Math for fun, sin(sin(z))=1
ฝัง
- เผยแพร่เมื่อ 8 ก.พ. 2025
- The trigonometric equation sin(z)=1 is fairly easy to solve but not sin(sin(z))=1. Here we will be using the complex exponential definition of sine, which is from Euler's formula e^(i*theta)=cos(theta)+i*sin(theta), to solve this equation. We will see sin(sinz)=1 actually has infinitely many complex solutions, just like how we solve sin(z)=2.
sin(z)=2 • Math for fun, sin(z)=2 got over 1M views recently. Thank you all! Enjoy solving sin(sin(z))=1
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blackpenredpen
I was so focussed on the board I didn't see the shirt until I saw a comment 😂Thanks for wearing it!
Tibees gang
😃 😃
😂
😂😂😂
Super 😂💞
You know its terrifying when he giggles time to time.
India: We use colors as variables.
Arabia: Well, we don't really want to mess with different pigments while doing math. We're just going to use letters.
Europe: The letters are not Christian enough, we will use our own latin and greek letters.
Blackpenredpen: 🙂
Europeans are the pioneers and father of maths . LETS GO EUROPE!!!
@@chronicsnail6675 stfu
@@chronicsnail6675 says the guy who uses indo-arabic numerals
@@chronicsnail6675 A stunning display of ignorance.
@@adrianfrauca8118 and?
For happy face^2 it should've been drawn as an actual square face, for the true immersion.
Wow! Didn’t think of it. Nice one.
@@blackpenredpen HOW sin(sin(sin(sin(sin(sin(...))))))=1 ???
what about the cartesian square of a happy face?
happy face^3:
a cube with happy face on every side
Square root of square face is happy face
My guy here rockin' that Tibees merch while solving these unholy equations
"This looks... yeah." Sums it up quite nicely.
You know its worse when he has more than 2 pens
Algebra: letters as variables
Trigonometric algebra: latin letters as variables
Blackpenredpen algebra: emojis
The video on sin(z)=2 was the first video I saw from your channel! I've been following you since that video was uploaded ;)
Thank you!
I love how you use basic examples to train all the little concerns and caveats that must be observed when solving a specific class of problems. Very effective and fun to watch, especially with emoji substitution 😄
6:15
and that’s why most people don’t have happy faces when they do maths
6:11*
Lol
7:05 Actually, ln(x+sqrt(x^2-1)) = arccosh(x), which can further simply the answer.
This only simplifies one branch of the answer, though.
Probably another branch ln(x-sqrt(x^2-1)) can be written as ln(-1)+arccosh(-x), but the domain might be tricky
actually, you can simply extract the +/- sign outside the ln. ln(x - sqrt(x^2 - 1)) = ln[(x - sqrt(x^2 - 1))(x + sqrt(x^2 - 1)) / (x + sqrt(x^2 - 1))] = ln[(x^2 - (x^2 - 1))/(x + sqrt(x^2 - 1))] = ln[1/(x + sqrt(x^2 - 1))] = - ln(x + sqrt(x^2 - 1)). So the +/- can be extracted out of the ln. So in the end it would be +/- arccosh(x).
We don't allow trig functions in the simplification... Otherwise the answer to the whole problem can be trivially reduced at step 2 to "z = arcsin(pi/2+2npi)"
@@spencergrogin1074except the input is outside the domain of arcsin
I’m surprised how I’m slowly starting to understand these types of videos as I learn. I still remember how I would not understand any statements in these videos a few years ago.
Though he is doing something very wrong, it seems reasonable
@@karryy01 what he did that was wrog?
Me: using letters as variable
Him:😃
Happy face, meet fish. Fish, meet happy face.
