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@Anonymous-df8it

4 ай бұрын

Or alternatively, sin'(x)=cos(x)/rad=cos(x)/(180/pi deg)=(180/pi)cos(x)/deg

If you admit sin'(x)=cos(x) then this formula is just a basic application of the chain rule.

@zh84

4 ай бұрын

The derivative of "sin x degrees" was a simple exercise in the chain rule that we had at school.

@henrymarkson3758

4 ай бұрын

Michael Penn did not assume sin'( x radians) = cos(x radians) He derived his result from first principles. In fact, using his method, you can prove for yourself that when x is expressed in radians, sin'(x) = cos(x). Left as an exercise to the reader.

@tcmxiyw

4 ай бұрын

I think Michael was trying to show us how much more complicated life would be if nobody had discovered/invented radian measure. Also, note that sin(t) and all its derivatives are pure numbers (do not have a unit of measurement) when t is a pure number (as it is when radian measure is used). The unit of measure on the derivative of sin(x degrees) is /degree (per degree). The unit of measure on the nth derivative of sin(x degrees) is (/degree)^n. Think of the ugly constants that will get even uglier as we take higher derivatives. Yikes! Thank the heavens for radian measure.

@pierrot31511

4 ай бұрын

@@henrymarkson3758 yes, that is why the first word of my comment is "if".

@BederikStorm

4 ай бұрын

Yes, but you need to first prove the first amazing limit. That's what he did

So, the reason for using radians to measure angles in calculus is similar to why we use e^x instead of another exponential function for exponential growth. Just like taking the derivative of any other exponential function, other than e^x, requires multiplying by a scaling factor, taking the derivative of trigonometric functions also requires a scaling if the derivative is taken with respect to an angle measured in anything other than radians.

@anlev11

4 ай бұрын

I thought the same thing

@evansaschow

4 ай бұрын

Not just similar but a special case of the same thing. Since cos(x)=(e^ix+e^-ix)/2, using degrees gets us cos(x°)=(a^ix+a^-ix)/2 where a=e^(π/180)

@Avighna

3 ай бұрын

Exactly!!!!

@livef0rever_147

3 ай бұрын

I think the reason is that, historically, the trigonometric functions were viewed as functions of circular arcs, that’s why they’re sometimes called “circular functions”. A problem to which the derivative of sine was applied was finding the sine given the arc. Using geometric reasoning one can arrive at the differential equation y’’(x)^2+y(x)^2-1=0 Where y, x are respectively a sine and its arc of the unit circle. This differential equation can be solved using infinite series. The particular solution given that when x=0, y=0 will be the series for sine.

Nice way to demonstrate why radian angles are preferable

@ikhu6042

4 ай бұрын

yes, the next step in the exploration would be to instead consider the derivative of sin(a * x degrees) -> a pi cos(a * x degrees) / 180 and then finding that a "natural" unit would be when a = 180/pi (giving us the conversion from degrees to this new natural unit of radians)

@jackychanmaths

4 ай бұрын

that's why I always wonder why schools need to teach degrees first and people need to use degrees radians is a far better unit for angles

@user-hp2dr5qc8p

4 ай бұрын

​@@jackychanmaths 1. 360 is a much "nicer" number than a mysterious 2π for kids to gobble down. 2. Degrees allow for more precision, and you can still kind of figure out how an angle "looks", in contrast to radians (what the hell is 1.012 rads?)

@jackychanmaths

4 ай бұрын

@@user-hp2dr5qc8p units like "drad", "crad", "mrad" can be used for more precision degree measure is already learnt after learning circumference so 2π should not be a problem

@p0gr

3 ай бұрын

absolutely not. for the derivative a factor is completely neutral. radians are preferable bc it gives a natural taylor series related to exp.

What's strange isn't that using degrees gives an unnatural result (anyone can see that 360 is an arbitrary constant) but that using *revolutions* would wind up with an arbitrary constant too. Revolutions seem like the obvious obvious unit for angles. Why does Math care so much about making us measure the distance around the edge of the unit circle? We don't even have to draw a circle at all to have an angle!

@junkgum

3 ай бұрын

Isosceles triangular arcs instead of just angles.

