Cosine of 1 degree from the ground up.

[@MichaelPennMath]

🌟🌟To try everything Brilliant has to offer-free-for a full 30 days, visit brilliant.org/michaelpenn. The first 200 of you will get 20% off Brilliant's annual premium subscription.🌟🌟

[@xbzq]

And that's a bad place to stop. Where's the rest? You stopped right before the end.

[@Anonymous-df8it]

That's not a good place to stop! Why didn't you substitute the sines and cosines of 3 degrees into the final equation?!

[@leif1075]

​@Anonymous-df8it EXACTPY WHY ISNT EVERYONE ELSE WONDERONG WHY THE HELL ARE THERE is in the answer..the cosine kf 1 is NOT IMAGINARY ..you agree with me that there shouldn't be i'a in the answer right.?

[@xbzq]

@@leif1075 There's not though. There are no imaginary numbers in the answer.

[@bosorot]

Without calculation, the engineer will say 1.

[@BIGWUNuvDbunch]

Lmao engineers in shambles

[@benjaminojeda8094]

Actually he Will Say 1/180, because 1° = 1/180

[@landsgevaer]

@@benjaminojeda8094 You forgot the cos? And the pi? An engineer showing off might say cos(x) = 1-½x² and for 1° = pi/180 = 3/180 = 1/60 you get 1-1/7200 = 1-1.5/10800 = 1-1.4/10000 = 0.99986 which is off by one in a hundred thousand.

[@BillSmithPerson]

Cosmologist will say somewhere between -1 and +1. That’s close enough.

[@trucid2]

@@BillSmithPerson And an astrologist will say that it means mercury is in retrograde.

[@moorsyjam]

I've seen a few graphical representations of sin or cos(a+b) and of (a-b) and they've always been clunky and awkward. That derivation at the beginning was smooth.

[@hiiistrex2838]

Yes!!! Never seen a more elegant proof of this

[@SimonTyler0]

You can justify the rectangle proof by thinking of compounding two rotations. Falls straight out of the unit circle.

[@KrasBadan]

I like the derivation with e^i(a+b) but this one is cool too. The triple angle was nice.

[@giansieger8687]

I always use the multiplication of two rotation matrices to derive the addition theorems. No geometrical thinking required and works really fast

[@willthecat3861]

@@SimonTyler0 I don't see how? (You can draw a rectangle around the points (x,0) and two other distinct points on the unit circle, so that those points touch the rectangle; but there won't, in general, be a 90 degree angle between the terminal arms, and the cord joining the two distinct points that are not (x,0). (see Penn's diagram, at least 1:18. The triangle containing angle beta, having base = 1, does not have a hypotenuse of 1 )

[@afuyeas9914]

As some others pointed out the last line essentially reads cos(1°)=cos(1°), ultimately showing there's no point in trying to derive a closed form for some angles. cos(1°) is a 48th degree algebraic number with a mammoth minimal polynomial. That said I had never seen the derivation of cos(a-b) in that manner, it was great to see

[@sleha4106]

48th degree algebraic number and mammoth minimal polynomial. Could you elaborate on these two

[@ceratos9588]

Cos(1) is not algebraic

[@leif1075]

Hiw have you seen the derivation od cosine (a-b) then I'm curious? Since youbsaid you'd never seen it that way? And you don't need triple.abgld formula for this. Je cam write 3 as 2 plus 1 and then 2 in turn is just. 1 plus 1 and use osulve angle formula tomsoove for cosine (1) after getting cosine (3)...anyone else think this?

[@GrouchierThanThou]

There is an error in this video: cos(18) = √(5 + √5) / 2√2 not (5 + √5) / 2√2. Note the additional square root in the numerator.

[@DrR0BERT]

I just caught that too.

[@bethhentges]

Yes

[@bethhentges]

And again at 15:20 he’s missing the square root if the numerator.

[@bethhentges]

And again at 15:33.

[@psymar]

Thank you. O wondered what was wrong, the value given in the video is >1 which I knew was wrong

[@shauas4224]

I remember calculating all cos and sin of degrees that are multiple of 3 in 9th grade. Then I spent like 3 days trying to figure why can't I calculate sin 1°

[@tednelson5277]

"... through the magic of complex numbers.." All very interesting but as mentioned, the end is quite chaotic.

