Thank you for the shout out! 😺

@blackpenredpen

5 жыл бұрын

Tibees thank you for the t-shirt! I love it!!

@Mnemonic-X

5 жыл бұрын

@@blackpenredpen Do you really think that you do math with this absurd imaginary unit?

@user-vr3qg7xp5s

5 жыл бұрын

+Rahul Can't blame him lol

@gregorystocker971

5 жыл бұрын

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.

@abutaha4977

5 жыл бұрын

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

@AndDiracisHisProphet

5 жыл бұрын

Engineers more like "three is small, so sin(3°) = 3"

@maxsch.6555

5 жыл бұрын

lmao

@PuzzleQodec

5 жыл бұрын

@@AndDiracisHisProphet (Which is pi/60.)

@NateROCKS112

5 жыл бұрын

@@PuzzleQodec and the Fundamental Theorem of Engineering states that pi=3 so we plug it in and we get 1/20" = 0.05"

@edwardhuff4727

5 жыл бұрын

@@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*

@pedroandrade8727

5 жыл бұрын

lol

@blackpenredpen

5 жыл бұрын

Hahahhahahahha!!!!!

@poutineausyropderable7108

5 жыл бұрын

@@pedroandrade8727 sounds like a MC villager trying to scam you into giving some emerald.

@MrCoffeypaul

5 жыл бұрын

That was fast!

@pyglik2296

5 жыл бұрын

I was genuinely frightened for he NEVER stops in his videos :)

3:03 "Of course, 1 is equal to 2" -BPRP 2019

@anshumanagrawal346

3 жыл бұрын

3:03 actually

@not_allen1107

2 жыл бұрын

1 = 2

At 11:05 you can almost hear the cogwheels turning in his head...

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%

@Tactix_se

2 жыл бұрын

💀💀💀💀

@RealFreshDuke

2 жыл бұрын

@@analog_joe No, it's pi = 3.

@ajety

2 жыл бұрын

@@analog_joe Dude it's just a joke

@glendenog9095

2 жыл бұрын

​@@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 :)

@odysseasv7734

2 жыл бұрын

@@glendenog9095 HAHHAAHHAHAHHAAHAHAH WHAT

"1+1+1 =3" We did it boys, an A in maths 👏👏👏

@ffggddss

5 жыл бұрын

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

@THE-BIG-JP-REILS

5 жыл бұрын

ffggddss I HAD THE SAME EXACT THOUGHT

@colleen9493

5 жыл бұрын

Math*

@captasticts8419

4 жыл бұрын

@@colleen9493 both is fine

@kaaiplayspiano7200

4 жыл бұрын

18-15=3

Bprp: *Runs math channel like a boss* Also bprp: *1=2*

@aravindmuthu95

3 жыл бұрын

I almost read as "Bprp runs meth" 😂😂

Really shows you even an expert has troubled moments

@Roescoe

2 жыл бұрын

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.

@blackpenredpen

5 жыл бұрын

Eg. M wow, what a comment! Thank you!!

Me on exams: 11:03

@rafaelv.t1403

5 жыл бұрын

im the opposite

@alinajib4788

5 жыл бұрын

@@rafaelv.t1403 but you're gay

@andrewalex3581

5 жыл бұрын

Rafael V.T ok

@AbhishekKumar-jg7gq

2 жыл бұрын

😁😁😁😁😁😁😁😁

11:02 Me during an exam

@blackpenredpen

5 жыл бұрын

Zackary자카리 Me during a video.

@keescanalfp5143

5 жыл бұрын

@@blackpenredpen, great. I can't but admire that

@astralchan

5 жыл бұрын

@@blackpenredpen SENPAI NOTICED ME ~~

@kaaiplayspiano7200

4 жыл бұрын

@@astralchan is it legal to say Japanese-originated words to a chinese person?

@kaaiplayspiano7200

4 жыл бұрын

@@keescanalfp5143~ I CAN

"1 is equal to 2" - bprp 2019 Btw. Great video

@blackpenredpen

5 жыл бұрын

Sigma 1 thank you!

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)

@blackpenredpen

5 жыл бұрын

Perihelion Orbit yea. You can also use that picture to prove that formula.

@themcscripter2111

2 жыл бұрын

I think he was proving the sum and difference formula using complex numbers

@MattMcIrvin

Жыл бұрын

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.

@epikherolol8189

Жыл бұрын

@@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

@Zain-nc1ww

Жыл бұрын

@@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

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.

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

@gutschke

2 жыл бұрын

"If your only tool is a hammer, ..." And to be honest, Euler's formula does make for a wonderful hammer.

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...

@gumanelson2007

5 жыл бұрын

And then hope your pencil is infinitely sharp and the angle is perfectly 3 and you measure the distance very accurately.

@darkseid856

5 жыл бұрын

@@gumanelson2007 never knew that one can make angles using a ruler .

@akunog2708

5 жыл бұрын

@@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.

@darkseid856

5 жыл бұрын

@@akunog2708 yea that was basically what i was saying

@gumanelson2007

5 жыл бұрын

@@darkseid856 after using a compass or protractor

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) 😊

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.

@blackpenredpen

5 жыл бұрын

sharmsma yup that’s coming up soon

@DavidvanDeijk

5 жыл бұрын

(pi/60) - (pi/60) ^ 3 / 6 is a better estimate, expanded the taylor series by one

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 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.

