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.
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 :)
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
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) 😊
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
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
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.
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 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.
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.
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
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°
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
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.
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°)
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 RUclips videos ever! Take my “Like,” Sir! 🤣🤣🤣🤣🤣🤣🤣
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.
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.
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
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
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.
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.
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.)
Another approach can be A=3 5A = 15 sin(3A+2A) = sin(5A) sin3Acos2A + sin2Acos3A = sin(5A) then expand sin3A ...and so on.. put the value .. and find out sinA .. Yeah I know old school and tedious but will save u sanity if solving 100 problems in an assignment.. Btw Great Approach👍
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 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 !
You can go even further and get the sin of 1 degree by applying the triple angle formula: sin 3θ = 3 sin θ − 4 sin^3 θ It involves solving a cubic in the form of Ax^3 + Bx + C = 0, but it does have a closed form solution.
hey, bprp, how do you like the idea to make a video with analysis of the sinuses from 0 to 45 degrees? Maybe video will be for a few hours, but i think it won't stop you)
@@rafciopranks3570 I agree with you but I have searched and I have not been able to find the exact answer in case I find it or manage to do it I will tell you but look what she gave me with an approximation of 3 decimal places I would show it to you in a photo but on youtube I don't know well jeje (50-5(50+5(50+5(50+(2310)½)½)½)½)½/10
@@andresfelipemunoz4417 It cannot be. If we had a way to express it in terms of the radicals we would be able to construct a regular 180 -gon which is proved impossible.
Fun fact: the 6 trig ratios of ANY multiple of pi/60 (3 degrees), for that multiple between 1 and 30, can be expressed in terms of nested radicals. All the other angles in between require you to take the cube-root of a complex number. An equivalent expression for sin(pi/60) is: (1/8)*[sqrt(10+5*sqrt(3)) - sqrt(2+sqrt(3))-sqrt(2*(2-sqrt(3))*(5+sqrt(5)))] You could probably calculate the cos(2pi/15). Answers (1/4)*sqrt(9-sqrt(5)+sqrt(30+6*sqrt(5)))
I asked my calc 1 professor in college to show me how to solve for sin or cos of any angle, she said she didn't know (?????). Didn't say anything about a taylor series or how there are ways to solve for it other than just looking at the unit circle or taking advantage of the difference function
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 :)
At 11:05 you can almost hear the cogwheels turning in his head...
Bprp: *Runs math channel like a boss*
Also bprp: *1=2*
I almost read as "Bprp runs meth" 😂😂
3:03 "Of course, 1 is equal to 2"
-BPRP 2019
3:03 actually
1 = 2
@@not_allen1107 There are four lights
Really shows you even an expert has troubled moments
processing processing, I felt like I was in the moment.
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
"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
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!!
He teaches very friendly. Even for a simple calculation, he explains very kindly. So I can understand whole topic. Thank you for your works!
Me on exams: 11:03
im the opposite
@@rafaelv.t1403 but you're gay
Rafael V.T ok
😁😁😁😁😁😁😁😁
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!
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.
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
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.
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 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...
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
Super fun video :) I love how you talk about angles like they are people
Bayan Mehr hahaha thank you
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
Congrats on 300K!!!
JZ Animates thank you!!!!!
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
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😮💨
Men you deserve 1million subscribers and you deserved the position of my professor in calculus
No one:
Minecraft Villager: 11:03
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
Mad props for not cutting the video when solving the problem.
I love the silence starting from 11:02 😂😂😂😂😂😂
This is evidence that mathematicians really don't mind that long walk for a short drink of water
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.
Completing that rectangle was lovely. Well done.
The video is old, but it contains valuable skills, and I benefited a lot from it. Thank you very much, Mighty Professor
Congrats for gaining 300k subscribers 👏
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°)
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 RUclips videos ever! Take my “Like,” Sir!
🤣🤣🤣🤣🤣🤣🤣
after watching this ... I'm ready to consume 3 large pizzas (with each slice's tip at 18 degrees wide)
nimmira hahaha nice!!
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.
One of the best video ever upload on RUclips.
Thanks you
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.
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 :]
11:30 when you gotta read your own notes because you forgot how genius you used to be
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
3:05 "One is equal to two"
Astronomical Units:
c = G = h = π = 1 = 2
Thats really fantastic....you give us passion to learn new things....you've new subscriber from Aleppo, Syria 💐🌸
Glad to hear! Thank you!
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!
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.
This pause was very nice, it gave me just enough time to figure it out
Nice of you to prove the angle addition and subtraction trigonometric identities.
Will you do multivariable calc vids ?
Or pde?
I'm gonna thank you for this video. So glad that I created a graph in desmos that has 120 point unit circle coordinates.
0.3M subscribers. Congrats!!!
Yale NG yay thank you!
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
It’s 4am, I have lessons at 9am, and idk why am I watching this now.