@MattMcIrvin

3 ай бұрын

The same argument from the video applies: if we care about the sine and the tangent, the squeeze theorem will tell us that both approach the arc length in the limit of small angles. And everything follows from there. Old slide rules took advantage of this in their trig scales. The output on a slide rule is generally scaled from 1 to 10 times some power of 10 (or, to put it another way, from 0.1 to 1 times the next power of 10), so the trig scales had the problem of not being able to represent sine or tangent values for small angles. But to the precision of a slide rule, when the sine or tangent is less than 0.1, you can approximate it pretty well by just the arc length on the unit circle. These scales were always marked in degrees. So they had another scale that just multiplied everything by pi/180 (well, pi/1.8 really, you always had to keep track of the power of 10 in your head with these things). And you could use that to approximate the sine or tangent of as small an angle as you liked. Or to convert any angle between radians and degrees. (With a really big, really good slide rule you could sometimes get three significant figures, so this approximation is pushing it on the very edge of that small-angle range, around 5.7 degrees. But it's not so bad.)

Using sin(h) is annoying since sinh is it’s own function 😂

@DeJay7

4 ай бұрын

True, but it's still completely unambiguous because we can't just write sinh meaning the hyperbolic sine, because there would be no argument to the function.

@CamEron-nj5qy

4 ай бұрын

Yes, ofc @@DeJay7

@pietergeerkens6324

4 ай бұрын

My super power is being able, from context, to distinguish sin h from sinh. 😉😜

@th3ta

3 ай бұрын

That's why i like the soviet notation of sh(x) and ch(x), and instead of having to pronounce them whole as hyperbolic sine or cosine, we pronounce them neatly as "shinus" and "chosinus"

A lot of people here missing the point. The "definition" that d/dx sin(x) = cos(x) is not fundamental. What Michael shows here is that to calculate this derivative, you NEED to first compute the limit as x --> 0 of sin(x)/x. Many Calc I courses have you use L'Hôpital's rule to compute this limit, but then you would need to first know the derivative of sin(x), which makes the argument circular. What is done on this channel is very careful math (mostly). If you want to see why your arguments for just replacing the degree sign with π/180 fail, try using L'Hôpital's rule to compute lim x --> 0 sin(x°)/x° by using that same trick. You'll get 1, which is incorrect.

@reckless_r

4 ай бұрын

1 is correct 😂

@phiefer3

4 ай бұрын

I've never seen a calc course use L'H to derive the derivative of sin, since as you say, it would require you to already know the derivative. In fact, most courses that I've seen introduce the derivative of sin before introducing L'H, and then specifically point out that L'H does not typically allow you to find new derivatives exactly for this reason. That said, the comments suggesting the chain rule are not missing the point, nor are they circular arguments. While not fundamental, the derivative of sin(x) for angles measured in radians is a known derivative, and there is nothing wrong with using a known derivative to find another derivative. Just as we can use the known derivatives of sin and cos along with the quotient rule to derive the derivative of tan, we can also use the known derivative of sin for angles measured in radians along with the chain rule to determine the derivative of sin for angles measured in degrees. Also your argument at the end regarding using L'H to calculate the limit is false. You've forgotten to account for units, while the result of the trig functions don't carry any inherent units, the angle in the denominator does. So the answer we get at the end is actually 1 with a radian unit in the denominator, when the limit we were looking for would have had the degree unit in the denominator. How do we convert radians back into degrees again? Oh yeah, by multiplying by 180/pi, but since the unit is in the denominator we end up with pi/180, same as in the video.

@carrotfacts

4 ай бұрын

@@phiefer3they said using L’H to calculate the limit of sin(x)/x, not to calculate the derivative of sin(x). Gotta read before you respond

@phiefer3

4 ай бұрын

@@carrotfacts I did read it, I'm not sure you fully read either of our posts. As he spoke both of the derivative of sin(x) as well as the limit of sin(x)/x. And the latter part of my post was specifically about the limit of sin(x)/x, I even specifically referred to the fact that the denominator is an angle and therefore has a unit associated with it.

@NotBroihon

4 ай бұрын

​@@carrotfactsyou're the one who's unable to read.

There's an example I like to do with my Calc. I students where we compute the height of a pyramid using trigonometry (with the length of the side and the angle of elevation). Then we use differentials to investigate what happens if we have an error in our angle measurement. I typically "forget" to convert the angle differential into radians to show the disparity and really drive the point home: calculus is done in radians.

I remember getting into a heated debate years ago on another channel with what appeared to be a all-knowing calculus1 student who tried to argue, quite viscerally, that sin(x °) or sind(x) was not differentiable, when it was clearly just a matter of differentiation with a simple chain rule. They also couldn't understand the fact that Radians is a SI derived unit that is dimensionless. I hope that student eventually passed calculus1 without embarrassing himself infront of his teacher or peers. Good day.

