Proof of the derivative of sin(x) | Derivatives introduction | AP Calculus AB | Khan Academy

Proving that the derivative of sin(x) is cos(x).

AP Calculus AB on Khan Academy: Bill Scott uses Khan Academy to teach AP Calculus at Phillips Academy in Andover, Massachusetts, and he’s part of the teaching team that helped develop Khan Academy’s AP lessons. Phillips Academy was one of the first schools to teach AP nearly 60 years ago.

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[phosus]

"Let's see if I can draw a relatively straight line"

*draws perfectly straight

[connorfitzgerald]

I love this guys voice and the way he accentuates on words. He just makes anything sound so interesting

[BackrubBucks]

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.

[cphillips]

Great proof, Sal! Thanks! I enjoy learning with Khan Academy; it’s a pastime not a sad-time.

[socialheretic]

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.

[jammesters]

Would like to ask as to whered you get the "cosx+sin∆x+sinxcos∆x" from sin(x+∆x)?

[mahdiyousef]

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!

[someone]

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

[mathlover]

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?

[MayowaLasisi-in]

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

[lil_weasel]

How does he single out the cosx like that and put it in front of lim

[davea]

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?

[vynneve]

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

[DDG]

Why is he not using h instead… what a weirdo l

[Ultamate-Jas]

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)

[vansf]

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) = <1 as Δx approaches 0 in your limit evaluation gives:
d/ dX (sinx) = as Δx approaches 0,
cos(x) { Lim (Δx)/Δx} - sinx{ Lim (sin(Δx)/Δx)} { Lim (sin(Δx) / (1+ cos(Δx))}
= cos(x) * (<1) - sinx { (<1) } { (0 / 1 + 1)
= < cos(x) - sin(x){ (<1)*(0) }
= < co(x) - 0 = < co(x)
Hence, the derivative of sin(x) < cos(x), but it is obviously not that d/dX(sin(x) = cos(x) as how you have claimed

Another way to understand that d/dx{sin(x)/x} ≠ ( not equal to) cos(x) is:
Suppose that sin(x) / x = 1 --> sin(x) = x . We can see that the case can happen only when the angle x is exactly equal to zero so that the length of the perpendicular y will become a point on the x axis of the unit circle, which means y = 0, while the arc length opposite to the angle x will also become another point overlapping the intersecting point of the circle and the x axis. It means that the angle x will become 0, and the arc-length will therefore also become O, and if the angle x = 0, then sin(0) = 0
---> As the angle x approaches 0, Lim (sin(x)/x = ( y/R) / S/R = y/S = 0/0 =1, which is impossible
Conclusion, there is obviously a flaw in the application of the "squeeze" theorem which claims that if sin(x)/x is between 1 and cos(x), then the limit of sin(x)/x = 1 as x approaches 0, while the fact is that 1 and cos(0) = x/ R = R/R = 1 are the same value. Therefore, you can never eve squeeze any other value within 1 if it is equal to 1 itself. In other words, you can squeeze a value within 1 only when it is < ( smaller than) 1.
Hence, the notion of the "squeeze" theorem here is absurdly or irrationally incorrect, and that the derivative of sin(x) is smaller than or not equal to cos(x) as defectively mathematically manipulated in the foundation part of calculus.

[rahulbathini]

What would this mean intuitively, does it mean that the tangent of a cosine function at different points varies as a negative sine function???

[johngodfreymalig]

I went here because of 2-dimensional physics' finding the optimal angle. And I am very, very confused.

[raphaelnash]

How does the squeeze theorem come into this? i.e. why does lim as dx --> 0 x sin dx/dx = 1?

[aarongeorgeson]

Thanks, great proof. My math book was super confusing when it came to this subject!