Сергей Мишин imaginary numbers are used in a ton of real world physics situations. The name makes them seem like someone made them up for no reason, but they are very legitimate.
That's from John Lennon, _The Beatles,_ "Come Together" (Abbey Road) - "One and one and one is three, "Got to be good lookin' cause he's so hard to see, "Come together "Right now, "Over me." Fred
this question is very easy using the fundamental theorem of engineering *sin x ≈ x* | x in radians | *π ≈ 3* using these we get the answer as 0.05 %error of 4.5%
@@ajety Jokes are supposed to be funny. Like how funny it is that the top rated comment is from a bunch of math fanboys who are so divorced from reality that they accept 3 degrees as an input for a special case solution without question even though angles (and all measurements) are analog values in every case with tolerances (aka limits), and instead of contemplating their own limitations are like "engineers are sooo dumb hyuk hyuk". Congrats you have solved sin(3), a contrived impossibly accurate degree reading, which was solved to the n'th digit long ago to actual usable digits, using the most convoluted and inefficient method. I'm not entirely sure who didn't get the "joke" here :)
I work as a teacher of control systems which involves a lot of different math subjects. Thank you for showing HOW TO TEACH STUDENTS. I like how you tell in detail mathematics. I really appreciate it.
Nice problem! I have 2 comments: 1. 23:52 it's easier to just say "divide the hypotenuse by sqrt(2) to get the leg so it's sqrt(3)/sqrt(2)" 2. 24:44 the second leg should be sqrt(3)/sqrt(2) not sqrt(3)/sqrt(3) 😊
For sin and cos of 15° couldn't you have also used the difference formula for sin and cosine? sin(45 - 30) = sin(45)cos(30) - cos(45)sin(30) cos(45 - 30) = cos(45)cos(30) + sin(45)sin(30)
I can never remember those formulas, but I can remember how to derive them with complex numbers. BPRP took a while to do it because he was being very explicit about writing out all the steps. For the 15 degree bit, I figured he was going to bring out complex numbers again to derive the half-angle formulas, which is definitely how I would do it, but he had a cleverer way.
@@MattMcIrvin what's there in remembering them, it's not even that hard, for me it's like if it's cos formula then all cos terms together+- sin terms together If it's sin formula then angles exchange
@@MattMcIrvin The way I was taught to derive the half angle formulas was to first derive cosine's double angle formula, then isolate cos(a), and plug (π/2)-a into the cosine half angle formula to derive the sine half angle formula cos(a+b) = cosacosb - sinasinb cos(a+a) = (cosa)^2 - (sina)^2 cos(2a) = (cosa)^2 - (sina)^2 cos(2a) = 2(cosa)^2-1 (cos(2a)+1)/2 = cosa^2 √((cos(2a)+1)/2) = cosa √((cos(2((π/2)-a)+1)/2) = cos((π/2)-a) √((cos(π-2a)+1)/2) = cos((π/2)-a) √((1-cos(2a))/2) = cos((π/2)-a) √((1-cos(2a))/2) = sin(a) I'm curious how you'd derive it with complex numbers; I've never seen that before
Wohoo I'm not the only one deriving the angle sum from Euler's formula! My professor thought me mental xD. Didn't subtract points but asked me if I'm slightly troubled that I find that simpler than geometric proofs xD
Digging through some of my old papers, I found where I ran this calculation years ago . I just ran the half angle formula on 30 degrees to get the sin and cos of 15 degrees. I love your construction to do it geometrically - never seen that before.
I like how this went back to your old video about special phi triangles! Also, I loved how there's such an elegant way to find an exact sine of an angle! Great job on the video.
I learnt a lot of special specials for the 1st time, though I knew sin (18) and sin (15) algebraically. Also the proofs of sin (a-b). Thank you, you are a special special teacher : )
I take my pen and ruler, draw a Triangle with one angle of 3 Degrees and an angle of 90 Degrees and then use the Definition of sin. I am a simple man...
@@darkseid856 It's easy to make relatively precise right triangles using a ruler if you know the length of the legs.. but yeah, since the length of the legs is kind of the goal it's not helpful here hehe. This is probably why ~easygoing~ considers himself a simple man.
Tibbes is awesome, glad that you shouted her out. On a similar note, another great video. I've had less time to watch them because of my final high school exams (GCSEs), but I'm excited to binge watch all of them after they finish. I'm sure that after your videos, I'll have no problem getting the top grade in my Maths exam =D
For a long time I looked for a channel like yours and when I found it it was better than I thought, friend you are the best, ahhhhh and by the way I will do the 100 integrals with you, hehe I already finished the derivatives but or my god I do not know how you resist so much standing time the truth I admire you very much
The result should be almost equal to pi/60. For small angles, sin x approximates x with x in radians. Converting 3 degrees to radians is just multiply with pi/180.