This is completely unrelated, but I was trying to figure out transistors earlier today since one of the bonus problems in my principles of electrical engineering textbook had them in an example of a monostable vibrator (I havent exactly seen a transistor before in problems... or real life... not even sure why it brought them up because the problems were about basics of DC RC circuits)
But apparently the way to calculate the voltage across transistors uses the Lambert-W function and I thought back to your "fish" videos you did on the Lambert-W function. Honestly I didn't know it had much use out of "math for fun".
Wow! I didn’t know!
2:52 "I'm just going to put down a happy face....because a fish is too difficult."
🤔 but a fish can be represented with just one letter: 𝛼
Surprised?
Nope got used to it
yes :D
Well well
@2C (02) Chan Kwan Yu
This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
I think it'd be cool if you quickly calculated the solution for n,m=0, to show what value for z that would present. I absolutely love these complicated problems that you keep showing!! Any plans to consider more AIME or even USAMO/IMO videos?
I wish I could see more videos like this. Complex numbers are my specialty, and I'd love to learn more about them.
Just recently saw that sin(z)=2 video of urs.Amazing. Love ur videos🤗
4:41 “Alright so it looks.... yeah”😭
your beard looks like a perfect binary tree
"We'll have to go to the complex world"
*[screams in agony]*
The complex realm is called complex for a reason. 😃
I remember u used this technique/Other way to write arcsin in ur sin(?)=2 vid. Amazing
Yea. This is a continuation video and also a little celebration (since sinz=2 got over 1M views recently).
@@blackpenredpenI was a follower of your channel since then I think, I really like the content of yours man.
Keep it up!
@@blackpenredpen
*BPRP please please please read this comment.*
Your videos are very amazing. I have a request, can you please please please make a video on what I have derived. I have derived a formula for sin inverse of x. The proof is as follow:
y=sin^-1(x)
sin(y)=x
e^(iy)-e^-(iy)=2ix
(e^(iy))^2-2ixe^(iy)-1=0
Using quadratic formula:
e^(iy)= ix+-√(1-x^2)
y= -iln(ix+-√(-(x^2-1))
y= -iln(i(x+-√(x^2-1)))
Using ln(ab)=ln(a)+ln(b)
y= -i(ln(i))-i(ln(x+-√(x^2-1)))
sin^-1(x)= π/2 - iln(x+-√(x^2-1))
To check this formula put x=2 and you will get:
sin^-1(2)=π/2-iln(2+-√3)
You have proved that sin(π/2-iln(2+-√3)=2 in one of your previous videos.
I also request you to put sin^-1(x)=π/180 and put formula of sin^-1(x) which I derived and solve for x so we will get value of sin(1°) or sin(π/180), I had tried to find value of sin(1) this way but I failed.
I hope you will make a video on this formula.
My name is Kathan Parikh and I am 16 years old.
And if you want one more golden equation which includes Phi,π,i,e and even Fibonacci series(All five in one equation) then just reply me so I will give my phone number and you can call me as it is difficult to type the equation, so I will be easily able explain the equation and it's proof to you by sending you a pic or on call.
@2C (02) Chan Kwan Yu
This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
6:10, there are like 26 letters in english alphabet, around a infinite many symbols and other stuff, and he choose a happy face 🤣🤣🤣🤣🤣
This is what I call fun! From now on I'll use pi*(4n+1)/2 instead of =)
Great! [ π(4n+1)/2, m є Z, n є Z ] is funnier (idk why I used [] )
z = arcsin(π/2)
Finally, emojis in math
x/0 should be the poop emoji :-)
Yhank you for that video. It was quite interesting! Even the solution was a little bit too "complex" :-)
I really love complex formula for sin(z), really fun to do
8:16 He's so done with it, lmao
The videos on the shorts channel (bprp fast) are so fast that seeing you teach this at normal pace feels very strange .
Two..?
Two...
And then it got intense when another pen approached the board
Now this video has 100k views. So do another video like this
Wow. Gears grinding on this one! Chapeaux.