@Avighna

3 ай бұрын

I hope that student learned that it was lmao 💀 x degrees = x pi/180 radians, so sin(x degrees) = sin(x pi/180 radians) It’s easy to differentiate this with the chain rule

@Nick-wh4jt

3 ай бұрын

@@Avighna I hope that student learned to stop removing units from calculations also 😂

This is due to the degree operator just multiplying x by 180/pi. The way to rectify is to just convert from radians to degrees at the end when you want the final form but unless its a physics class its just an extra step for the math student to mess up on for no real benefit

@cameronspalding9792

4 ай бұрын

The degree operator is multiplying by pi/180 not 180/pi

@tcmxiyw

4 ай бұрын

I can’t count the number of times my physics professors said something like “since x is small, sin(x) is approximately equal to x.” This is not true when x is measured in degrees. When physicists analyze periodic behaviors, they absolutely use sine and cosine as they are defined using radian measure. This is done because there is no conversion factor when you want to use another unit of measure within sine, say sin(t) where t is in seconds. If the sine function is expecting a degree measure instead of radian measure then you have to convert from seconds to degrees before applying the sine function. (There is no conversion factor because radian measure is a ratio of lengths and is thus a pure number, I.e. has no unit of measurement.) Pick out just about any calculus based physics text. If they use the formula d/dx(sin(x))=cos(x), then they’re absolutely using radian measure.

@cameronspalding9792

4 ай бұрын

@@tcmxiyw The degree operator is effectively like multiplying by pi/180 (the number of radians in one degree)

@tcmxiyw

4 ай бұрын

@@cameronspalding9792 Degree is a unit of measurement, not an operator. The phrase “number of radians” is effectively meaningless because radian is not a unit of measure. If x is given a radian measure, then the correct usage is “x”, not “x radians”. Radian measure is a pure number (has no unit of measurement because radian measure is the ratio quantities with the same units of length, where the units “cancel” and the result has no unit of measurement-I.e. a pure number). A more correct phraseology would be something like “the angle whose radian measure is 1 is equivalent to an angle of 180/pi degrees.” I know this is somewhat pedantic, but if you’re doing a units analysis, and a variable that is inside the sine function finds its way outside of the sine function, using degree measure will mess up your units analysis, as I’ve illustrated in my example, and elsewhere. You cannot depend on the sine function to erase your errors when you think first in radian measure, then in degree measure and finally in radian measure again. If you start off thinking in radian measure, as you most certainly are if you use cos as the derivative of sin, and in an intermediate step you express the argument of sine expressing the argument in degrees, you have made an error. It will be a persistent error if you calculate a value at that point and your calculator is in radian mode. Unfortunately this kind of error is covered up by the sine function because the intermediate values are not calculated and the beginning and end expression have the correct values when thinking in radian measure. Covered up or not, they are errors.

@cameronspalding9792

4 ай бұрын

@@tcmxiyw let me rephrase it: if you see a number with a degree symbol, you treat that symbol as a constant that is equal to pi/180

You could have done that simply using the chain rule. (sin(x°))'=(sin(x•1°))'=(sin(x•pi/180))'=pi/180•cos(x°)

@jneedle92

4 ай бұрын

Was definitely thinking that the whole time!

@TheEternalVortex42

4 ай бұрын

It’s kind of interesting to do it the way in the video because it doesn’t use radians at all. So you can imagine if we started out doing calculus with degrees in the first place this is the result we would get

@harrymetu2746

4 ай бұрын

Indeed! I agree. Using chain rule by writing sin(x°) as sin(x(π/180)) is more of a quick short cut rather than a rigorous proof using standard principle definitions. It can actually be done easily that way because you already know that the derivative of sin(y) is cos(y) for any radian measure input y, which was basically gotten using the popular and pretty much essential limit, sinx/x as x--> 0, via squeeze theorem, where x is a radian measure. So similarly, assuming we wanted to determine the derivative of sin(x°), say not knowing anything about radians, or the relationship between it and degrees, we'd also have to compute this limit sin(x°)/x° as x--> 0 along the way, which can also be done via squeeze theorem. So the way Michael did it is a much more formal and rigourous way than just making a substitute then using chain rule. Then again, asking why he didn't just do it that way, well of course he probably knows but then everything would be much shorter and the video could end in like 2 mins. Also some who don't really get it could be left confused, especially if you haven't really thought of why radians is used instead of degrees in the first place, and never tried doing it with a degree measure input. Although, I see that using chain rule would be a quite clear way to most who are pretty familiar with it already. I feel that such a way would be better if this had been on a yt shorts video.

@BrianGriffin83

4 ай бұрын

Of course, but you first need to prove that if x is in radians dsin(x)/dx=cos(x) (which means radians need to be defined beforehand). It can be done by following the exact same reasoning of prof. Penn but without that π/180 factor. This is how my high school teacher proved it to my class.