M. Shebl lol. I actually thought about it and it shouldn’t that bad. I could just do the same procedure when I constructed the 15-75-90 special special right triangle.
BPRP typically blasts through complex integral calculus, leaving melted markers and white boards in his path. Viewers lag, struggling to follow his genius. BPRP hits geometry. Brain: “Halt and catch fire.” One of the best TH-cam videos ever! Take my “Like,” Sir! 🤣🤣🤣🤣🤣🤣🤣
Can we do this without triangels 1]for 18° For this equation sin(X)=cos(4X) X=18° satisfies the eq where 4*18=72=90-18 We know cos(4X)=2cos^2(2X)-1 =2(1-sin^2(x))^2-1 Let y=sinx Then y=1+8y^4-8y^2 8y^4-8y^2-y+1=0 This eq has 4 solutions but one of them is sin18 8y^2(y^2-1)-(y-1)=0 (y-1)(8y^2(y+1)-1)=0 y=1 is a sol but not sin 18 cuz sin90=1 8y^3+8y^2-1=0 8y^3+4y^2+4y^2-1=0 4y^2(2y+1) + (2y-1)(2y+1)=0 (2y+1)(4y^2+2y-1)=0 y=-0.5 is a sol but not sin18 cuz sin 210=-0.5 4y^2+2y-1=0 y=(-2±sqrt(16+4)) /(2*4) =0.25(-1±sqrt(5)) Two solutions but we have one +ve solution and we know sin 18 is +ve Then sin 18° =0.25(-1+sqrt(5)) Cos 18° =sqrt(1-sin^2 (18)) =sqrt(1-(6-2sqrt(5))/16) =sqrt(5-sqrt(5))/2sqrt(2) 2]for 15° Cos 30=2 cos^2(15)-1 Cos15=sqrt((1+sqrt(3)/2)/2) =sqrt(4+2sqrt3)/2sqrt2 =sqrt(3+2sqrt3+1)/2sqrt2 =sqrt((sqrt3)^2+2(sqrt3)+1))/2sqrt2 =sqrt( (sqrt3+1)^2 ) /2sqrt2 =(sqrt3+1)/2sqrt2 Sin15=sqrt(1-cos^2(15)) =sqrt((1-sqrt(3)/2)/2) =sqrt(4-2sqrt3)/2sqrt2 =sqrt(3-2sqrt3+1)/2sqrt2 =sqrt((sqrt3)^2-2(sqrt3)+1))/2sqrt2 =sqrt( (sqrt3-1)^2 ) /2sqrt2 =(sqrt3-1)/2sqrt2 3] finally sin 3°=sin (18°-15°) =sin18°cos15°-cos18°sin 15°
Ok I get your point, and I also had that in my mind. But then the geometric proof is what gives maths so much fun. Indeed I would expect Mr Chao (aka bprp) to show the geometric proofs for the formulas for compound angles, namely sin(A±B), cos(A±B) and indeed tan(A±B).
@@peterchan6082 I like to do the geometric derivation of sin(A+B), but that one is all you need. You can use oddness of sine/evenness of cosine, sinA=cos(90-A) and vice versa, and tan=sin/cos to get all the others from there :)
@@Gold161803 Not quite enough. There are more to be desired. I already have several other geometric proofs of the compound angle formulas (some are simpler and even more beautiful than the one presented here) . . . indeed I've even done one for tan(A±B) from scratch, without the need to resort to sin(A±B)/cos(A±B)
@@peterchan6082 well yeah, I know there are several lovely proofs of all of them, I'm just saying you can also just derive them all from sine of a sum if you'd rather be boring like me :p
oh, just approximate square roots the archimedes way. you know, dividing, squaring and adding some bunch of numbers and taking days to just get 7 decimals of precision.
Only one I can remember from top of my head is double angle formula for cosine. Every other one I need I always derive before I use them. It keeps your wits as it is finicky to get all those cosines and sines and what not correct. Helps you keep everthing tidy.
Dude you're so amazing. I really appreciate all of your work. I'm 14 years old and I really like math. I never liked how it's explained on schools, it seemes really basic to me and doesn't give a chance to us math enthusiasts to go further. Thanks to people like you, I get to learn more about my passion, which is math. People often see math as a hard thing which involves tons of numbers, but in reality, thanks to people like you, I realised it's really about cool concepts an ideas. Keep up the good work, bprp, because what you're doing is amazing and for some of us, a lifechanger. Sorry for grammar mistakes, I'm spanish.
i strongly agree with every single thing you said. i think teachers should make the students *interested* in the subject and show its actual beauty. bprp must be an excellent teacher that i would LOVE to have. P. S. i'm Ukrainian and i thought Europe had better education, but i can't see any difference though... guess we are screwed ¯\_(ツ)_/¯
@@artemetra3262 Yeah you're right. People won't be interested if you just show formulas without proving them. Math is about concepts and ideas. We all really need to work on fixing education for next generations.