The simplest case n=m=0 gives rather nice and tidy solution Z = π/2 - i*ln(π/2 ± √(π²/4-1)) which is quite close to just π/2-i
ah yes me at late in the evening pretending to understand these kind of maths
Actually it was basic
a little late but in your final formula you do not treat the case when inside the "ln" the value is negative (ln of a negative value is undefined). This case happens for n
ln of a negative value is just ln(abs(x))+i*pi it isn't undefined
@@jsjsjjsud9556 the equation exp(z)=-x has an infinite solutions : ln(abs(-x))+(2k+1)*i*pi with k integer. So there is no clear extension of the ln function on the real negative. Just writing ln(abs(-x))+i*pi is a shortcut and there are interesting exercises using this abuse to lure the reader
Well, that was my guess to begin with. It's quite intuitive.
Ah ! With this t-shirt, I finally understand : you want a beard as long as Tibee's hair !
😆
Fun fact: my beard is growing at a logarithmic rate.
@@blackpenredpen legend says if you live to infinity years old your beard size will approach a constant called pen’s constant
The infamous C and R axes
😂
Love the Tibees shirt!!
Thanks, Carter.
Can you do IITJEE math questions? Those are terrifying when you read them but are fun to solve and give you tons of views
"Technically, should have written pi m." ...
pi m...
as in... Dr Peyam?
XD
That’s deep
LMAO that's not anything new...!
z is Arg, so we can z+ 2πl is the answer (l is integer)
I understood all of that and was able to follow along, but, if I were given that problem to solve from scratch I'd not have a chance.
Loving these Math Content 😊
The sin(?)=2 vid is such a good video, lol
Love the Tibees shirt!
BPRP: 2:58
also BPRP: 3:19
I think it would have been a lot better if you cancelled out the 1/2 before substituting :)
I have been subscribing u for 3 years , providing that there are infinite unsolved questions and I am slightly less than being as good as u so I should create a math channel on TH-cam, should not I?
You should surely I will support it
You should
If you have nothing better to do and are confident in your teaching skills, then go for it! :D
10:39: He: Very Nice
Me: 😫
Fun fact: complex nunbers are not taught anymore in greek high schools
@@tzonic8655
Unfortunately 😪😪😪
I am greek and I indeed was not taught complex numbers because when I was at the last grade of high school, complex numbers had stopped being taught already for 2 years... However thanks to uni and youtube videos I think I have a decent understanding of complex numbers!
@@geosalatast5715
Είναι κρίμα Ένας τόσο ωραίος τομέας των μαθηματικών να διδάσκεται μόνο στο πανεπιστήμιο...
@@geosalatast5715 yeap,m2! Next semester i have to choose between discrete math or arithmetic analysis(αριθμητική ανάλυση δεν ξέρω αν είναι έτσι στα αγγλικά) or complex analysis.complex analysis looks so interesting but I'm not sure yet
Does this have more solutions than sin(z)=2 or its the same infinite?
Good question! I believe they are both “countable” infinity.
I'm so happy the only part I didn't get was the intrusion of (e^iz - e^-iz)/2i and of course the ln log, my weakness.
e^(i·z) = cos(z) + i·sin(z) for all complex z. If you substitute z |-> -z, you derive e^(-i·z) = cos(z) - i·sin(z). Subtract the second equation from the first, and this results in 2·i·sin(z) = e^(i·z) - e^(-i·z). Since 2·i is not equal to 0, you can divide, obtaining sin(z) = (e^(i·z) - e^(-i·z))/(2·i), and this is where that substitution in the video came from. In fact, this formula is often taken to be the definition of sin(z) for all complex z.
@@angelmendez-rivera351 Oh, so that's where it comes from, thanks for breaking it down for me
How about:
sqrt(happyface^2-1)=sqrt(sin(z)^2-1)=sqrt(-cos(z)^2)=sqrt(i^2cos(z)^2)=i*cos(z) then rewrite in exp form and then try to workout z. Note this was just a quick thought. Probably doesnt take some stuff in to account like multiple branches of solutions. However I though perhaps there was a way to get a bit nicer form. Not sure though, didnt work anything out. The form and considering trig identities just made me think of this.
