There's something up-front about his occasional hesitations and backups. It shows that he's not presenting the material by rote, but he's thinking it through again along with the student. That helps a lot. Also the informal intuitive "proofs" are enlightening.
Concepts conveyed in the pauses he takes. And, the seeking of answers from students pretending he has no idea about what it would be! :) Great Prof. Can't get enough of his lectures.
Totally agreed..For example in this lecture he says,,,’as when they as usual cut the pizza in 6 pieces’.. he perfectly know that they never cut pizza in 6 pieces, usually 8.. he wants you to focus on the area in that particular moment of the lecture,, he does this in almost all lectures and that’s make the different between an avarage professor and a great professor as he is.
I had to come back to this video. I'm going on 60, and after surviving long periods of plagues, aka schools, literally plagued by the presence of incompetent would-be teachers on both state and private schools' payrolls, this gentleman, is probably THE most lovable and competent maths teacher i ever encountered. No hard feelings, Sal Khan & friends. Thank U, professor Strang.
Thank you, Professor. You illustrate mathematical principles so beautifully, and with humor. You are a gift to humanity. May God bless and keep you always.
Prof Strang is an excellent teacher. Every time I've read one of his textbooks or seen his videos he explains things in a way that makes it seem obvious and intuitive. I learned linear algebra from his text. I wish my college profs had been like him. Of course, I didn't go to MIT... which I think must cost $$$....
Absolute beauty. Prof. is able to make a person like me, without much math background, understand such important math subject. This material can transform lot of lives across world. Thank you. Now it is clear, why MIT students and products are of such high standard.
loved the presentation. Best explanation on the derivatives of sin(x) and cos(x) I found so far :D. My only problem would be that I found the last proof at the end about the limit with the cosine a bit rushed.
Professor Strang. I love your lectures…I can say finally I can understand the meaning of calculus but I’ve never seen a whole pizza cut in 6 pieces, sometimes 4,, more usually 8..YOU ARE THE BEST!
Omgoodness. He is totally explaining trigonometric identies! I pretty much memorize the formula but this makes so much sense why it is that way! -_- the world is getting brighter!
Min 10:48 If you are asking why the heck that tangent line is 45 degrees, here is the answer. In the graph of sin vs rad, what happens when sin X / X goes to Zero.............They tend to have a ration of 1 as Dr. Gilbert proves, soooooo, what angles has an isoceles triangle when one of the three angles is straight????, I mean one side of the triangle is going to be Sin X and the other side Rad X; the angles subtended by equal sides in a triangle are equal, then each angles is л/4 or 45°.
Thank you for explaining this. My college math class only equips you with the tools to be a human calculator, the instructors do no explain any background as to what we are doing and why we are doing it. Knowing the system and the reason makes mathematics easier to understand and to remember, rather than just memorizing the derivative of sin is cos..
The very last step in the proof is given only a very cursory discussion. I'd call it hand-waving, except I acknowledge Strang isn't being rigorous on purpose for the sake of the bigger picture. The proof that the limit as delta x approaches 0 of (1 - cos delta x)/(delta x) = 0 is not complete. Giving a diagram and suggesting that (1 - cos delta x) seems to approach 0 as delta x approaches 0 is not enough. If you are curious to see this last result proved, you can see a proof that the limit is 0. For example, here: th-cam.com/video/CcNmx8XKf5M/w-d-xo.html Key steps: lim (1 - cos Δx)/Δx = lim (1 - cos Δx)(1 + cos Δx)/[Δx(1 + cos Δx)] = lim (1 - (cos Δx)^2)/[Δx(1 + cos Δx)] = lim (sin Δx)^2/[Δx(1 + cos Δx)] = lim (sin Δx)/Δx * lim (sin Δx)/(1 + cos Δx) = 1 * lim (sin Δx) / lim (1 + cos Δx) = 1 * 0/(1 + 1) = 1 * 0 = 0
Great lecture; here's a funny part: "Let me explain why we use radians... We need to use the natural way... the natural number is two pi... it can't be helped... It's the right one to use. 7:20"
"2pi is the number to use. it can't be helped, radians are the right way to measure an angle" wow that's so funnyawesome the way he says that! Such lucid _connection of concepts_ -- that's EXACTLY how you understand this stuff. EXACTLY. Though I'd say that since 2pi is indeed the number to use, it deserves a better name: Tau :)
Great presentation. The slope of a curve at the maximum is necessarily zero,because it's horizontal at the curve (Physics), if the cos function get its maximum when x takes the value zero, Of course, the slope of the cos function at this point (x) is zero. I think it's more easy like that.