[@Simon-fg8iz]

The last step is circular reasoning and doesn't actually compute anything. You just got back the Euler's formula you started with, after all that work: the last expression is just expansion of Re(e^iθ)=Re(3√e^3θ). We know from the fact that there are 3 real roots, that you are dealing with "casus irreducibilis", meaning you don't have an closed-form algebraic expression in real domain. In fact, this case reduces to trig functions when solving cubic equations.

[@MarcoMate87]

It's not circular reasoning, because Michael computed cos(3°), showing it is an algebraic number with a closed-form algebraic expression (even if the expression he gives us is wrong, as the correct closed form is cos(3°) = √(5 + √5) / 2√2). Thus, even cos(1°) has a closed-form algebraic expression, only this form involves extracting cube roots of complex non-real numbers.

[@Simon-fg8iz]

@@MarcoMate87 I meant the entire derivation could be skipped, e^i1°=3√(e^i3°)=3√(cos(3°)+i sin(3°)) and taking the real part is one step, no need for first applying the triple angle formula, do a bunch of reduction steps and then arrive back to complex numbers you started with.

[@aniruddhvasishta8334]

The derivation at the beginning was really cool and I haven't seen it before, but if you were going to derive the triple angle formula from Euler's identity anyways, you could've just used that to derive the subtraction formula too.

[@thepaperperson_gay]

everyone knows how to do that, I'd rather see something I haven't seen yet

[@leif1075]

Why not solve without Euler or triple angle at all? Just break up cosine of 3 into cosine(1+2) and then cosine 2 is just cosine (1+1) use double.angle and you're done..see what I mean?

[@aniruddhvasishta8334]

@@leif1075 It's been a while since I watched this video but isn't that kind of what he did, just in reverse?

[@mairc9228]

also using eulers formula and picking the principle cube root will yield ((cos 3º + i*sin 3º)^1/3+(cos 3º - i*sin 3º)^1/3)/2=(e^(i*pi/180) + e^(-i*pi/180))/2 which is precisely the complex expression for cos(pi/180) or cos 1º

[@pancito3108]

how can cos18° be (5+√5)/2√2? √5 is more than 2, and 2√2 is about 3, so we end up with approximately 7/3, which is more than 2, but cosine ranges from 0 to 1.

[@Jeity_]

How is the last expression any useful? That's the same thing as writing cos(1°) = Re( e^(3°i/3) ) which does not give you an explicit formula (unless you know how to explicitly compute third roots in the complex numbers without using polar coordinates).

[@NeiroYT]

Solving this Re( e^(3deg*i/3) ) with given cos(3deg), sin(3deg) is really looks harder than school stuff given in this video

[@marsgal42]

120 = 2^3 * 3 * 5, so a 120-gon is constructible and the formulas for sin(3) and cos(3) only have square roots in them. 360 = 2^3 * 3^2 * 5. Two factors of 3, so a 360-gon isn't constructible. The final formula for cos(1) has a cube root in it. Neat the way it all fits together.

[@GreatComposer1]

How can I read about the information you just wrote?

[@marsgal42]

@@GreatComposer1 Do a search for "constructible number" or "constructible polygon". See also "Fermat Prime". Gauss (who else?) proved many of the initial results.

[@thomaszhang4765]

@@GreatComposer1You might look in to constructible numbers and compass and straightedge construction. Or if you are wondering about the technique she uses to discern whether a number is constructible or not: she uses the prime factorization of the n-gon to determine whether a number is constructible or not. If any number p other than two occurs multiple times (that is to say has a power) then the solution contains a radical of degree p. This is because constructible numbers can be obtained by the addition, subtraction, multiplication, division, and square root operation.

[@borisjerkunica4442]

love to see the simplified answer once the actual values of sin3 and cos3 are substituted in. Not convinced that its easy to get to an answer where there are no i's. In fact maybe there is no answer that is only based on radicals plus the usual operations.

[@leif1075]

How can there NOT be..cosine of 1 us real, not complex so there must beca way to write without i

[@thebushmaster0544]

ayoooo ive been wondering about this for quite a while but ive never gotten around to doing it! thank you!!!

[@beniborukhov9436]

I really appreciate how you derived the angle formulas you used here. I'd learned them as just facts passed down which made me feel like I didn't really understand why they worked.

[@roberttelarket4934]

0:03 a good place to start.