I've always been a little annoyed at radians, they felt pretty pointless in pre-calculus, but after seeing this video I have a far bigger appreciation for them. Maybe I won't complain as much anymore... Great video, thanks for sharing this!

These visuals bring your videos a whole new charm

On of my more memorable facepalm moments as a physics TA was when one of my students used the "small angle approximation" [sin(x) is approximately equal to x], but measured x in degrees. "x is 5 degrees, so sin(x) = 5." Uh, no!

@MyOneFiftiethOfADollar

4 ай бұрын

If you react to your students with a "facepalm", consider resigning and working in the real world. You appear to be one of those "teachers" who are diminishing the reputation of a once respected occupation.

@randy-x

4 ай бұрын

Wait but why is that wrong? Is 5 degrees not a small angle?

@bradwilliams7198

4 ай бұрын

@@randy-x It is, but sin(x) can never be >1 for a real value of x. So if your approximation leads you to believe that sin(x) = 5, you either did something wrong (like measure x in degrees) or you're outside the region where the approximation is accurate. BTW, I did not embarrass anybody, just wrote a comment on the written homework that the angle needs to be measured in radians.

@randy-x

4 ай бұрын

@bradwilliams7198 ah ok, yeah that makes sense. Thank for the explanation!

@MattMcIrvin

3 ай бұрын

@@randy-x The right way to do it would be "x is 5 degrees, so sin(x) is approximately 5 degrees converted to radians (5 * pi/180, about 0.087)."

Michael, don't you think it should be sin(x°)/x, not sin(x°)/x°? Because I can have y=x° and my limit would be sin(y)/y which is 1. And if you treat x° as a change of variable then it would be x° = pi/180 * x, which still makes the second expression equal to 1 in limit as x -> 0. Also because if you plug in some numbers, you want to have sin(30°) = 1/2, but then 30° = 30 in the denominator. Maybe that's a units problem, but the radians are so hardcoded in my brain I cant interpret radians and degrees as something like grams

a lot of people pointed out the chain rule, but you can also multiply the differentials in numerator and denominator by 1° so that you just take the derivative wrt. the argument of the sine. so the answer is identical to cos(x°)°

@p0gr

4 ай бұрын

careful, d/dx sin(x°)=cos(x°)° but d/dx° sin(x°)=cos(x°).

@demenion3521

4 ай бұрын

@@p0gr that is why i said you multiply both differentials by 1° and not only the dx. otherwise you obviously change the result

Hi Dr. Penn! This is a great proof of the derivative of sine.

Instructive geometrically, but result follows quickly from d/dx(sin x degrees) = d/dx(sin x PI/180 radians) = (PI/180 )cos( x degrees), note this is chain rule applied to linear function x PI/180 radians

Good video as always. Something I've always wondered but have been unable to find an answer, is where do the theorems for the similarity of triangles come from?

@juliavixen176

3 ай бұрын

Euclid

A different point of view why we are using radians for angles is the following. Angle in radians is unitless quantity. Angle in degrees is not unitless quantity. Any "suitable" function can be expressed in a form of power series; e.g. sin(x). It is possible only for functions with unitless independent variable. In the other case, you will not be able perform summation because each term of the power series has different unit, e.g. degree^0, degree^1, degree^2, ... .

@Nick-wh4jt

4 ай бұрын

Radian is a SI derived unit that is dimensionless. It is a standard unit People get this wrong also because it is assumed the use of radians especially in calculus, the unit is omitted from results and calculations.

@vladimirlinhart4486

4 ай бұрын

@h4jt Yes, rad is not a base unit of the SI. It is derived unit with dimension of m/m = 1. So it is dimensionless quantity with a formal name radian. As can be seen from web: en.wikipedia.org/wiki/International_System_of_Units

@Nick-wh4jt

4 ай бұрын

⁠@@vladimirlinhart4486I’m not asking you I’m telling you. Radians is indeed a SI derived unit that is dimensionless. Where 1 Radian = 1 The definition of unit is 1 1 Radian = 180 °/π = approx 57.3 ° Radians is assumed in calculus hence why it’s omitted from calculations. This is why student get it wrong On another note, if you get your information from wiki then you’re also going to get it wrong, wiki is not an acceptable source of information. I suggest you recheck.