1 radian is defined as the angle at which the arc of the circle subtended (the curve on the circle cut off by the angle) has a length of the radius. So describing angles in terms of radians gives you the neat ability to also speak of the "length" of that angle (which is actually the length of the arc subtended on a circle of radius 1) because you would just multiply the angle (in radians) by the radius (1) and you would get the arc length of that angle (which would be equal to theta)
I understand how to find the area of the piece of pizza by the following: let's stretch the arc corresponding with theta until it reaches the perpendicular line in order to get a right triangle whose area: 1/2 θ
The sine of x is the number you get when you make x the angle in a right triangle and find the opposite over hypotenuse of that triangle. That is the simplest definition of sine.
(1) He uses that (cosh-1)/h=(cosh-cos0)/h=slope near x=0 of cos(x), so approaches 0, when h approaches 0. (2) cos(h)/h=1/(near 0)=infinity when h approaches 0. Be aware!:)
+Stas Milev Prof. Strang has a number of complete courses available (as well as an online textbook) at MIT OpenCourseWare: 18.06SC Linear Algebra, 18.085 Computational Science and Engineering I, 18.086 Mathematical Methods for Engineers II, RES.18-001 Calculus Online Textbook. See ocw.mit.edu for the course materials.
The way he does the explanation confuses. But after you explained the algebraic manipulation it is clear enough. I was also not very satisfied with his proof that the slope at the maximum is zero - of course it is obvious that at the maximum the slope is zero, but I wish there was some other way of proving it. Thanks for your effort.
Oh that puzzled me too, but I got it now: The whole expression is: sin(x)cos(dx)+cos(x)sin(dx)-sin(x) If we ignore the middle part for a second, we have: sin(x)cos(dx)-sin(x) = sin(x)(cos(dx)-1) Let us do the same with algebra: ab-a = a(b-1) Try to multiply and see that they are the same. It is the same thing happening with the trigonometric function.
Alright , ever wonder why value of sin(45°) = 0.70 and sin(90°) = 1 , I mean from 0° to 45° it is 0.70 and 45° to 90° it is just 0.29. infact from 0° to 30 is 0.5 and 30° to 45° is 0.20.. just do this recursively and see what happens ... 30°-> 15° -> 12.5° -> 6.025° -> ...-> 0 This has to do with the geometry of the unit circle ! 0 to 90 - curve of the circle is increasing with decrease in rate . 90 to 180 - decreasing with increase in rate .
◇ for delta I don't get why in the trig identity for sin of the sum professor wrote sin(x)(cos(◇x)-1) I checked on the calc for a very small ◇x and it indeed approaches 0, but how he deduced it
great lecture! I can't believe i can learn calculus before i spend $700 on the actual class this coming fall. (UTstudent) 1 small mistake: Sin(A+B)= sinAcosB+cosAsinB. (not minus)
So the Cos is not only asymptotic approaching y=1 but is limited in space getting there because it drops at x=0 ? this is really amazing and I have to say mind-blowing. So what is the maximum value before it drops again? It has to be a finite number yet we cant get to it!?
I did not also understand how he got (Cos dx - 1) i.e. in the second part of the first term which is Cos dx, how did he get (Cos dx-1) when he subtracted Sin x ??? This is at the time 25:25 in the video lecture. I'll be grateful if someone could explain it to me step-wise !!
it's a useful ratio that helps you keep a tag on your right triangle, and find an unknown length. If you fix 2 sides and one angle, you know the whole triangle. In the "odd case" that one of the two unknown angles is 90deg, you can solve the triangle in seconds. The adjacent length = hyp cos(theta), and the opposite length = hyp sin (theta) Basically, sine and cosine are useful information about a right triangle.