[@nrrgrdn]

😂

[@iansamir18]

Isn't that last equation just the definition of cosine? (e^{ix} + e^{-ix})/2 and obviously [e^{i*3°}]^(1/3) = e^{i*1°}

[@iansamir18]

I see others have commented the same

[@iansamir18]

That's why there's no cos(x/3) and sin(x/3) formulas, they cannot be derived

[@Alan-zf2tt]

Excellent - very impressed with original triangles dipping into e to the eye theta and back again *** all in less than 23 minutes! *

[@nosuchthing8]

Very close to 1

[@1495978707]

12:40 what makes this triangle special is that the smaller angle is half of the other two which allows for the bisection trick, which are the same, and they have to add up to 180. So the smaller angle is a fifth of 180.

[@juandesalgado]

12:50 For a moment, I thought the Golden Ratio was making a surprise entrance. (It kind of did: x = 1/phi.)

[@vladimir10]

You're right, friend. It sure did, cause we're dealing here with 5-fold symmetry when calculating trig values for 36°. And this symmetry pretty much always leads to the golden ratio in some way. Isn't that the reason for having 5-fingered palm?

[@juandesalgado]

@@vladimir10I don't know about you, but I have opposing thumbs :) (J/K don't kill me)

[@mickkorrawit2386]

You should have substituted radical expression of cos3, sin3 to make the most satisfying result

[@ruilopes6638]

I love that up to the last calculation of cos 1° everything is constructible with compass and straight edge. But then those pesky cube roots stop the magic

[@wes9627]

It is mind boggling having to take cube roots of complex numbers to get real numbers.

[@trueriver1950]

There's a Numberphile video where trisecting an angle is shown to be possible if you are also allowed to fold the paper.

[@ruilopes6638]

@@trueriver1950 yeah, it’s not an impossible construction. There are several always, like neusis, and mechanical methods. But I have a fascination with the simplicity and scope of the classical starlight edge and compass constructions.

[@bethhentges]

At 22:10 he says “This number is close to one-half,” but he meant to say, “This number is close to one.”

[@landsgevaer]

Sneaky. But, if arbitrary roots of complex numbers are allowed, then I could just write cos(1°) = (⁹⁰√i + ⁹⁰√-i) / 2 and call it a day...

[@nHans]

Exactly what I was thinking. If one's going to invoke de Moivre's formula and also allow-as you said-arbitrary roots of complex numbers, one might as well get the final result much more directly: (cos 1° + i sin 1°)^90 = cos 90° + i sin 90° = i => cos 1° + i sin 1° = i^(1/90) (cos 1° - i sin 1°)^90 = cos 90° - i sin 90° = -i => cos 1° - i sin 1° = (-i)^(1/90) and so on, giving the result that you've mentioned.

[@zkolonits]

Great! I just wonder, how to calculate that complex cubic root at the end - without using sin1° ?

[@Anonymous-df8it]

There's an algorithm for cube roots by hand

[@alexandergroysman2357]

cos(18) can not be 5+sqrt(5)/2sqrt(2) because its bigger than 1.

[@namanhnguyen7933]

he missed the square root sign on the numerator

[@lapaget1]

Like several viewers, I liked the first half of the video. However, I was unconvinced with the second half. In real situations, an engineer would use the famous approximation cos x ≈ 1-x*x/2 when x approaches 0. With x=1°=pi/180 radians, it’s much easier and much faster to use this approximation, cos 1° ≈ 1-pi*pi/(2*180*180) to the result with a precision where the first 8 digits are correct.

[@rugbybeef]

Wow, I 𝘳𝘦𝘢𝘭𝘭𝘺 wish my geometry teacher had gone through this example. The more linear algebra oriented strategy where you use the configuration to derive the identities finally lets me comprehend these relationship. I took math through Calc IV and have math minor. I work in biostatistics professionally and literally do statistical programming for a living. I am far from math averse. I still hated geometry and resented it as a subject. I struggled with proofs and completeness. While I understood the principles, I couldn't rote memorize the identities effectively and did poorly. In calculus the course was built up systematically and allows me to use its reasoning and principles daily. These examples are revealing geometry. There is in fact a way for me to understand it.

[@AtariDays80]

The other 2 solutions (around -1/2) are 121 degrees and 241 degrees.

[@SimonClarkstone]

I had hoped that was the case. They are the other things that triple to 3 degrees.