@Nick-wh4jt

4 ай бұрын

@@vladimirlinhart4486in fact I don’t think you read my first comment properly. Read it again. You were saying above that radians is not a unit. How is it not a unit? 1 is the definition of unit. Why do you insist on arguing? Do you argue with your mathematics professor as well? Maybe you think I don’t know what I’m talking about is that it? Look here, if you and everybody else are going to omit units from your calculations then you’re going to get incorrect results and I can tell you that quite categorically

@prestonbecker8784

3 ай бұрын

@@Nick-wh4jt Are you really going to give people a hard time because they used the word "unitless" instead of "dimensionless"?

sin(xº) = sin(x * π/180) and so sin'(xº) = sin'(x * π/180) = cos(x * π/180) * π/180 = cos(xº) * π/180

@64videosgunner

4 ай бұрын

*180/pi

@Didymus888

4 ай бұрын

@@64videosgunner how so?

@64videosgunner

4 ай бұрын

@@Didymus888 assuming x is in radians, x degrees would be 180x/pi

@Didymus888

4 ай бұрын

@@64videosgunner no x is in degrees

@64videosgunner

4 ай бұрын

@@Didymus888 didn’t u differentiate between x and x degrees?

Please let me know if I am wrong and why, but couldn't we use the resulta for the derivative of sin x in radians, write the angle as 180 over pi times the angle in radians to get the angle in degrees and apply the chain rule to get the correct derivative?

@ciaopeople9664

4 ай бұрын

Yes that's one line.

@cameronspalding9792

4 ай бұрын

sin(x degrees) = sin(x*(pi/180) radians), d/dx(sin(x degrees)) = d/dx(sin(x * pi/180 radians)) = pi/180 * cos(x * pi/180 radians) = pi/180 * cos(x degrees)

@EphemeralEphah

4 ай бұрын

Yeah, this seems to be a very roundabout way of achieving the same result. Either way, to me these both seem to just be stating that sin(x) is defined in terms of radians, which doesn't really explain why this is true.

@TheEternalVortex42

4 ай бұрын

The approach Michael uses in the video actually doesn’t use the concept of radians at all.

@joelklein3501

4 ай бұрын

All we assumed in this excercise in the trigonometric definition of sin, cos, and tan. We did not assume any other prior knowledge, and developed a formula for the derivative using degrees since if all what we knew about these functions came from trigonometry, then we probably would have been accustomed to working in degrees. Maybe we wouldn't even know about the existance of radians This method, although in radians instead of degrees, was what my calculus teacher used in order to prove that (sinx)' = cosx Therefore I think this video does motivates for defining radians. If any formula would be carrying the ratio (π/360) in some way, we might as well define radians to get rid if this tiresome term

In fact it doesn't matter if you use radians or degrees - so long as everything is kept in terms of sin(x), cos(x) or tan(x). When the angle itself starts to appear in the equations as a variable in its own right, that's when you need to start switching.

Totally was going into this video expecting a semantic arguement (I dont know why). Now Im left with my mind blown and searching my memory for if I have ever given a problem in degrees. Its been a few years since I taught calculus so I cant remember

The limit of sin(theta degrees)/(theta degrees) I think was confusing me since I would have expected it to be 1 under a basic u-substitution as the degree operation is just a constant multiple, until I realized that the degrees was being dropped off of 180 when finding the limit. Since originally the area of the circle slice is theta degrees / 360 degrees, the limit is pi/(180 degrees) not pi/180.

To this day I am still curious as to why π acts so much like a fundamental mathematical constant. It's the ratio of a circle's perimeter to its diameter, why would it have any special property (like for example e) in the world of calculus? Because, sin(x), on one hand, is a trigonometric function, therefore it is obviously related to π. On the other hand, it's the solution to a very interesting differential equation, y(x) = -y"(x) = y(4)(x) and y'=-y(3)(x) (a sort of four-cycle function of two functions, y and y'). More than anything, I would expect this differential equation to give a complex function (since i's four-cycle property is similar), but it's actually trigonometric, and so when expressed with an infinite Taylor series, somehow plugging x = π in that infinite sum mess gives exactly 0.

@juliavixen176

3 ай бұрын

Generally, the solutions for ordinary second-order linear differential equations is going to involve exp(), sin(), cos()... which, as you might be aware from complex analysis, are all related as summarized by exp(iπθ) = cos(θ) + i•sin(θ) ... because vector multiplication is rotation... and imaginary numbers are vectors.

@MattMcIrvin

3 ай бұрын

Complex exponentials and trig functions are basically the same thing. That multiplication by i and differentiating trig functions both give you a four-element cycle is not a coincidence. That differential equation is what physicists call a harmonic oscillator, like an idealized mass on a spring. And a harmonic oscillator, and one coordinate of a mark on a rotating wheel, give you the same motion. One way to think about it: if you're at the top of a rotating wheel, you're moving to the side. If you're at the side, you're moving up or down. So taking the derivative of your motion is a lot like moving 90 degrees around the wheel (up to a constant factor that is the kind of thing discussed in this video!) Do it four times, you get back to where you started. That's the connection to the circle. Multiplying unit complex numbers is also moving around a circle.