I enjoy watching your channel. Thank you. About 40 years ago I was shown a problem. Calculate the sine of 13 degrees. I haven't seen a good solution yet.
I have solved the value of sin(1°) . I have leveraged the information from you about sin(3°) and applied the formula about sin(x/3) exactly as you did with sin(10°)
Mmm I tried this way and i got a fifth order polynomial. Let A=3°, x=sin A, y=sin (5A), then y=16*(x^5) - 20*(x^3) + 5x. Problem: there Is no Solve formula for 5th order polynomial (Abel's theorem). So, i had to watch the video xD
I know this is an old video that the Almighty Algorithm just recommended to me so my apologies for the late comment. I guess you won't see this anyway but if you do, I'll say that it was a really cool video. It's not that I was losing sleep over the exact value of sin(3°) but it was fun to see it developed. One comment about that one instant where you "buffered" for a good number of seconds and towards the end of the video you said that maybe you should have prepared better. I'll say: don't. It was really satisfying that even folks with advanced math knowledge don't always see everything right away. So, kudos for not having this edited out! I love your channel!
I am from India. Your explanation is really awesome. It's very nice. I haven't words for appreciation. Awesome awesome awesome.........................
Another approach (and this was done about 1000 years ago) is to find the value of sin and cos of 18 degrees. For that you use regular pentagon with side 1. Then by the same difference formula you can find sin12 because 12=72-60. After that just use the half angle formula twice. For people asking about sin of 1 degree, after finding sin of 12 degrees, you use triple angle formula, solve the corresponding cubic equation to find sin4. After that use the half angle formula twice and get sin1 !
At 16:20 you could also have used that your value for x solves x^2 = 1 - x, and therefore (x/2) ^2 = x^2/4 = (1-x)/4. Then the simplification under the root sign becomes a bit easier and/or faster.
i spotted circular reasoning at the proof of the angle sum formula: in order to prove euler's formula you need to know the derivative of sinx, which requires sin(a+b). So you can't use the result to prove the base. If you know any proof of the euler's formula without the derivative of sinx, please inform me
By knowing the sin and cosin of 3° we can also get sin and cosin for every angle multiple of three. For example sin(117°) = sin(120°-3°) = sin120°×cos3° - cos120°×sin3°. If you were to use the cubic formula on that equation you got a long time ago for the sin of 10 degrees (8x^3-6x+1=0 ; x=sin10°) we could then do the following: sin(7°) = sin(10°-3°) = sin10°×cos3° - cos10°×sin3° sin(4°) = sin(7°-3°) = sin7°×cos3° - cos7°×sin3° sin(1°) = sin(4°-3°) = sin4°×cos3° - cos4°×sin3° Then using sin^2(θ) + cos^2(θ) = 1 we can get the cosin of 1°. Knowing sin(1°) and cos(1°) we can use sin(α+β) = sinα×cosβ-cosα×sinβ and every other related formula to get the sin and cosin of every angle expressed by an entire amount of degrees.
I was watching this and wondering if the sine and cosine of any whole number of degrees was algebraic. But I poked around on Wolfram Alpha and realized that of course it is, because e^i*(one degree) = e^i*(pi/180) = (-1)^(1/180), so any sum of degrees can be expressed algebraically in terms of integer roots of -1. (Wikipedia says that defined trig function values of all rational multiples of pi are algebraic, which would incorporate all integer degrees. That is not to say they are *constructible* numbers, but I guess bprp just proved that trig functions of the multiples of 3 degrees are constructible?) (Edit: Yes, he did. Apparently any angle of a*pi/b degrees is constructible if and only if, in simplest form, b is a product of *unique* Fermat primes and a power of 2, and 3 degrees is pi/(3*5*2^2). 1 degree is not since its prime factorization has two 3s in it.)
@blackpenredpen you could have substituted 5θ=90, then breaking 5θ=3θ+2θ we get 2θ=90-3θthen apply sine func. on both side we get sin2θ=cos3θ After some calculation we get... 4(sinθ)^2+2sinθ-1=0 Thus without any geometrical application and scratches 😃you can evaluate the value of sin18... P.S. forgot to mention θ=18 and BTW huge fan of your supreme integrals. Keep Posting such integrals with new techniques!!!
Thank you for the shout out! 😺
Tibees thank you for the t-shirt! I love it!!
@@blackpenredpen Do you really think that you do math with this absurd imaginary unit?
