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Without calculation, the engineer will say 1.

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.

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

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

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.

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.

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°

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

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

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.

0:03 a good place to start.

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

Great video Michael! It felt like a true journey.

Just beautiful....now do it from 1° to 90°

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...

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.

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

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

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

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

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.

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.

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.

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

Very well explained.

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).

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.

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

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.

this guy's tenacity never fails to amaze

22:14

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

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

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.

A bit chaotic at the end but very interesting

SuppaCool explanation :)!!!

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

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.

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

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!

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.

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.

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.

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!

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

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

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

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

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.)

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

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?

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?

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

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)

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?

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

I liked the end of the video )

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

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

this exercise could be an entire test

Important reminder 🙂

finally a video i can understand 😭😭😭

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

Degrees are a really arbitrary unit, aren't they

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

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

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

beita ahahah. look up the transcription.

"...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?

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.

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°}

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).

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?

What is "haypanjus"?

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?

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

Can you do sin 5°

This is math from the ground up 😊

The derivation that Michael mentioned: _A much more friendly approach to the cubic formula_ (kzfaq.info/get/bejne/hbRifLB2qMXUhHk.html)

Wow , real beaut

A wedding cake.

approximately 1

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

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.

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

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?

I did it a lot back

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

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.

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

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°.

you didn't put the degrees symbol

@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))

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

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

huh? isnt cos(a-b) supposed to be cos a cos b + sin a sin b, i know it's adding them together here so it dosen't matter but still

I have a cosine. he lives in norway.

just use a ruler and a protractor.

Meu Deus, tava tão feliz entendendo tudo até chegar na parte que aparece e^i kkkkkk