[@hasanjakir360]

At 22:00, cos3° + isin3° = e^(iπ/60) cos3° - isin3° = e^(-iπ/60) cube root of these would be e^(iπ/20) and e^(-iπ/20) They expand and add up to 2cos20, so x is cos20 which is cos(3600/π)° which is not cos1°.

[@Phylaetra]

Interestingly, the cosine (and sine) of 360/17 degrees also has an exact expression (though much more complex). As do the angles for any of the constructable n-gons. Now - this is a bit of a pain to go through, but I appreciate working out exact values. I once tried constructing a table and it's pretty easy to get the principal angles of : 15, 18, 30, 36, 45, 54, 60, 72, 75, and 90. But - once you start looking at differences and sums (to get say 3 from 18-15 and 6 from 36-30 etc.) the expressions become very messy and it's very difficult to get through it all and be sure that you didn't make a mistake somewhere in your calculation. This led me to wonder just how tables were constructed - and it is a cool process : (1) calculate the sine and cosine for the smallest angle you will use, say 1 degree or 15 minutes or whatever. This is done by taking the appropriate radian measure and using the first couple or three terms of the Taylor series. For such a small angle, the series converges very quickly and you can easily get 10 decimal places of precision. (2) Starting with the sum formula: sin( a + b ) = sin(a)cos(b) + cos(a)sin(b) sin( a - b ) = sin(a)cos(b) - cos(a)sin(b) So: sin( a + b ) + sin( a - b ) = 2sin(a)cos(b) : or sin( a + b ) = 2sin(a)cos(b) - sin( a - b ) Setting b equal to our increment, say 15', we get: sin( a + 15' ) = 2sin(a)cos(15') - sin( a - 15' ) Notice that we are multiplying by 2cos(15') each time; sin(a) and sin(a-15') are the two previous entries in the table, we have already calculated them. For cosine (derived similarly) the formula is: cos( a + 15' ) = 2cos(a)cos(15') - cos( a - 15' ) Note - this does create a cumulative error as you go through the process (you must have some error since you are truncating or rounding at some point). It's good to be able to check with the exact values that you can calculate in another way and 'reset' your table values. you can also work backwards (towards 0 from 15 degrees, say) then your error grows in the opposite direction and you can get an idea for how far off you are (or do more complex error analysis). it's all fun though!

[@hunterlouscher9245]

Was bummed but unsurprised that you had to break out Euler for the trisected angle formula. I always use Euler to find/tailor whatever angle formula I need, but sometimes I feel a little bit of shame relying on pure algebra to do what the Greeks could do drawing circles in the sand and "truly" having an intuition for it.

[@trueriver1950]

Using the triple angle formula, you could get directly from cos 45 to cos 15. However that will not have the handy short cut where Michael took the root close to one.

[@leif1075]

Yea isn't using Eulers formula overkill..youndont even need triple abglem formula at all right..just rewrite 3 degrees as 1 plus 2 and then rewrite 2 as 1 plus 1 and solve for 1 plus (1+1) degrees using just double angle formula ..see what I mean?

[@Alo762]

Michael, you should really do this again because the cos(18) error. It messes everything else after it.

[@hacatu]

Surprisingly, exact representations actually have some applications ... well kind of. I spent a long time trying to work out how to exactly represent sin and cos of 1 degree when solving project euler problem #180. Ultimately, I found that just representing sin and cos of integer numbers of degrees as polynomials modulo the minimal polynomial for e^i(pi/180) was the best approach. If f(x) is this minimal polynomial and a is e^i(pi/180), then we can represent say cos of 1 degree as (x + 1/x)/2, since e^it = cos t + isin t and e^-it = 1/e^it. Note that we can reduce (x + 1/x)/2 mod f(x) to a polynomial, but it would have degree 47 so I won't write it out. But when computing with these numbers we would have to have the computer write it out. Plugging in e^i(pi/180) for x would give us the numerical value for cos of 1 degree

[@makmidov]

this channel is jewel on utube. I learned so many new things from prof. Penn!

[@koendos3]

Great video Michael! It felt like a true journey.

[@ed.puckett]

Thank you for your videos. I appreciate your thoroughness of explanation, and the interesting topics.