+Rahul
Can't blame him lol
Сергей Мишин imaginary numbers are used in a ton of real world physics situations. The name makes them seem like someone made them up for no reason, but they are very legitimate.
Hi Tibees
Mathematician: sin(3°)=((sqrt(5)-1)(sqrt(6)+sqrt(2))-2(sqrt(3)-1)sqrt(5+sqrt(5)))/16
Physicist: sin(3°)=pi/60
Engineer: sin(3°)=0
Engineers more like "three is small, so sin(3°) = 3"
lmao
@@AndDiracisHisProphet (Which is pi/60.)
@@PuzzleQodec and the Fundamental Theorem of Engineering states that pi=3 so we plug it in and we get 1/20" = 0.05"
@@AndDiracisHisProphet Use that and the bridge falls down. sin(3°)=sin(3π/180)=sin(π/60)≈π/60
11:03 *thought download begins*
12:07 *thought download complete*
lol
Hahahhahahahha!!!!!
@@pedroandrade8727 sounds like a MC villager trying to scam you into giving some emerald.
That was fast!
I was genuinely frightened for he NEVER stops in his videos :)
Bprp: *Runs math channel like a boss*
Also bprp: *1=2*
I almost read as "Bprp runs meth" 😂😂
At 11:05 you can almost hear the cogwheels turning in his head...
3:03 "Of course, 1 is equal to 2"
-BPRP 2019
3:03 actually
1 = 2
@@not_allen1107 There are four lights
"1+1+1 =3"
We did it boys, an A in maths
👏👏👏
That's from John Lennon, _The Beatles,_ "Come Together" (Abbey Road) -
"One and one and one is three,
"Got to be good lookin' cause he's so hard to see,
"Come together
"Right now,
"Over me."
Fred
ffggddss I HAD THE SAME EXACT THOUGHT
Math*
@@colleen9493 both is fine
18-15=3
this question is very easy using the fundamental theorem of engineering
*sin x ≈ x* | x in radians |
*π ≈ 3*
using these we get the
answer as 0.05
%error of 4.5%
💀💀💀💀
@@analog_joe No, it's pi = 3.
@@analog_joe Dude it's just a joke
@@ajety Jokes are supposed to be funny. Like how funny it is that the top rated comment is from a bunch of math fanboys who are so divorced from reality that they accept 3 degrees as an input for a special case solution without question even though angles (and all measurements) are analog values in every case with tolerances (aka limits), and instead of contemplating their own limitations are like "engineers are sooo dumb hyuk hyuk". Congrats you have solved sin(3), a contrived impossibly accurate degree reading, which was solved to the n'th digit long ago to actual usable digits, using the most convoluted and inefficient method. I'm not entirely sure who didn't get the "joke" here :)
@@glendenog9095 HAHHAAHHAHAHHAAHAHAH WHAT
Me on exams: 11:03
im the opposite
@@rafaelv.t1403 but you're gay
Rafael V.T ok
😁😁😁😁😁😁😁😁
Really shows you even an expert has troubled moments
processing processing, I felt like I was in the moment.
He teaches very friendly. Even for a simple calculation, he explains very kindly. So I can understand whole topic. Thank you for your works!
I work as a teacher of control systems which involves a lot of different math subjects. Thank you for showing HOW TO TEACH STUDENTS. I like how you tell in detail mathematics. I really appreciate it.
Eg. M wow, what a comment! Thank you!!
Nice problem! I have 2 comments:
1. 23:52 it's easier to just say "divide the hypotenuse by sqrt(2) to get the leg so it's sqrt(3)/sqrt(2)"
2. 24:44 the second leg should be sqrt(3)/sqrt(2) not sqrt(3)/sqrt(3) 😊
"1 is equal to 2"
- bprp 2019
Btw. Great video
Sigma 1 thank you!
11:02 Me during an exam
Zackary자카리
Me during a video.
@@blackpenredpen,
great.
I can't but admire that
@@blackpenredpen SENPAI NOTICED ME ~~
@@astralchan is it legal to say Japanese-originated words to a chinese person?
@@keescanalfp5143~ I CAN
For sin and cos of 15° couldn't you have also used the difference formula for sin and cosine?
sin(45 - 30) = sin(45)cos(30) - cos(45)sin(30)
cos(45 - 30) = cos(45)cos(30) + sin(45)sin(30)
Perihelion Orbit yea. You can also use that picture to prove that formula.
I think he was proving the sum and difference formula using complex numbers
I can never remember those formulas, but I can remember how to derive them with complex numbers. BPRP took a while to do it because he was being very explicit about writing out all the steps.
For the 15 degree bit, I figured he was going to bring out complex numbers again to derive the half-angle formulas, which is definitely how I would do it, but he had a cleverer way.