It is indeed confusing that he is saying that he is subtracting sin(x), but he is not really doing any subtraction at all. All he is doing is the manipulation I gave in my other answer. It is also confusing that he writes -sin(x) on the other side of the equals sign when it clearly belonged to the end of the right hand side, but only before he manipulated the -sin(x) into the sin(x)(cos(dx)-1) expression. After the manipulation he should just have left it out.
no i learned in my school in ap calculus class its not supposed to have a - 1 basically what its supposed to be (cos deltax)/(deltax)=0 and (sin deltax)/(deltax)=1)
Some time derivative by first principle makes difficult to understand its logical way of deducing the derivative result. for example while arriving result for derivative of cosine(x) and derivative sine(x) by first principle approach or logic deducing the result are completely different. For derivative of sine(x) approach or logic is different when dx tends to zero And For derivative of cosine(x) approach or logic is different when dx tends to zero. So different logic for same principle for just to the result is not correct method. Any way limits tends to zero does not have logical approach to get result We know derivative of sine and derivative of cosine but if we does not knew derivative of some other function before hand then derivative by first principle is not correct method.
I understand "limit of sin theta over theta is 1 as theta approaches zero" by the following way: In unit circle, as the arc (theta) becomes smaller and smaller, the perpendicular (sine) fits better and better with the curvature of the arc so that, henceforward a given limit, both will be virtually the same. Is it correct?
Hmmm. I like this guy but we are talking ghosts of departed quantities here. The limit of (cos delta x - 1)/delta x as delta x tends to 0 is of the indeterminate form 0/0. Further work is needed to ascertain that the limit is in fact 0.
can someone tell me why did he subtract the cosx*sinΔx (from sin(x+Δx)) part instead of adding it? by arc adding formula we have sinx*cosΔx + sinΔx*cosx. i hope there a really good explanation or he just did a mistake
I'm sorry, I have a question. why did he use "theta" as the length of the curve? I think it's used to show how much the angle was taken, not for the length of its curve. thanks. :)
The fact is the triangle is out of the circle. so adjacent side here is equal to 1 (the adjacent side is the radius so 1) and opposite side is still opposite side, when you do tan = opp/adj, here you will have tan = opp/1 so tan = opposite side, so the opposite side is tan. I guess, it's how i see it. The area will be adj*opp/2 = 1*tan/2
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There's something up-front about his occasional hesitations and backups. It shows that he's not presenting the material by rote, but he's thinking it through again along with the student. That helps a lot. Also the informal intuitive "proofs" are enlightening.
Concepts conveyed in the pauses he takes. And, the seeking of answers from students pretending he has no idea about what it would be! :) Great Prof. Can't get enough of his lectures.
Totally agreed..For example in this lecture he says,,,’as when they as usual cut the pizza in 6 pieces’.. he perfectly know that they never cut pizza in 6 pieces, usually 8.. he wants you to focus on the area in that particular moment of the lecture,, he does this in almost all lectures and that’s make the different between an avarage professor and a great professor as he is.
I had to come back to this video. I'm going on 60, and after surviving long periods of plagues, aka schools, literally plagued by the presence of incompetent would-be teachers on both state and private schools' payrolls, this gentleman, is probably THE most lovable and competent maths teacher i ever encountered. No hard feelings, Sal Khan & friends. Thank U, professor Strang.
I really like how he goes over why we need to do something, and a small history of that thing. It helps a LOT
I am a math. Teacher , and i really appreciate these videos of professor Strang .
d(Gilbert Strang)/dx= Brilliance
S Koolwal then integral of brilliance is Gilbert strang ❤️❤️
Doesn't make sense mathematically.
Thank you, Professor. You illustrate mathematical principles so beautifully, and with humor. You are a gift to humanity. May God bless and keep you always.
Prof Strang is an excellent teacher. Every time I've read one of his textbooks or seen his videos he explains things in a way that makes it seem obvious and intuitive. I learned linear algebra from his text. I wish my college profs had been like him. Of course, I didn't go to MIT... which I think must cost $$$....
Absolute beauty. Prof. is able to make a person like me, without much math background, understand such important math subject. This material can transform lot of lives across world. Thank you. Now it is clear, why MIT students and products are of such high standard.
Dr. Strang is the Mr. Rogers of mathematics
Lol
loved the presentation. Best explanation on the derivatives of sin(x) and cos(x) I found so far :D. My only problem would be that I found the last proof at the end about the limit with the cosine a bit rushed.
In high school, I couldn't pass pre- algebra. Now I'm amazed at how I understand perfectly, these mathematics MIT courses.
Professor Strang. I love your lectures…I can say finally I can understand the meaning of calculus but I’ve never seen a whole pizza cut in 6 pieces, sometimes 4,, more usually 8..YOU ARE THE BEST!
