if applied correctly. You should seek epsilon-delta definitions to understand how limits truly work. I wish you a happy Real Analysis class
Hi
You mention that in other videos you will complete the proof. In which of your videos do you show the complete proof. You might provide a reference number or name for a videos that are used in subsequent subjects.
I found it under the Squeeze Theorem topic. Thanks.
good call though
5:15 bruh just use l'hopital rule
Are you using small angle approximations at the end?
xd
no
For those of you who want to know the proof, it's not explained in this video.
I went here because of 2-dimensional physics' finding the optimal angle. And I am very, very confused.
So. Did u pass
Why is he not using h instead… what a weirdo l
underwhelming
Thank you