@@MattMcIrvin what's there in remembering them, it's not even that hard, for me it's like if it's cos formula then all cos terms together+- sin terms together
If it's sin formula then angles exchange
@@MattMcIrvin The way I was taught to derive the half angle formulas was to first derive cosine's double angle formula, then isolate cos(a), and plug (π/2)-a into the cosine half angle formula to derive the sine half angle formula
cos(a+b) = cosacosb - sinasinb
cos(a+a) = (cosa)^2 - (sina)^2
cos(2a) = (cosa)^2 - (sina)^2
cos(2a) = 2(cosa)^2-1
(cos(2a)+1)/2 = cosa^2
√((cos(2a)+1)/2) = cosa
√((cos(2((π/2)-a)+1)/2) = cos((π/2)-a)
√((cos(π-2a)+1)/2) = cos((π/2)-a)
√((1-cos(2a))/2) = cos((π/2)-a)
√((1-cos(2a))/2) = sin(a)
I'm curious how you'd derive it with complex numbers; I've never seen that before
Wohoo I'm not the only one deriving the angle sum from Euler's formula! My professor thought me mental xD. Didn't subtract points but asked me if I'm slightly troubled that I find that simpler than geometric proofs xD
"If your only tool is a hammer, ..." And to be honest, Euler's formula does make for a wonderful hammer.
Digging through some of my old papers, I found where I ran this calculation years ago . I just ran the half angle formula on 30 degrees to get the sin and cos of 15 degrees. I love your construction to do it geometrically - never seen that before.
You could also do Sin(45-30)! (Not factorial)
I like how this went back to your old video about special phi triangles! Also, I loved how there's such an elegant way to find an exact sine of an angle! Great job on the video.
I learnt a lot of special specials for the 1st time, though I knew sin (18) and sin (15) algebraically. Also the proofs of sin (a-b). Thank you, you are a special special teacher : )
No one:
Minecraft Villager: 11:03
Super fun video :) I love how you talk about angles like they are people
Bayan Mehr hahaha thank you
I take my pen and ruler, draw a Triangle with one angle of 3 Degrees and an angle of 90 Degrees and then use the Definition of sin.
I am a simple man...
And then hope your pencil is infinitely sharp and the angle is perfectly 3 and you measure the distance very accurately.
@@gumanelson2007 never knew that one can make angles using a ruler .
@@darkseid856 It's easy to make relatively precise right triangles using a ruler if you know the length of the legs.. but yeah, since the length of the legs is kind of the goal it's not helpful here hehe. This is probably why ~easygoing~ considers himself a simple man.
@@akunog2708 yea that was basically what i was saying
@@darkseid856 after using a compass or protractor
Tibbes is awesome, glad that you shouted her out. On a similar note, another great video. I've had less time to watch them because of my final high school exams (GCSEs), but I'm excited to binge watch all of them after they finish.
I'm sure that after your videos, I'll have no problem getting the top grade in my Maths exam =D
I love the silence starting from 11:02 😂😂😂😂😂😂
11:30 when you gotta read your own notes because you forgot how genius you used to be
Men you deserve 1million subscribers and you deserved the position of my professor in calculus
For a long time I looked for a channel like yours and when I found it it was better than I thought, friend you are the best, ahhhhh and by the way I will do the 100 integrals with you, hehe I already finished the derivatives but or my god I do not know how you resist so much standing time the truth I admire you very much
The result should be almost equal to pi/60. For small angles, sin x approximates x with x in radians. Converting 3 degrees to radians is just multiply with pi/180.
sharmsma yup that’s coming up soon
(pi/60) - (pi/60) ^ 3 / 6 is a better estimate, expanded the taylor series by one
Amazing, my bicolor pen friend... It's amazing how with a "few" roots and triangles, you can express sines and cosines in a closed form. Good work!
(In funny, annoyed tone) No! Prove it all geometrically! :cat:
M. Shebl lol. I actually thought about it and it shouldn’t that bad. I could just do the same procedure when I constructed the 15-75-90 special special right triangle.
Congrats on 300K!!!
JZ Animates thank you!!!!!
BPRP typically blasts through complex integral calculus, leaving melted markers and white boards in his path. Viewers lag, struggling to follow his genius.
BPRP hits geometry. Brain: “Halt and catch fire.”
One of the best TH-cam videos ever! Take my “Like,” Sir!
🤣🤣🤣🤣🤣🤣🤣
Completing that rectangle was lovely. Well done.
One of the best video ever upload on TH-cam.