8:15 lol
Why don't you use the simplified quadratic formula for even b coefficient?
i love your t-shirt!! tibees
U got me at the first minute not gonna lie
Holy moly! I thought this was impossible.
Simple sin x ~= x so you have sin(x) which is also x so x=1
Cqfd
So it's like a warped 2d lattice of points in the complex plane!
6:12
Regret at its peak
It was a bit messy ... but i understood... thanks again !
inb4 this question comes out for my finals in 3 weeks
I don't even know the first step but it's seems like very cool solutions so here's a like
Something very serious is going on when blue pen is involved.
One can only imagine
God I am such a nerd for enjoying this, but non-nerds will never see the beauty in this accomplishment.
What about sin(sin(sin z))=1?
No, thanks lol
Just so that we're clear, the sin(z) here (and the related cos(z) which can be defined similarly) are NOT the usual sin and cos with real domains. For example we know the usual sin and cos are bounded but this is not the case with sin and cos with complex domains. As they have the same name, it might mislead some. Sure, sin(z) and cos(z) act as the usual sin and cos for real inputs but that stops when the inputs have imaginary parts.
When sin(z) and cos(z) are defined like this, it pays to look at their series (which is their actual definition as functions with complex domains and ranges anyway because exp(z) is just a short way of writing Σz^n/n! )
When we write their serieses, they look the Taylor serieses of the real sin and cos but with z instead of x. This is what makes all the difference. Since we are defining sin(z) or cos(z) as serieses (by writing them in terms of exp(z)), we have to rederive all the usual formulae like sin(A+B) and cos(A+B) like we would with the real series. I can't emphasize this enough, with different domains come different properties. If two different domains have similar properties then that's good but that shouldn't be taken for granted. Not all properties cross over to different domains necessarily.
It's the same for e^(imaginary). This means nothing in the usual sense. What we're doing here is taking the real series definition of exp(x) and extending it to the complex numbers. exp(x) which is also written as e^x has nothing to do with exp(z). So trying to make sense of e^z is futile because exp(z) is actually Σz^n/n! from where we derive properties like exp(A+B)=exp(A)*exp(B) for complex inputs. Just so that we're driving this point home, have a look at this Dr. Peyam video th-cam.com/video/l61k97RFsfs/w-d-xo.html
Does e^(Matrix) make any sense? No it clearly doesn't. This is the case with e^z. Whether the input be real, complex or even a matrix, exp(something) actually is Σ(something)^n/n! if "something" has a notion of being raised to natural powers and can be divided by scalars. When it does satisfies these requirement, claiming e^(something) is the same as e^x doesn't hold. This should be evidently clear when we take somethings to be matrices
I understand your point, but saying e^z has no meaning in the usual sense is silly, because there is no "usual sense" in the first place: we work with R and C equally as commonly in mathematics. Besides, I would argue that before we even get there, we need to define what e^x even means for real irrational x, because technically, this also has no meaning a priori.