[@bot24032]

you could get that last part much faster if you use cos1°= 1/2(e^(iπ/180)+e^(-iπ/180)) and express the two e^... expressions as (e^(3...))^(1/3)

[@The_Commandblock]

bro be using radians and degrees in the same equation

[@bot24032]

@@The_Commandblock do you want me to put degrees into the exponent instead

[@a52productions]

I always wondered how they made tables of sines and cosines before Taylor series and complex exponentials were well-known. This answers that nicely! (With the exception of the triple angle formula, but you can also derive that using angle-sum formulas instead of Euler's formula.) While others are right that exponentials or approximations would lead to the same answer much faster, I think it's interesting and informative to do things the old-fashioned way once-in-a-while. I've learned a lot from this!

[@wes9627]

We can't take cube roots of most complex numbers without using Euler's formula, and in this case we have to know cos1° and sin1° in order to compute cos1° and sin1° from the cube root equations. We have gained nothing.

[@neilreynolds3858]

The table of trig functions in Ptolemy are chords. They didn't have sines or cosines or any of the others back then. I like finding rational values for chords. The one for the heptadecagon is a bit messy.

[@juanpablosimonetti147]

Me encantó la demostración inicial!!! La voy a usar en mis clases, muchas gracias.

[@goodplacetostop2973]

22:14

[@reubenmanzo2054]

Sin and cos of 3 degrees could be simplified to make the numbers a little bit nicer. When simplified, we get: sin(3)={[sqrt(3)+1]sqrt[3-sqrt(5)]-[sqrt(3)-1]sqrt[5+sqrt(5)]}/8 cos(3)={[sqrt(3)+1]sqrt[5+sqrt(5)]-[sqrt(3)-1]sqrt[3-sqrt(5)]}/8

[@jarikosonen4079]

Very well explained.

[@reeb3687]

for the function f(n) = sin(π/n), lim k->+inf f(k)/f(2k) -> 2

[@huguesbornet1211]

To avoid the many radicals in the cubic equation’s real solutions, one can always use Vieta’s strategy: normalize any equation to X^3+3/4 X +Q =0 The 3/4 allow precisely the use of all trig identities for triple angle. No radicals but reciprocal functions of sin, sinh, cos will appear.

[@TomFarrell-p9z]

Good video. Now I wonder: Can you generate the trig tables, including for cos(1 deg), without using complex numbers? (In my high school math curriculum, about 50 years ago, we learned trig before an introduction to complex numbers.)

[@landsgevaer]

And the answer to that is no. If you doubt that, we should have a duel. 😉 Unless you introduce special functions of course. Like cos. Commonly, we express cubic roots in terms of trigonometric functions, not the other way round.

[@charleyhoward4594]

this guy's tenacity never fails to amaze

[@bobbyheffley4955]

A cubic equation with three real and irrational roots will have its solutions expressed with irreducible cube roots of complex numbers.

[@sleha4106]

how can you say that

[@felipe-n9t]

Using a degree two Taylor polynomia(1-x²/2) to approximate the value much more easily, you get a result that is off by 0,000000386% of the original value, the first 9 decimal places are the same in both approaches.

[@Phylaetra]

Is that an exact value though? :)

[@Phylaetra]

Even closer for the sin - and there are other approaches to use if you want to create a table, though I think it is useful to know the exact values for major angles.

[@neilreynolds3858]

I got it to 18 places using excel. That should be sufficient for most calculations but they must have been using log tables to 20 places in France around 1810. That's impressive when you know everything was generated by hand.

[@CommanderdMtllca]

Hey I have a potentially dumb question: why don’t you do the checklist of proofs like you did 2ish+ years ago? For someone who stopped at Calc III, this level of detail was always incredibly helpful.

[@Happy_Abe]

Couldn’t we use the triple angle formula to find cos(45) from cos(15) too?

[@trueriver1950]

In principle, but you don't get the nifty short cut off knowing that the error of that equation is near one. And it's a cubic so it's less effort to stick with quadratics as long as possible

[@FadkinsDiet]

Why is the fiest part done geometrically, but the last part is done with complex numbers and algebra? Is it because of the same reason my geom4teacher told me that there is no formula for sine and cosine of 1/3 of an angle?

[@florisv559]

Yes. It was proved quite a while ago that ruler and compasses constructions can always be translated into at most quadratic equations. Trisection of an angle OTOH is equivalent to solving a cubic equation with three different real roots, and the general formula for that, found by Tartaglia, gives cubic roots of a complex number. The only way to solve this is by using the trigonometric formula that you started with. It's like trying to calculate the square root of 2 by just dividing 2 by some number - only, you have to divide by the number that you are trying to find.