Thanks you
Congrats for gaining 300k subscribers 👏
Can we do this without triangels
1]for 18°
For this equation sin(X)=cos(4X)
X=18° satisfies the eq where 4*18=72=90-18
We know cos(4X)=2cos^2(2X)-1
=2(1-sin^2(x))^2-1
Let y=sinx
Then y=1+8y^4-8y^2
8y^4-8y^2-y+1=0
This eq has 4 solutions but one of them is sin18
8y^2(y^2-1)-(y-1)=0
(y-1)(8y^2(y+1)-1)=0
y=1 is a sol but not sin 18 cuz sin90=1
8y^3+8y^2-1=0
8y^3+4y^2+4y^2-1=0
4y^2(2y+1) + (2y-1)(2y+1)=0
(2y+1)(4y^2+2y-1)=0
y=-0.5 is a sol but not sin18 cuz sin 210=-0.5
4y^2+2y-1=0
y=(-2±sqrt(16+4)) /(2*4)
=0.25(-1±sqrt(5))
Two solutions but we have one +ve solution and we know sin 18 is +ve
Then sin 18° =0.25(-1+sqrt(5))
Cos 18° =sqrt(1-sin^2 (18))
=sqrt(1-(6-2sqrt(5))/16)
=sqrt(5-sqrt(5))/2sqrt(2)
2]for 15°
Cos 30=2 cos^2(15)-1
Cos15=sqrt((1+sqrt(3)/2)/2)
=sqrt(4+2sqrt3)/2sqrt2
=sqrt(3+2sqrt3+1)/2sqrt2
=sqrt((sqrt3)^2+2(sqrt3)+1))/2sqrt2
=sqrt( (sqrt3+1)^2 ) /2sqrt2
=(sqrt3+1)/2sqrt2
Sin15=sqrt(1-cos^2(15))
=sqrt((1-sqrt(3)/2)/2)
=sqrt(4-2sqrt3)/2sqrt2
=sqrt(3-2sqrt3+1)/2sqrt2
=sqrt((sqrt3)^2-2(sqrt3)+1))/2sqrt2
=sqrt( (sqrt3-1)^2 ) /2sqrt2
=(sqrt3-1)/2sqrt2
3] finally sin 3°=sin (18°-15°)
=sin18°cos15°-cos18°sin 15°
Woah nice sol
my brain be like😮💨
This is evidence that mathematicians really don't mind that long walk for a short drink of water
The video is old, but it contains valuable skills, and I benefited a lot from it. Thank you very much, Mighty Professor
Mad props for not cutting the video when solving the problem.
Will you do multivariable calc vids ?
Or pde?
after watching this ... I'm ready to consume 3 large pizzas (with each slice's tip at 18 degrees wide)
nimmira hahaha nice!!
Thats really fantastic....you give us passion to learn new things....you've new subscriber from Aleppo, Syria 💐🌸
Glad to hear! Thank you!
For cos15 and sin15, couldn't you just use compound formula again...cos(45-30) and sin(45-30)?
@Blackpenredpen
Ok I get your point, and I also had that in my mind. But then the geometric proof is what gives maths so much fun.
Indeed I would expect Mr Chao (aka bprp) to show the geometric proofs for the formulas for compound angles, namely sin(A±B), cos(A±B) and indeed tan(A±B).
@@peterchan6082 I like to do the geometric derivation of sin(A+B), but that one is all you need. You can use oddness of sine/evenness of cosine, sinA=cos(90-A) and vice versa, and tan=sin/cos to get all the others from there :)
@@Gold161803
Not quite enough. There are more to be desired. I already have several other geometric proofs of the compound angle formulas (some are simpler and even more beautiful than the one presented here) . . . indeed I've even done one for tan(A±B) from scratch, without the need to resort to sin(A±B)/cos(A±B)
@@peterchan6082 well yeah, I know there are several lovely proofs of all of them, I'm just saying you can also just derive them all from sine of a sum if you'd rather be boring like me :p
厉害！But my abacus doesn't have the square root function, so I'm still unable to calculate the exact value.
oh, just approximate square roots the archimedes way. you know, dividing, squaring and adding some bunch of numbers and taking days to just get 7 decimals of precision.
0.3M subscribers. Congrats!!!
Yale NG yay thank you!
3:05 "One is equal to two"
Astronomical Units:
c = G = h = π = 1 = 2
11:03 .. A magical moment of thought .. See how your mind works :)
Like it
Can someone explain to me why he paused
@@leonhardeuler6811 to think
Exactly a perfect relation between complax analysis and real number , i love them ❤❤❤❤
Nice of you to prove the angle addition and subtraction trigonometric identities.
Only one I can remember from top of my head is double angle formula for cosine. Every other one I need I always derive before I use them. It keeps your wits as it is finicky to get all those cosines and sines and what not correct. Helps you keep everthing tidy.