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)
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
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)
The Euler formula is much harder to prove than trig identities, bro!
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
You know it’s serious when he becomes blackpenredpenbluepen
every triangle is special for me
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
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
[Cos(x)+isin(x)]^n=cos(nx)+isin(nx
Find formula cos (nx)
For n is integer.
I should learn for my exams right now :D Here we go again bois!
Laci yeah mine is next week :(
Love the way you base this proof on (1) = (2)
Take a shot every time he says "of course".
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.)
mistake reversing 2 and sqrt 3 in the triangle at 23:33?
Another approach can be A=3
5A = 15
sin(3A+2A) = sin(5A)
sin3Acos2A + sin2Acos3A = sin(5A)
then expand sin3A ...and so on.. put the value .. and find out sinA ..
Yeah I know old school and tedious but will save u sanity if solving 100 problems in an assignment..
Btw Great Approach👍
Actually you can simply use approximation with limits
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.
11:03
RUclips when the ads buffer
20:18
RUclips when the actual video buffers
I am from India. Your explanation is really awesome. It's very nice. I haven't words for appreciation. Awesome awesome awesome.........................
Very nice to use Euler's formula and geometry 💕
Sergio H yea!!
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)
6:20 thanks for a new way of proving angle difference. It blew my mind.😀🔥
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 !
In my head using small angles i worked out pi/60 to be around 0.05236 so i will watch and see how close i was
The fact that we're all here watching how this man calculates such a random number for 30 min straight just fascinates me
Just draw a right triangle with degrees 3° and 87° and divide opposite of 3° by hypotenuse
Why the the equation of line (altogether at infiniy):0x+0y+c=0? Is this even possible?
The length of the base of the triangle (x) is the reciprocal of the golden ratio.
It arises from this video than sin/cos/tan of any multiple of 3° can be written purely in terms of positive square roots.
Exactly a perfect relation between complax analysis and real number , i love them ❤❤❤❤
This Video is 33 minutes long and 3 years old and has 353,753 views and is about finding sin(3).... Amazing
You can go even further and get the sin of 1 degree by applying the triple angle formula: sin 3θ = 3 sin θ − 4 sin^3 θ
It involves solving a cubic in the form of Ax^3 + Bx + C = 0, but it does have a closed form solution.
Is it just me or did you do 10 / 2 = 8 at 17:26 ?
It was a 16 on the bottom
@@blackpenredpen Oh, I don't even want to think about how tired I must've been to not realise that
Bro you can just use the trigonometric formula to get the value of cos 15⁰ which is
Cos(45⁰-30⁰)=cos45⁰×cos30⁰+sin 45⁰ × sin 30⁰
To get 75 15 90 triangle just draw equilateral triangle and extend its height by the side of this equilateral triangle
hey, bprp, how do you like the idea to make a video with analysis of the sinuses from 0 to 45 degrees?
Maybe video will be for a few hours, but i think it won't stop you)
Trigonometry really flares up my sinuses
Хмм, действительно интересная задумка
I have a question, it is possible to calculate the sine and cosines of one degree???????????but that is exact or has an approximation with roots
Came here to ask the same question
Friend i mánaged to solved this question whit an approximation using roots but it is somewhat long , do you want to see my answer ?
Actually, I'm sorry but I'm only interested in the exact answear. Or at least a proof that it's impossible to find one.
@@rafciopranks3570 I agree with you but I have searched and I have not been able to find the exact answer in case I find it or manage to do it I will tell you
but look what she gave me with an approximation of 3 decimal places I would show it to you in a photo but on youtube I don't know well jeje (50-5(50+5(50+5(50+(2310)½)½)½)½)½/10
@@andresfelipemunoz4417 It cannot be. If we had a way to express it in terms of the radicals we would be able to construct a regular 180 -gon which is proved impossible.
1-x, what are you waiting for so long?
Fun fact: the 6 trig ratios of ANY multiple of pi/60 (3 degrees), for that multiple between 1 and 30, can be expressed in terms of nested radicals. All the other angles in between require you to take the cube-root of a complex number. An equivalent expression for sin(pi/60) is:
(1/8)*[sqrt(10+5*sqrt(3)) - sqrt(2+sqrt(3))-sqrt(2*(2-sqrt(3))*(5+sqrt(5)))]
You could probably calculate the cos(2pi/15). Answers (1/4)*sqrt(9-sqrt(5)+sqrt(30+6*sqrt(5)))
sin(3) on calculator:
0.14112000806
blackpenredpen:
*33 minutes and 15 seconds of description*
I asked my calc 1 professor in college to show me how to solve for sin or cos of any angle, she said she didn't know (?????). Didn't say anything about a taylor series or how there are ways to solve for it other than just looking at the unit circle or taking advantage of the difference function
An elegant way to combine euler formula and trigonometry 👍