I think the utility of radian measure is that radian measure, being the ratio of lengths, is a pure number (by which I mean it has no unit of measure, like feet, inches, seconds). Similarly, sine, cosine, and their derivatives are pure numbers, if x is also a pure number. It’s possible that in sin(x), x is not a pure number, say x is measured in degrees, but even in this case the sin(x) is a pure number. It’s helpful, when x is in degrees, to consider what the unit of measure on the derivative of sin(x) would be: /degree (per degree). Similarly, if x is in seconds, then the unit of measure for the derivative of sin(x) is /sec. Now it’s straightforward to define periodic behavior, e.g. let f(t)=a*sin(t) where a is measured in cm and t in sec. f(t) is in cm, f’(t) is in cm/sec, and f’’(t) is in cm/sec^2. Imagine how much more difficult this would be if the sine, cosine were defined with their arguments in degrees. Think about the cyclic behavior we see with successive differentiations of sine and cosine when x is measured in radians. If they were defined in terms of degrees, all of the higher derivatives have those annoying constants. It looks weird when we first see it, but radian measure is natural. (BTW radian is not a unit of measure. It is correct to write “x” rather than”x radians”. Think of the latter as shorthand for something like “x is an angle in radian measure.”)

@Nick-wh4jt

4 ай бұрын

I have seen people get this incorrect many times and I’m going to tell you that Radian is indeed a SI derived unit that is dimensionless. Where, 1 Radian = 1 The definition of unit is 1 1 Radian = 180 °/π = approx 57.3 ° It is also assumed the use of Radians especially in calculus1 hence why the SI DERIVED UNIT is often OMITTED from results and calculations, this is why inexperienced students don’t yet understand the concept and get it incorrect

@tcmxiyw

4 ай бұрын

@@Nick-wh4jt Thank-you for this information. I don’t think it changes my point that the dimensionless character of radian measure contributes to its usefulness.

@Nick-wh4jt

4 ай бұрын

@@tcmxiywwhen you refer to things such as “pure number” then your complete understanding does come into question. Radian is a SI derived unit.

@tcmxiyw

4 ай бұрын

@@Nick-wh4jtI used a term for which I gave a definition, and, as far a I can tell, used that term in a logically consistent manner throughout my comment. Your response seems to have little other purpose than to insult because I didn’t use your preferred term. I admit that I was not up to date with the current state of SI, not having to refer to it since my undergraduate science experiences over 50 years ago, and I expressed my appreciation for your information about its current state. I will even own that I don’t completely understand my own discipline, which is why I strive to expand the boundaries of my understanding (and even to explore in other disciplines). I do, however, celebrate your complete understanding.

@Nick-wh4jt

4 ай бұрын

@@tcmxiyw you finally admitted that you were not familiar with the SI derived unit, which is what was brought to your attention in the first comment. Do you now concede that there was a purpose to my comment.

2:31 You should've used the COMPLEX UNIT CIRCLE, it would've made this explanation much easier.

Well sin(x°)=sin(pix÷180) and that should be ez to calculate

@MyOneFiftiethOfADollar

4 ай бұрын

as EZ as easy is to spell

Or alternatively, sin'(x)=cos(x)/rad=cos(x)/(180/pi deg)=(180/pi)cos(x)/deg

Sometimes I need an arc length and radians helps me do that where degrees cannot. If people want to measure angles using radians that is they prerogative. I am thinking it would be easier since circles are defined by the radius to use radians as a standard of measure. I do not really need any convincing. Another way to measure angles is by using spread but that is for some reason less debated. I assume either a double standard or because it is less prevalent. Both radians and spread are natural units because they define the angle and lengths we are measuring. It is only right to favor natural units as intuition tells us that a natural unit will eliminate unnecessary conversions and calculations saving us time. This is just a like a logical conclusion based on the parameters of the argument. Unless arbitrary units provide us something more such as solving a problem arising from using natural units like catastrophic cancelation, then using natural units would make calculations easier.

To convert to radians you multiply by pi and divide by 180 (pi/180) So sin(x degrees) = sin(pi/180 radians) By the chain rule that is pi/180 * cos(pi/180 radians) = pi/180 * cos(x degrees)

Why the convoluted way? Sin(x degree) = sin(x*pi/180), so d/dx sin(x degree) = (pi/180) * cos(x degree)? I.e. the function gets stretched out horizontally by (180/pi) so the derivative is (pi/180) smaller.