[@garyknight8966]

The tangent of 3 degrees in terms of phi=golden ratio is sqrt(4-p*sqrt 3-sqrt(3-phi) / sqrt(ditto with + signs) .

[@garyknight8966]

,, the p*sqrt 3 there means phi*sqrt(3)

[@denelson83]

13:43 - Error. That value on the right is the _square_ of cos 18°.

[@gp-ht7ug]

A bit chaotic at the end but very interesting

[@jdsahr]

fantastic proof the angle sum and difference formulas! Whatever I saw in highschool 48 years ago was forgettable --- this one will stick.

[@tiborgrun6963]

Looking at cos(15°) and sin(15°) I wonder: Is there some kind of field extension of the rationals where going from (sqrt(6)+sqrt(2))/4 to (sqrt(6)-sqrt(2))/4 is some kind of conjugation? And how would that conjugation be defined when there are multiple square roots?

[@Qermaq]

Here's a problem you may like. Start with a (regular, throughout) 2-simplex, or an equilateral triangle. A 2-simplex has facets that are 1-simplices (line segments). Assume all edges of all the n-simplices are of identical length throughout. Let a be the distance from the center of a facet (vertex in this case) of a 1-simplex to its center, and let b the distance from the center of a facet (edge in this case) of a 2-simplex to its center. If we then make this 2-simplex into a 3-simplex we find c is the distance between the center of a facet (face) to the center of the 3-simplex. Etc. etc. For a 10-simplex, express j in terms of i.

[@rugbybeef]

Hey, it would be helpful if you were to use YayText to italicize your variables if you are using 𝘢 and 𝘣. As they are now, they blend confusingly into the text looking like the syntactic determiner "a". It left me asking "a what?" Or alternatively, if you dont want to use YayText or UnicodePad for italics, x and y are much less ambiguous when used in plain text. Respectfully, just a suggestion. 😄

[@rugbybeef]

Given that you are using 𝑛-simplices and also that you pluralize simplex as simplices, I would strongly suggest UnicodePad which provides access to all the mathematical symbols available in the full Unicode glyph set. You can set favorites to easily copy and paste as I suspect this is not your only YT comment involving things like ∠𝘢𝘣𝘤₁₋₁₀ etc

[@ouwebrood497]

Wow, I didn't realize it is that easy to prove the cos and sin of difference angles.

[@jamesfortune243]

Excellent and didactic content. Proposed problem: What kind of shape is the steady state U shape carved out by the perpendicular cross section of a glacier?

[@Qermaq]

Michael, thanks for doing geometry and trig videos. I don't know why people don't like them but I sure appreciate it.

[@itzmrinyy7484]

cosine of 18 degrees here at 14:23 is a bit off, there should be a square root over the numerator

[@channelbuattv]

sin 1° and cos 1° are algebraic (they are roots to polynomials of degree 48)

[@nHans]

As an engineer, I'm very happy to see a math professor do trig calculations using degrees. 😎 Everybody-including engineers, architects, machinists, carpenters, high school students and their geometry teachers-uses degrees for practical tasks like measuring and drawing angles. Ironically, for doing trig calculations, I've always used radians-except for really trivial calculations involving the basic trig identities. In fact, radians are mandatory, particularly when θ (or x), the variable denoting the angle, appears both inside and outside trig functions. For example: Taylor/Maclaurin series, Euler's formula, limits, sinc function etc.

[@pierreabbat6157]

I'm a surveyor, and I write angles in degrees, minutes, and seconds on plats, but in software, I compute angles as 2^-31 turn. Why not 2^-32 turn? Two reasons: the delta (surveying jargon for subtended angle) of an arc ranges from (-360°,360°) which my software represents as [-2147483647,2147483647], and the same type also represents area on a sphere. However, this unit is about 18 mm on the surface of the earth, so for lat/lon, I use radians internally, using 8-byte floats.

[@nHans]

​@@pierreabbat6157 Fascinating. We build tools to assist us-in this case, digital computers-but end up adapting to their limitations. I take it, however, that quantizing delta the way your software does is not a universal practice across surveyors? Presumably, others could be using 64-bit signed integers, thereby achieving much higher precision? Out of morbid curiosity-how does your software behave when presented with a delta of -2,147,483,648?