That's a brilliant explanation -- rigorous yet not overly formal. Thank you!
25:17
sin(x+Δx) = sinx cosΔx + cos x sinΔx
If you know something very well you can teach it. This man knows.
Profesor Gilbert Strang presentations are amazing, thank you professor!
Omgoodness. He is totally explaining trigonometric identies! I pretty much memorize the formula but this makes so much sense why it is that way! -_- the world is getting brighter!
Min 10:48 If you are asking why the heck that tangent line is 45 degrees, here is the answer. In the graph of sin vs rad, what happens when sin X / X goes to Zero.............They tend to have a ration of 1 as Dr. Gilbert proves, soooooo, what angles has an isoceles triangle when one of the three angles is straight????, I mean one side of the triangle is going to be Sin X and the other side Rad X; the angles subtended by equal sides in a triangle are equal, then each angles is л/4 or 45°.
Thank you for explaining this. My college math class only equips you with the tools to be a human calculator, the instructors do no explain any background as to what we are doing and why we are doing it. Knowing the system and the reason makes mathematics easier to understand and to remember, rather than just memorizing the derivative of sin is cos..
Thanks a lot Dr. Strang! I'm from Uzbekistan!
Great stuff. I learned that symbolism isn't enough to understand A = A;
Concepts related to other concepts is the holy grail.
To get the derivative of cos x, just write cos x as sin (pi/2-x) and use the chain rule.
The very last step in the proof is given only a very cursory discussion. I'd call it hand-waving, except I acknowledge Strang isn't being rigorous on purpose for the sake of the bigger picture.
The proof that the limit as delta x approaches 0 of (1 - cos delta x)/(delta x) = 0 is not complete. Giving a diagram and suggesting that (1 - cos delta x) seems to approach 0 as delta x approaches 0 is not enough.
If you are curious to see this last result proved, you can see a proof that the limit is 0. For example, here:
th-cam.com/video/CcNmx8XKf5M/w-d-xo.html
Key steps:
lim (1 - cos Δx)/Δx
= lim (1 - cos Δx)(1 + cos Δx)/[Δx(1 + cos Δx)]
= lim (1 - (cos Δx)^2)/[Δx(1 + cos Δx)]
= lim (sin Δx)^2/[Δx(1 + cos Δx)]
= lim (sin Δx)/Δx * lim (sin Δx)/(1 + cos Δx)
= 1 * lim (sin Δx) / lim (1 + cos Δx)
= 1 * 0/(1 + 1)
= 1 * 0
= 0
just L'hopital :)
the limit in the video is cos Δx - 1, not 1 - cos Δx, btw
Matthew Iversen you are not understanding the video btw , and he is right . You aren’t
Great lecture; here's a funny part:
"Let me explain why we use radians... We need to use the natural way... the natural number is two pi... it can't be helped... It's the right one to use. 7:20"
"2pi is the number to use. it can't be helped, radians are the right way to measure an angle" wow that's so funnyawesome the way he says that! Such lucid _connection of concepts_ -- that's EXACTLY how you understand this stuff. EXACTLY.
Though I'd say that since 2pi is indeed the number to use, it deserves a better name: Tau :)
"Well...you know...that's a nice number, 360." Stay awesome, Dr. Strang!
Now this is how people should be taught the subject!
Prof strang you re a fabulous teacher. Because of people like you i still have faith in huma race
Great presentation.
The slope of a curve at the maximum is necessarily zero,because it's horizontal at the curve (Physics), if the cos function get its maximum when x takes the value zero, Of course, the slope of the cos function at this point (x) is zero.
I think it's more easy like that.
Love watching/learning from your vids, love your style! Thanks Dr!
1 radian is defined as the angle at which the arc of the circle subtended (the curve on the circle cut off by the angle) has a length of the radius.
So describing angles in terms of radians gives you the neat ability to also speak of the "length" of that angle (which is actually the length of the arc subtended on a circle of radius 1) because you would just multiply the angle (in radians) by the radius (1) and you would get the arc length of that angle (which would be equal to theta)
watching his lectures help me develop a habits of always turning up the speed when watching youtubes. :)
I understand how to find the area of the piece of pizza by the following: let's stretch the arc corresponding with theta until it reaches the perpendicular line in order to get a right triangle whose area: 1/2 θ
Excellent presentation Sir .Thanks with sincere regards. Rahul.Rohtak Haryana India
The sine of x is the number you get when you make x the angle in a right triangle and find the opposite over hypotenuse of that triangle. That is the simplest definition of sine.