Dude you're so amazing. I really appreciate all of your work. I'm 14 years old and I really like math. I never liked how it's explained on schools, it seemes really basic to me and doesn't give a chance to us math enthusiasts to go further. Thanks to people like you, I get to learn more about my passion, which is math. People often see math as a hard thing which involves tons of numbers, but in reality, thanks to people like you, I realised it's really about cool concepts an ideas. Keep up the good work, bprp, because what you're doing is amazing and for some of us, a lifechanger. Sorry for grammar mistakes, I'm spanish.
i strongly agree with every single thing you said. i think teachers should make the students *interested* in the subject and show its actual beauty. bprp must be an excellent teacher that i would LOVE to have.
P. S. i'm Ukrainian and i thought Europe had better education, but i can't see any difference though... guess we are screwed ¯\_(ツ)_/¯
@@artemetra3262 Yeah you're right. People won't be interested if you just show formulas without proving them. Math is about concepts and ideas. We all really need to work on fixing education for next generations.
@@diegomullor8605 That's why I endorse Aops. Check them out at aops.com
They've been a life changer for me!
I enjoy watching your channel. Thank you. About 40 years ago I was shown a problem. Calculate the sine of 13 degrees. I haven't seen a good solution yet.
I have solved the value of sin(1°) . I have leveraged the information from you about sin(3°) and applied the formula about sin(x/3) exactly as you did with sin(10°)
11:04 - I like my math videos like I like my Jerry Springer videos: Raw and Uncut.
Edward Sanville
I once put “raw footage” in my title but YT demonetized it.
@@blackpenredpen filthy youtube
Great! Let's find the relationship between sin(3°) and sin(15°) and construct the one-fifth angle formula!
Mmm I tried this way and i got a fifth order polynomial. Let A=3°, x=sin A, y=sin (5A), then y=16*(x^5) - 20*(x^3) + 5x. Problem: there Is no Solve formula for 5th order polynomial (Abel's theorem). So, i had to watch the video xD
@@Drk950 16x^5 - 20x^3 + 5x = 0
x(16x^4 - 20x^2 + 5) = 0
The root x = 0 is extraneous, ignore that
16x^4 - 20x^2 + 5 = 0
Let w = x^2
16w^2 - 20w + 5 = 0
Which is a quadratic equation :]
This pause was very nice, it gave me just enough time to figure it out
every triangle is special for me
The Euler formula is much harder to prove than trig identities, bro!
11:00-12:05
You know it’s serious when he becomes blackpenredpenbluepen
I'm gonna thank you for this video. So glad that I created a graph in desmos that has 120 point unit circle coordinates.
I know this is an old video that the Almighty Algorithm just recommended to me so my apologies for the late comment.
I guess you won't see this anyway but if you do, I'll say that it was a really cool video. It's not that I was losing sleep over the exact value of sin(3°) but it was fun to see it developed.
One comment about that one instant where you "buffered" for a good number of seconds and towards the end of the video you said that maybe you should have prepared better. I'll say: don't. It was really satisfying that even folks with advanced math knowledge don't always see everything right away. So, kudos for not having this edited out!
I love your channel!
6:20 thanks for a new way of proving angle difference. It blew my mind.😀🔥
Love the way you base this proof on (1) = (2)
I am from India. Your explanation is really awesome. It's very nice. I haven't words for appreciation. Awesome awesome awesome.........................
You tried to sneak in the true proof of the angle addition formula with the boxes, and you thought we wouldn't notice!
noahtaul hahahaha yea
11:03
TH-cam when the ads buffer
20:18
TH-cam when the actual video buffers
Nobody:
Me after reading the title: WHOT¿ ARE YOU SERIOUUS!?!?!?
blackpenredpen: 0:01
#maths4fun
Write 0:01 instead 0:00 doesn't work
Take a shot every time he says "of course".
Wow this is really nice thanks for sharing this...
maths- physicshub thank you!!!
It’s 4am, I have lessons at 9am, and idk why am I watching this now.
Also after calculating sin and cos of 45 and 30 why not just subtract?
Everyone knows that 5+5+5=15 but I know that 45-30=15.
That is kind of what he did. Just... geometrically
Are we not able to to 45 degrees divided by 15 degrees or is that not allowed?
@@christianalbina6217 boi thats not how it works ! (As much as I know it doesn't )
if you want to do it with algebra you can, he did it with geometry
You could do this to find the values of sin15 and cos15 but you would need to use the angle difference identities again
Very nice to use Euler's formula and geometry 💕
Sergio H yea!!
Time to grab some popcorn 🍿
I should learn for my exams right now :D Here we go again bois!