[@trueriver1950]

Meteorologists use degrees too 😅

[@Phylaetra]

Some European countries use the grad (1/100th of a right angle) as part of the decimalization left over from the French Revolution. In the military, we used the 'mil', which is 1/1600th of a right angle - chosen because it is very close to 1 milliradian, and gives us the very useful 1 meter subtends 1 mil at 1 kilometer. For DMS, you can use 1 inch subtends 1 minute at 100 yards or 1 foot at 1200 yards.

[@pierreabbat6157]

@@nHans Depends on what you're asking it to do. If you ask for the sine or cosine, it'll just give it to you. If you ask for the offset of the midpoint of the arc, that's infinity. To find out how it'd draw such an arc, I'd have to try it. Such an arc looks like an infinitely long line with a segment removed. Segments, arcs, and spiralarcs (arcs of Euler spirals, which are used for highway and parkway design and in Bezitopo for drawing contours) are stored as start point and end point, delta(s), and other data (they also have vertical curves, which are low-degree polynomials).

[@arthur.v.babayan]

SuppaCool explanation :)!!!

[@Alan-zf2tt]

Just wondering ... what differences are between, say, this way of obtaining a trig function value and, say, a limited finite and/or infinite way using, say, Maclaurin series or other approximation methods. Perhaps on aspects of computational ease, accuracy when these sort of things are done by hand rather than by calculator or computer?

[@marsgal42]

This result is mathematically exact. A Maclaurin series can tell you, for example, that sin(45) is about 0.707. It can't tell you that it's exactly 1/sqrt(2) with all the insight that follows.

[@Alan-zf2tt]

@@marsgal42 yes I understand this but I was talking to an engineer and he said one over root two means nothing to him. His instrumentation can accept numbers up to so many decimal places. It got me wondering how did people make such numerical methods practical with sufficient accuracy to be doable. Thank you for your reply

[@isaaclearningtominecraft4751]

Wait... Shouldn't an arbitrary numbered root of a complex number be forbidden in a "closed for of sine certain degrees"? Otherwise why we are not just saying that the answer is 30th root of 1/2+sqrt(3) i/2 with imaginary part depressed by averaging with its conjugate?

[@jimschneider799]

@14:00: 5 + sqrt(5) > 5, and 2*sqrt(2) < 4, so (5 + sqrt(5))/(2*sqrt(2)) > 5/4 > 1, so this cannot be the cosine of a real number. The correct answer is sqrt(5 + sqrt(5))/(2*sqrt(2))

[@miguels1271]

Awesome video! I guess this way of thinking (heuristic, geometric) was the only way available before calculus and Taylor series were invented/discovered.

[@nothingbutmathproofs7150]

Dr Penn, there is no x, so how can you set x = cos 1?????

[@iradeourum]

What is "haypanjus"?

[@Ed1414One]

Can you do sin 5°

[@mtiganik]

I liked the end of the video )

[@Bodyknock]

An interesting possible variant way to get from 1° to 3° might be to set x=sin(1°) and note that then sin(-1°) = -x since sin is an odd function. Likewise, if you set y=cos(1°), then cos(-1°) = y as well since cos is an even function. From there though you can note that sin(3°) = sin(2° - (-1°)) and cos(3°) = cos(2° - (-1°)). If you then expand those both out you get: sin(3°) = sin(2°)cos(-1°) - cos(2°)sin(-1°) = sin(2°)cos(1°) + cos(2°)sin(1°) cos(3°) = sin(2°)sin(-1°) + cos(2°)cos(-1°) = cos(2°)cos(1°) - sin(2°)sin(1°) But now notice that sin(2°) and cos(2°) in those formulas can both also be reduced to formulas involving just sin(1°) and cos(1°) using the well known formulas for sin(2x) and cos(2x): sin(2°) = 2sin(1°)cos(1°) cos(2°) = 2cos(1°)² - 1 Which means you can replace reduce all the occurrences of 2° in the formulas above with simple functions using 1° instead, leaving you with sin(3°) and cos(3°) both equal to functions only involving sin(1°) and cos(1°). But as per the video you can already derive closed forms for sin(3°) and cos(3°), meaning you've got two equations with two unknowns (namely sin(1°) and cos(1°)) and so you can solve for both. I'm not going to bother working on expanding everything out here in this comment, I'll just note that it looks doable from here and I don't think the solution would be more "complicated" than the final one in the video (they're both kind of gnarly looking in the end). But I did think it was interesting to note that, by making use of -1° in those difference formulas and taking advantage of sin and cos being odd and even functions you can sort of "build up" to higher integer multiples of 1°.