@@angelmendez-rivera351 Yes you're correct. It's that _you, me and BPRP_ know what e^z really means but the average viewer might not. I didn't for example. When I learnt e^iπ is -1, I chalked it up to mathematical mysticism and went on with my life without ever learning why this was the case until 3B1B started his lockdown course where he addressed this issue. This is anecdotal but I'm pretty this is true for many. It shouldn't be binding but I think with a viewership as big as BPRP, there comes a certain responsibility of making sure your viewers know what stuff is without going too much into the details (which would be more befitting of an complex analysis course anyway)
Again I agree with you but we already know this which is why we find this trivial and frankly, annoying. This isn't true of first-time viewers however. For them, rationals and irrationals rank more or less the same in terms of their mysticism (a few cranks claim irrational numbers aren't a thing but that's pretty much it) but complex stuff isn't "clear" (clear as in they don't know how to make sense of something^i and so on) to them as they're used to R. The issue here isn't one of math as it is one of communication. I don't think everything is being communicated clearly here which is why I decided to write the original comment
@@pbj4184 You are right. It is a communication issue, and an issue with how the subject is taught. Personally, I am an advocate of the 3b1b philosophy, that the notation for real-valued exponents and complex-valued exponents should instead use the notation exp(r·t) instead of e^(r·t). I think this would help reduce confusion, because it would more clearly communicate that the "exponential function" is really a function that agrees with functions of the form e^q for rational values q, but which actually operate on an extended domain, and whose definition is not based on repeated multiplication (because it cannot be), but rather on a functional equation, or a differential equation, or a limit of sequences (the most useful one being the Maclaurin series). I know this approach does have its disadvantages under the current paradigm, but I think it is less misleading, and an approach like this would certainly lead to much less mysticism around the subject of the exponential function.
Part of what complicates the topic too is how the subject of function itself is taught. I have major, MAJOR gripes with how functions are taught, and from what I can tell, there are too many people, even in graduate school, who cannot define a function correctly, and I can't blame them, because schools do a horrible job with this. The idea that people need to really grasp is that a function is define not just by how it maps its inputs to outputs, but also by which sets it relates. If I write "f(x) = e^x," this is not meaningfully a function, even though it uses the functional notation of the output. The reason is because I have no idea what set my x's belong to, and I have no idea what set my e^x's belong to. This also makes it difficult to establish whether this "f" I have is injective and surjective: I need to know my domain and codomain. People may say: "but isn't the domain just the set of real numbers R?" No, because I have not specified the domain, and the fact that my function sends x to e^x does not tell me anything about the domain or codomain. I could choose my domain to be the set of natural numbers, and the codomain to be the set of real numbers, and this would tield a valid function with the same mapping. I could choose my domain to be the empty set if I wanted to. "e^x" has no "inherent" domain. This is what people really need to understand. So f: {1, 2} -> R, x |-> e^x and g: {2, 3} -> R, x |-> e^x are different functions, even though the outputs are the same for elements their domains share. This is because what defines a function is not solely how it relates inputs to outputs, but also the sets these belong to.
I mention this because explaining that f : C -> C, z |-> exp(z) is a different function than g : R -> R, x |-> exp(x) is going to be difficult for people who probably had the wrong definition of what a function is.
You should have gone a little further and shown that 'n' cannot be between -.152 to -.508 (about)...and since no integers lie within that range, THEN you can deduce that it works for all integers.
Could use arcosh for the second logarithm?
In min 3:17, Didn't you forget a minus for the multiplicative inverse of I? 🤔
P.D. I like your videos are very interesting and educative. 😁
multiplicative inverses dont have a minus, those are for additive inverses, np, alot of ppl in my class are sometimes confused about it too
@@SesquipedaliaThe multiplicative inverse of i in particular is equal to its additive inverse.
Proof:
1/z = -z
1 = -z²
z² + 1 = 0
z = ±i
Thus, 1/i = -i (and -1/i = i, which is equivalent)
Nobody:
Math students:
Math teachers:
Math professors:
Blackpenredpen: :)
X Æ A-12
😆
Who is nobody? You mean everybody?
Hello bprp..please tell about sin^-1(2) + cos^-1(2) =?
arccos(z) = π/2 - arcsin(z) for all complex z ==> arcsin(z) + arccos(z) = π/2 for all complex z ==> arcsin(2) + arccos(2) = π/2.
You actually became much more funnier
Apparently it’s also asin(pi/2) and -asin(pi/2)+pi
Shouldn't there also be another variable (lets say s since n and m are already used) which represents the period of the outer sin function?