Did you actually state the area of the small triangle is less than or equal to the area of the large triangle?

@shishkabob984

3 ай бұрын

If the ray he started with is on the x axis then they are equal to each other

@tolkienfan1972

3 ай бұрын

@@shishkabob984 I'm taking about the statement he made. It's funny because you'd assume that the phrase "the smaller triangle" implies "has a smaller area". It sounds like a truism

I haven’t watched the video yet, but my best guess to this answer is that radians, unlike degrees, have no units so it’s more convenient. For example, in physics, if you want to convert from angular to linear velocity, that is quite easy since radians have no units. All you would need to do is multiply the radial displacement by the angular velocity and it will give you m/s. But since degrees have units, it’s less convenient and less conventional to use degrees.

I'm just a bit confused of the method. Why didn't you just analysed sin((π/180)x) ? It's just a linear equation to solve like 180X⁰ = πx radiant in terms of degree so X⁰ = (π/180)x

If e wasn’t e, if pi wasn’t pi… of course, if you introduce a constant factor in things, like pi/180, then it’ll get annoying at some point. The math still works but it looks less elegant.

Some quick ways to get the formula for the derivative of cos(x°) using trigonometric identities, continuing from where Michael stopped- 1st way- sin²(x°) + cos²(x°) = 1 0 = (1)' = (sin²(x°) + cos²(x°))' = 2*(π/180)cos(x°)sin(x°) + 2(cos(x°))'cos(x°) cos(x°)[(cos(x°))'+(π/180)sin(x°)] = 0 This should be true for all x°, therefore (cos(x°))'+(π/180)sin(x°) = 0 (cos(x°))' = -(π/180)sin(x°) 2nd way- sin(x°) = cos(90°-x°) cos(x°) = sin(90°-x°) (cos(x°))' = (sin(90°-x°))' = -1(π/180)cos(90°-x°) = -(π/180)sin(x°) 3rd way- sin(2x°) = 2cos(x°)sin(x°) cos(2x°)= cos²(x°)-sin²(x°) 2*(π/180)cos(2x°) = 2(π/180)[cos²(x°)-sin²(x°)] = 2(π/180)cos²(x°)+2(cos(x°))'sin(x°) (cos(x°))'sin(x°) = -(π/180)sin²(x°) This should be true for all x°, hence (cos(x°))' = -(π/180)sin(x°)

@joelklein3501

4 ай бұрын

Btw, 1st and 3rd way gives the identity (cos(x°))' = -(π/180)sin(x°) Besides on a descrete set of points x° = 90° + 180°k, k in Z Since at those points cos(x°) is 0 so the desired equality doesnt necessarily have to hold at those points How do we know that this identity holds for those points as well? One when to do it is by applying L'Hôpital's rule of differentiation to the limit of the derivative at those points. For example: (cos*90°))' = lim [cos(x°)-0]/[x°-90°] = x-->90° (by L'Hôpital) Lim [cos(x°)]'/[1] = x-->90° Lim [-(π/180)sin(x°)] = -(π/180) x-->90° Therefore cos(x°) is continuously differentiable, and (cos(x°))' = -(π/180)sin(x°) For all x°

I think saying sin(x°) is a bit confusing. sin is a function that takes a real value as an input, and we interpret this value as an angle in radian. But this is purely an interpretation, the input is a real value. To denote the fact that we are using different unit of mesure, we should use instead another function, lets call it for example sin°(x), and not sin(x°).

We Can Just Substitute θ° = 180θ/π so, d/dθ(sin θ°) = d/dx(sin(180θ/π)) = 180cos(180θ/π)/π = 180cosθ°/π

@jeffimber7152

3 ай бұрын

I think so........my problem (as a kid) was that it was not really intuitive as to why radians was the "natural" unit for angles, instead of say "revolutions" or even "diameters" or something. I think.......the squeeze theorem in order to show that lim_{x->0} (sin(x)/x) = 1 only works cleanly and with no constant correcting factor when the angle measure is in radians.

By the chain rule, this is cos(x°)*°

Why not analyze this more simply as a composition? Y = Sin (f(x)), f(x) = x/360 * 2 * pi?

@Avighna

3 ай бұрын

Yes, that would indeed be a simpler way to do this. The point of this video is also to present the classical proof for the derivative of sin(x) being cos(x), just done for degrees instead of radians.