(1) He uses that (cosh-1)/h=(cosh-cos0)/h=slope near x=0 of cos(x), so approaches 0, when h approaches 0. (2) cos(h)/h=1/(near 0)=infinity when h approaches 0. Be aware!:)
it'd be so great to have not just the highlights but the full course by prof Gilbert.
+Stas Milev Prof. Strang has a number of complete courses available (as well as an online textbook) at MIT OpenCourseWare: 18.06SC Linear Algebra, 18.085 Computational Science and Engineering I, 18.086 Mathematical Methods for Engineers II, RES.18-001 Calculus Online Textbook. See ocw.mit.edu for the course materials.
The way he does the explanation confuses. But after you explained the algebraic manipulation it is clear enough. I was also not very satisfied with his proof that the slope at the maximum is zero - of course it is obvious that at the maximum the slope is zero, but I wish there was some other way of proving it. Thanks for your effort.
Oh that puzzled me too, but I got it now:
The whole expression is:
sin(x)cos(dx)+cos(x)sin(dx)-sin(x)
If we ignore the middle part for a second, we have:
sin(x)cos(dx)-sin(x) = sin(x)(cos(dx)-1)
Let us do the same with algebra:
ab-a = a(b-1)
Try to multiply and see that they are the same.
It is the same thing happening with the trigonometric function.
Just a 14 year old passing by, and trying to understand. This video really helped, arigato🙇
I will say in the simple words "Good Job Professor".
Thank you MIT!
Strang is GREAT at summarizing. As for me at least. You should listen to him, when you ALMOST KNOW)
Alright , ever wonder why value of sin(45°) = 0.70 and sin(90°) = 1 , I mean from 0° to 45° it is 0.70 and 45° to 90° it is just 0.29.
infact from 0° to 30 is 0.5 and 30° to 45° is 0.20.. just do this recursively and see what happens ... 30°-> 15° -> 12.5° -> 6.025° -> ...-> 0
This has to do with the geometry of the unit circle !
0 to 90 - curve of the circle is increasing with decrease in rate .
90 to 180 - decreasing with increase in rate .
All love to MIT💜
he is a great teacher
◇ for delta
I don't get why in the trig identity for sin of the sum professor wrote sin(x)(cos(◇x)-1)
I checked on the calc for a very small ◇x and it indeed approaches 0, but how he deduced it
What a wonderful lecture
great lecture! I can't believe i can learn calculus before i spend $700 on the actual class this coming fall. (UTstudent)
1 small mistake:
Sin(A+B)= sinAcosB+cosAsinB. (not minus)
So the Cos is not only asymptotic approaching y=1 but is limited in space getting there because it drops at x=0 ? this is really amazing and I have to say mind-blowing. So what is the maximum value before it drops again? It has to be a finite number yet we cant get to it!?
tonyrosam You are confusing asymptotic with tangent.
at 21:09 when prof gave example of pizza .He said," the area is 1/6th of 2pi",.Which equals pi/3 , so the area of sector should be pi/3 , isn't it?.
Thanks MIT for this great stuff.
24:33 where can I find the demonstration of that formula?
Using the factor formulae makes life easier.
I did not also understand how he got (Cos dx - 1) i.e. in the second part of the first term which is Cos dx, how did he get (Cos dx-1) when he subtracted Sin x ??? This is at the time 25:25 in the video lecture. I'll be grateful if someone could explain it to me step-wise !!
This is the point where sine theta is equal to cos theta becomes when the sides are equal as sine theta /theta is maximum
Beautiful explained.
At 26:09 if ∆x is assumed zero, won't we end up dividing by zero which is not possible?
Very detail information and great job.
it's a useful ratio that helps you keep a tag on your right triangle, and find an unknown length. If you fix 2 sides and one angle, you know the whole triangle. In the "odd case" that one of the two unknown angles is 90deg, you can solve the triangle in seconds. The adjacent length = hyp cos(theta), and the opposite length = hyp sin (theta)
Basically, sine and cosine are useful information about a right triangle.