Laci yeah mine is next week :(
Another approach (and this was done about 1000 years ago) is to find the value of sin and cos of 18 degrees. For that you use regular pentagon with side 1. Then by the same difference formula you can find sin12 because 12=72-60. After that just use the half angle formula twice.
For people asking about sin of 1 degree, after finding sin of 12 degrees, you use triple angle formula, solve the corresponding cubic equation to find sin4. After that use the half angle formula twice and get sin1 !
That cliffhanger had me rolling! Had to be the longest awkward pause in youtube history! Haha!!
This calculation was so much fun.
Thx
Thank you for all efforts God bless you
At 16:20, there's no need to develop the square. The equation on the left gives x^2=1-x and the red square root becomes sqrt(1-(1-x))=sqrt(x)
Very very nice method.. . u are best teacher
10:00 it is similar to the compass and ruler construction of a pentagon
cos(72°) = (√5-1)/4
from here it is easy to find (x)
[Cos(x)+isin(x)]^n=cos(nx)+isin(nx
Find formula cos (nx)
For n is integer.
19:43 "everybody is 60 degrees inside"
Not me. I'm dead inside
so 0?
@@ashtonsmith1730 yes pretty much
11:03
Literally me on my test two days ago.
You are a nice teacher !!! Good luck
At 16:20 you could also have used that your value for x solves x^2 = 1 - x, and therefore (x/2) ^2 = x^2/4 = (1-x)/4. Then the simplification under the root sign becomes a bit easier and/or faster.
The fact that we're all here watching how this man calculates such a random number for 30 min straight just fascinates me
Those were some badass triangles, no doubt! Great vid, thanks
Because we have sin(3), we can use that formula to find sin(6) because of sin(3+3), meaning we can find sin(any multiple of 3)
i spotted circular reasoning at the proof of the angle sum formula: in order to prove euler's formula you need to know the derivative of sinx, which requires sin(a+b). So you can't use the result to prove the base. If you know any proof of the euler's formula without the derivative of sinx, please inform me
cos(a+b) and sin(a+b) can be proven using the dot and the cross product, respectively
i'm totally taking "we know one is equal to two" out of context
Loved this! Please make more pure math content like this!
By knowing the sin and cosin of 3° we can also get sin and cosin for every angle multiple of three.
For example sin(117°) = sin(120°-3°) = sin120°×cos3° - cos120°×sin3°.
If you were to use the cubic formula on that equation you got a long time ago for the sin of 10 degrees (8x^3-6x+1=0 ; x=sin10°) we could then do the following:
sin(7°) = sin(10°-3°) = sin10°×cos3° - cos10°×sin3°
sin(4°) = sin(7°-3°) = sin7°×cos3° - cos7°×sin3°
sin(1°) = sin(4°-3°) = sin4°×cos3° - cos4°×sin3°
Then using sin^2(θ) + cos^2(θ) = 1 we can get the cosin of 1°.
Knowing sin(1°) and cos(1°) we can use sin(α+β) = sinα×cosβ-cosα×sinβ and every other related formula to get the sin and cosin of every angle expressed by an entire amount of degrees.
I was watching this and wondering if the sine and cosine of any whole number of degrees was algebraic.
But I poked around on Wolfram Alpha and realized that of course it is, because e^i*(one degree) = e^i*(pi/180) = (-1)^(1/180), so any sum of degrees can be expressed algebraically in terms of integer roots of -1.
(Wikipedia says that defined trig function values of all rational multiples of pi are algebraic, which would incorporate all integer degrees. That is not to say they are *constructible* numbers, but I guess bprp just proved that trig functions of the multiples of 3 degrees are constructible?)
(Edit: Yes, he did. Apparently any angle of a*pi/b degrees is constructible if and only if, in simplest form, b is a product of *unique* Fermat primes and a power of 2, and 3 degrees is pi/(3*5*2^2). 1 degree is not since its prime factorization has two 3s in it.)
An elegant way to combine euler formula and trigonometry 👍
The engineering way: sin(3)~0 😂
Nice video, I really enjoyed.
Or use small angle approximation and say that sin(3 degrees) ~ 3 degrees ~ pi/60, which has an error of only 0.05%
You should do more videos on Feynman’s method bprp! I love all your videos tho.
sin(3) on calculator:
0.14112000806
blackpenredpen:
*33 minutes and 15 seconds of description*
@blackpenredpen you could have substituted 5θ=90, then breaking 5θ=3θ+2θ we get 2θ=90-3θthen apply sine func. on both side we get
sin2θ=cos3θ
After some calculation we get...
4(sinθ)^2+2sinθ-1=0
Thus without any geometrical application and scratches 😃you can evaluate the value of sin18...
P.S. forgot to mention θ=18
and BTW huge fan of your supreme integrals. Keep Posting such integrals with new techniques!!!