Let BPRP's solution equal J:
to get all of the values of z, z = J + 2πs
Which is to say:
z = π ( 4m + 1 ) / 2 - i ln [ ( π ( 4n + 1 ) / 2 ) ± √ [ (π ( 4n + 1 ) / 2 )^2 - 1 ] ] + 2πs
Simplify by factoring out pi :
z = π ( 2s + ( 4m + 1 ) / 2 ) - i ln [ ( π ( 4n + 1 ) / 2 ) ± √ [ (π ( 4n + 1 ) / 2 )^2 - 1 ] ]
where n ∈ Z, m ∈ Z, s ∈ Z
I think so! Just a correction however, s would be accounting for the *inner* sine functions period, as n is already accounting for the outer sine function’s (the outer sin is collapsing all possible values for sin(z), which bprp denotes with 2n(pi), whereas the inner sin function is collapsing all possible values of z itself, which you denoted with 2s(pi).
Does the solution plot to a point or to some interesting pattern? Too lazy to do this myself.
I don’t understand the whole math but doesn’t (e^{iz} -e^{-iz})/2i used in Gaussian problems or am I over thinking this?
8:50 at this point the happy face 😀is now becoming 😭
8:51 2 Pie Em? emmm... Payem! Payem! calling Payem!
easy. inverse sine (pi/2)
sin sin sin sin sin x = 1
Lol. I pass.
@@blackpenredpen this is an imo compendium question ... I am not joking
@2C (02) Chan Kwan Yu
This formula will give you principal solution. If you want other solutions you can add 2πn. It will give you infinitely many solutions.
@2C (02) Chan Kwan Yu actually I am sending this same comment from last 20 videos so bprp will read but I think km he's not able to notice this comment so many other comments. I hope he reads this comment.
@2C (02) Chan Kwan Yu and this is not the same formula. You can get sin inverse of any number you want with this formula. For example he found sin inverse of π/2 in this video and he took so much time. But with my formula sin inverse of π/2 can be found in some seconds.
2*pi*n
2 pi en
2 pen
black pen
red pen
:0
Is it possible to have a term inside ln() to be negative?
Seeing ln as a real function, no
When working in the complex version tho, yes it is
He even made a video on that so you can check it out if you like
@@reeeeeplease1178 i'll clarify my question.
let 🙂=(4n+1)π/2.
Is it possible for ln(🙂±[🙂^2-1]^0.5) to be complex for some integer n?
Answer: NO. B/c i think we always have
🙂±[🙂^2-1]^0.5>0
@@orenfivel6247 ok ok got what you mean this time xD
maybe in n is negative, like n=-5 (since n is an integer)
if you now use the negative sqrt then you would have ln(neg - pos) = ln(neg + neg) = ln(neg)
Dont know how to do the smily face on my pc so let me use x insted :D
If we are talking positive sqrt and negative n then (as you indicated) x < 0 < sqrt(x^2 -1) => x^2 > 0 > x*sqrt(x^2 -1) (ineq. flips because x is neg)
This shows that for negative smily face, the argument is negative (for positive smily face, the whole term should be positive tho)
i just checked it with grapher and it seems like the function doesn’t intersect y = 1 at all. what’s the deal? i plugged y = sin(sin(x)) in desmos
There is no real solution, only complex solutions.
try to evaluate Gamma (1/3).
You missed infinite answers
After all, remembering that we are inside the sin function, at the end you must add +2πk
K being an integer
Very good video
thumbnail excited me to here
Well, I'll just stick to cos(cos(x))=1 :D
That stress at the end as ya were running out of space 😂
Yeah Bro Love Your Maths problems... Love from India....
Me same India
Here I was thinking that I could solve it by looking at the thumbnail...how wrong I was...
Well, it is solveable by staring if the question was asking only for real solutions.
Sin after sin
I have endured
Yet the wounds I bear
Are the wounds of COMPLEX ANALYSIS
does this prove than if I commit two sins in a row, then that still counts as one?
Easy!
arcsin(arcsin(1))
«I don't know if you use happy face in math, is very hard»😂