This year in geometry I learn about Sin Cos Tan and delta

Silly question for somebody smarter than me - probably nearly all viewers of this channel: There are the "normal" trig functions, cos, sin etc. And there are hyperbolic. And parabolic. Are the elliptic equivalents, say for an ellipse with axis a=1 and 0

@juliavixen176

3 ай бұрын

Yeah, Jacobi's elliptic functions are probably the closest thing to what you're asking. Weierstrass also defined some more complicated elliptic functions. (I assume you're already familiar with the parameterized versions of the standard trig function like sinp(p,z), cosp(p,z), tanp(p,z), etc.)

Dear Professor! When we use radians, we have no problem with the dimensionality in the limit - units are divided by units. But there is a problem with degrees - units are divided into degrees and this is not good. You have lost the degrees of 180 in your formulas

You could have just said than sin(x radians) = sin((pi/180)x degrees) and then done chain rule

Sin'(a*x)= acos(x), x° = x rad * 180 / phi

why not use x° = x rad x pi/180 = x rad x A ? = xA

@KeimoKissa

4 ай бұрын

Because that's not particularly interesting...

Change of variable + chain rule, easy

that's a good place to s indeed

good

And that is the first derivative. Think of the second one , and then on the Taylor series center at zero. A nightmare

Whenever I come across this I have to think about da^x/dx

Making sinx/x = 1 when x=0 doesn't seem like a big win compared with the disadvantage of littering the place with "2pi". Choosing the radian still seems like an arbitrary choice to me. What great advantage to the radian am I not seeing?

@ConManAU

4 ай бұрын

Since (sin x)’ = cos x and (cos x)’ = -sin x, working in radians means that sin and cos are solutions to the differential equation y’’=-y, which makes them fundamental to equations of simple harmonic motion among other things. If you don’t want ugly factors showing up everywhere then imagine if the defining equation for SHM was y’’=-pi^2/32400 y

13:38

Step 1: convert degrees to radians. Step 2: chain rule ez

And that’s a good place to stop using degrees and start using radians!!!

👍

8:53 I find it interesting that sin((1.8÷10⁶)° )≈ π/10⁸. Numberphile did a video on this. => sin((180/10⁸)°)/(180/10⁸)≈π/180 ...🤔

No, calculate the derivative of sin(ax) knowing the derivative of sin(x).

people saying "it's just the chain rule, why not do it that way!" are missing one of the key ideas of the video. But my question is, can anybody provide a nice intuition for lim sinx/x = 0, but lim sinx/x in degrees is *not* zero? That's what my mind finds most unintuitive. Particularly by that picture he drew, I don't see how that area "in the limit" is nonzero, in fact all of the areas look like they should be zero as the the angle approaches 0...!

@gerryiles3925

4 ай бұрын

The limit of sin x/x is not 0, it is 1...

So much bullshit to show: sin'(ax) = a cos(ax) Noticing y = x 180/π is the transformation from radiants to degree

this is plain wrong. you're playing fast and loose with the degree sign (which just means times pi/180), while the whole point is the difference between d/dx sin(x°)=cos(x°)° and d/dx° sin(x°)=cos(x°). of course sin(x)/x goes to 1, no matter what coefficients you place on x as long as its the same on both - you should know this! you tricked yourself by working sloppyly, dropping and reintroducing the degree symbol inconsistently. what you actually show is sin(x)/x goes to pi/180°=1. when you compute the differential you again remove the whole point together with the degree sign. you need to consider (sin((x+h)°)-sin(x°))/h without degree in the denominator. this is what will in the end give you the actual factor of pi/180, in your case its again pi/180°=1. the degree sign is admittedly a weird way to write a coefficient. if its easier for you, just define a:=pi/180 and write d/dx sin(ax) instead of d/dx sin(x°). its the same and will of course throw the factor a to the outside by the chain rule or, as you try but fail, by first principles. i am disappointed in you.

Why so complicated? Just use the definition of degree and the chain rule...

@chair7728

4 ай бұрын

Ah yes. Calculate derivative of sine assuming an already known fact about the derivative of sine

@HJKey

4 ай бұрын

@@chair7728 No, you calculate the derivative of sin(ax) knowing the derivative of sin(x), being x the angle in radians. Bye.

@chair7728

4 ай бұрын

@HJKey You are completely missing the point of this video, you should not assume you are given what the derivative of sine in radians is, of course Michael also knows the chain rule and that the derivative of sin(x) is cos(x) with x in radians...

@HJKey

4 ай бұрын

@@chair7728 Groundbreaking.

First of all, I liked the ad placement as you got your diagram ready. Thank you for covering this derivative. I did the radian version by definition a couple of months ago and people loved it. It is a great proof to see. I was surprised how much attention it got. Your presentation was really good. I liked your visuals, both drawn and inserted. Your explanation was excellent as well. Cheerful Calculations 🧮.