[@dalex641]

You still will end up with cubic equation.

[@Bodyknock]

@@dalex641No doubt, but you get there along a different pathway. (In fact whatever answer you get must be equal to the one in the video.)

[@ceratos9588]

Cos(1) is not algebraic. I dont know what you are suggesting at the end, but there is no such thing as a closed formula for cos(1).

[@Ensign_Cthulhu]

"...and that's a good place to stop, UNLESS you are building sine and cosine tables to many significant figures, in which case the number-crunching has just begun." So how did they handle that back then?

[@neilreynolds3858]

Very carefully? I've done some of it by hand just to see what it was like. You quickly go blind and get cabin fever. I collect old books of tables that were done by hand just to remind myself how easy we have it.

[@iradeourum]

beita ahahah. look up the transcription.

[@avaraportti1873]

Degrees are a really arbitrary unit, aren't they

[@nHans]

True. From the standpoint of pure math, the degree-which is 1/360th of one rotation, or equivalently, 1/90th of a right angle-is an arbitrary unit to measure angles. There is nothing special about those fractions. After all, pure math doesn't involve itself with the rotation of the Earth around the Sun. In fact, even the everyday units that we use for measuring physical quantities are arbitrary from the purely scientific perspective. But that's a good thing. It means that long before math and science existed as fields of study, practical people went ahead and invented numbers and units, and started _counting_ and _measuring_ instead of merely guessing. Paraphrasing Lord Kelvin, that's how science got its start. We wouldn't have the radian without the degree. We wouldn't have the Planck Length without the cubit.

[@neilreynolds3858]

I blame it on the Babylonians.

[@ivanlazaro7444]

I do notunderstand the last part. How the rest of the roots are complex (the 1st one but multiplied by a root of unity) if he said before that the polynomial has 3 real roots?

[@landsgevaer]

The other roots definitely do not equal the first root multiplied by a cube root of unity. The cubic roots in the final expression would each pick up such a factor, but their imaginary part would then cancel when adding them.

[@adamnevraumont4027]

It is clearly sqrt(1 - 1 degree^2) as sin x is x.

[@nosuchthing8]

A string theorist would demand a billion bucks, 20 years of tests in a particle accelerator, then conclude we can't say because who knows how many dimensions of space and time exist, and the unknown curvature in the unknown calabi yau space makes it unknown.

[@gamingzeraora443]

approximately 1

[@bethhentges]

At 19:10 he repeatedly says “theta” when he means “degrees.”

[@Kavukamari]

this exercise could be an entire test

[@marc-andredesrosiers523]

Important reminder 🙂

[@demenion3521]

from a physicist's point of view cos(1°)=1-π²/180²/2 up to an error of less than 10^(-8)

[@guerom00]

Does that mean there's a closed formula for all integer number of degrees? If we have 1°, we can use a formula for double angle, triple angle, quadruple angle, etc... via Euler's formula?

[@trucid2]

There's probably a closed form solution to any rational multiple of 360 degrees.

[@punditgi]

This is math from the ground up 😊

[@mcalkis5771]

It's a pity that there's no closed form of this expression that doesn't contain the imaginary unit.

[@mrosskne]

just use a ruler and a protractor.

[@gibbogle]

cos(18) > cos(15)??

[@ВикторПоплевко-е2т]

you didn't put the degrees symbol

[@philstubblefield]

The derivation that Michael mentioned: _A much more friendly approach to the cubic formula_ (kzbin.info/www/bejne/hX-VeoB7eMSjiqc)

[@Afra-p9l3]

Please give it a try : N=12345678910111213......2024.(here 1 to 2024 are written in a row).Now the question is what is the remainder if N is divided by 2025?

[@jfpeltier]

Is the answer 324 ?

[@rainerzufall42]

sqrt(2) was written sqrt(5) twice...

[@Hakuna_Mataha]

Of course, the golden ratio popped up along the way.

[@sashagornostay2188]

I did it a lot back