It is indeed confusing that he is saying that he is subtracting sin(x), but he is not really doing any subtraction at all. All he is doing is the manipulation I gave in my other answer. It is also confusing that he writes -sin(x) on the other side of the equals sign when it clearly belonged to the end of the right hand side, but only before he manipulated the -sin(x) into the sin(x)(cos(dx)-1) expression. After the manipulation he should just have left it out.
Muito obrigado MESTRE
no i learned in my school in ap calculus class its not supposed to have a - 1 basically what its supposed to be (cos deltax)/(deltax)=0 and (sin deltax)/(deltax)=1)
It's distributed, he takes it off of both sides.
why the area of triangle is 1/2 tan(theta) which is 1/2*a/b? Shouldn't the area is 1/2*a*b?
I got it, in this case, b=1.
Some time derivative by first principle makes difficult to understand its logical way of deducing the derivative result.
for example while arriving result for derivative of cosine(x) and derivative sine(x) by first principle approach or logic deducing the result are completely different.
For derivative of sine(x) approach or logic is different when dx tends to zero
And For derivative of cosine(x) approach or logic is different when dx tends to zero.
So different logic for same principle for just to the result is not correct method.
Any way limits tends to zero does not have logical approach to get result
We know derivative of sine and derivative of cosine but if we does not knew derivative of some other function before hand then derivative by first principle is not correct method.
God, this was beautiful.
Whoa great explanation, very thorough.
This is at time around 25:05
I Love his class
loved this lecture, thank you.
20:15 Pie = π lol
wish my professors explained it as good as u
I understand "limit of sin theta over theta is 1 as theta approaches zero" by the following way:
In unit circle, as the arc (theta) becomes smaller and smaller, the perpendicular (sine) fits better and better with the curvature of the arc so that, henceforward a given limit, both will be virtually the same.
Is it correct?
That's such a great way to think about it! That helps me understand it so much better.
Great video...
That opening was amazing
Bravo, bravo!
Hmmm. I like this guy but we are talking ghosts of departed quantities here. The limit of (cos delta x - 1)/delta x as delta x tends to 0 is of the indeterminate form 0/0. Further work is needed to ascertain that the limit is in fact 0.
Wow...its really amazing, Thank you sir!!😊👍✨💞
Thank you professor Strang!!!!
The whole pie is pie :)
great explanation
can someone tell me why did he subtract the cosx*sinΔx (from sin(x+Δx)) part instead of adding it? by arc adding formula we have sinx*cosΔx + sinΔx*cosx. i hope there a really good explanation or he just did a mistake
thank god he realized it was wrong 😅
um okay now I have a legit reason to skip class cause I'll actually be interested in watching to this professor from now on :)
Nicely done.
Professor Gilbert "Ok" Strange.. jokes aside LOVE THESE LECTURES
10:00 that's what i needed
Can anybody give me the youtube playlist of this lecture series by the same professor?
TH-cam playlist: th-cam.com/play/PLBE9407EA64E2C318.html. Best wishes on your studies!
@@mitocw thank you. 🙂
WAIT A MINUTE !!!!! That's a plus sign!
26:20
There was a mistake on his trigonometric identity. 25:02
Erratum
Sin(x+y)= sinxcosy + cosxsiny
24:55 It should be sin(a + b) = sin a cos b + cos a sin b
ERIC Sheng he later corrected it
I'm sorry, I have a question. why did he use "theta" as the length of the curve? I think it's used to show how much the angle was taken, not for the length of its curve. thanks. :)
you are right, theta(angle) = theta(rad) * 57.3, these two theta are in differenct unit.
Cos(x)/x
Best job 🕉️
Thank you so much! Very much obliged. Now I see it.
Remarkable!
great stuff except for the use of area in the inequality. such a needless and almost misleading method of proving the tan(theta) inequality.
fantastic
@19:00 area of triangle = .5 sin x cos; tan = sin/cos .. why the area of the triangle = tan !!!
The fact is the triangle is out of the circle. so adjacent side here is equal to 1 (the adjacent side is the radius so 1) and opposite side is still opposite side, when you do tan = opp/adj, here you will have tan = opp/1 so tan = opposite side, so the opposite side is tan. I guess, it's how i see it. The area will be adj*opp/2 = 1*tan/2
Hahaha. I just learnt that the pizza area of unit circle is a half of theta. This is cool.
great sir
sin (x+h)-sin x=2cos (x+0.5h)sin 0.5h , where h is delta x.
Sin(x+dx) = SinxCosdx PLUS CosxSindx !