Completing that rectangle was lovely. Well done.

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

Check out Toby's (Tibees) Channel: kzread.info/dash/bejne/X3Wtp5OQoMrVY7Q.html Purchase your Klein Bottle Kitty t-shirt here crowdmade.com/collections/tibees/products/tibees-toby-unisex-t-shirt Check out Chester's Channel: kzread.info/dash/bejne/q36d0rZwmtnAgdo.html Thanks! 100/(1-x)

@nuklearboysymbiote

5 жыл бұрын

Ok but to prove Euler's formula you NEEDED to already know the compound angle formulae. But here u r USING Euler's to show the compound angle… CIRCULAR REASONING

@obama2150

5 жыл бұрын

at 17:30 i think it would be better to sqrt the 16 to 4and the bottom. this would've made it easier later on at 30:34 since both partheses would have denominator of 8sqrt2 to make it much easier to simplify? since both sides will have a 4 and a sqrt2 gr8 vido m8 tho.

@hustler3of4culture3

5 жыл бұрын

Without watching I'm gonna say π /60.

@hustler3of4culture3

5 жыл бұрын

Ok this is better than my guess

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°

@lambda2693

2 жыл бұрын

Woah nice sol

@unidentifieduser5346

Жыл бұрын

my brain be like😮‍💨

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!

Exactly a perfect relation between complax analysis and real number , i love them ❤❤❤❤

I love the silence starting from 11:02 😂😂😂😂😂😂

Super fun video :) I love how you talk about angles like they are people

@blackpenredpen

5 жыл бұрын

Bayan Mehr hahaha thank you

No one: Minecraft Villager: 11:03

One of the best video ever upload on KZread. Thanks you

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

0.3M subscribers. Congrats!!!

@blackpenredpen

5 жыл бұрын

Yale NG yay thank you!

Congrats on 300K!!!

@blackpenredpen

5 жыл бұрын

JZ Animates thank you!!!!!

Congrats for gaining 300k subscribers 👏

Great! Let's find the relationship between sin(3°) and sin(15°) and construct the one-fifth angle formula!

@Drk950

3 жыл бұрын

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

@elias69420

2 жыл бұрын

@@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 :]

3:05 "One is equal to two"

@pietergeerkens6324

4 жыл бұрын

Astronomical Units: c = G = h = π = 1 = 2

厉害！But my abacus doesn't have the square root function, so I'm still unable to calculate the exact value.

@juxx9628

11 ай бұрын

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.

[Cos(x)+isin(x)]^n=cos(nx)+isin(nx Find formula cos (nx) For n is integer.

Mad props for not cutting the video when solving the problem.

Loved this! Please make more pure math content like this!

Awesome video bprp! You've really worked it out!

Men you deserve 1million subscribers and you deserved the position of my professor in calculus

(In funny, annoyed tone) No! Prove it all geometrically! :cat:

@blackpenredpen

5 жыл бұрын

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.

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.

Nice of you to prove the angle addition and subtraction trigonometric identities.

Thats really fantastic....you give us passion to learn new things....you've new subscriber from Aleppo, Syria 💐🌸

@blackpenredpen

5 жыл бұрын

Glad to hear! Thank you!

The video is old, but it contains valuable skills, and I benefited a lot from it. Thank you very much, Mighty Professor

You know it’s serious when he becomes blackpenredpenbluepen

you can also proof the cosine difference formula too and use sin(45 deg - 30 deg) and cos(45 deg - 30 deg) to find sin(15 deg) and cos(15 deg)

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.

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 !

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.

11:05 Fatal error

@blackpenredpen

5 жыл бұрын

Arthur Fonseca I knew it was 1-x but I forgot the reason lolll

This pause was very nice, it gave me just enough time to figure it out

I'm gonna thank you for this video. So glad that I created a graph in desmos that has 120 point unit circle coordinates.

Fun fact: We can only find the exact real-world value of the sine of something if it is a multiple of 3° (or is devided by a power of 2) If you have ever stumbled upon the triple-angle formula, and tried to reverse it, you know that trying to get sin(x) from sin(3x) gives you a third degree equation to solve. If you plug it in the Cardano-formula, you will always get a solution in the complex world, we cannot get a third-angle-formula in the reals, like we have a half-angle-formula. For the people interested: sin(3x)=3*sin(x) - 4*sin^3(x) gives sin(x)^3 - 3/4*sin(x) + 1/4*sin(3x)=0 Plugging it in the Cardano formula for sin(x), and simplifying, we get: sin(x) = 1/2 * [ cbrt( - sin(3x) + i * cos(3x)) + cbrt( - sin(3x) - i * cos(3x))] There will always be a number in the form a+bi under the cuberoot, and trying to plug it into Euler's formula will just give back that sin(x) = sin(x) If anyone knows how you'd get an exact soley real value for sin(1°), for example, please enlighten me.

@nuklearboysymbiote

5 жыл бұрын

If what u say is true, we can only bound sin(1°). That means get closer and closer approximations for the range of it, like what Chester did for sin(10°) on his channel

@ciberiada01

3 жыл бұрын

I don't know if it's only me, but as far as I know sin10° = (17427 − √3·√29·√2149867)/21600 and then I go with sin(10° − 9°) But personally I prefer sin10° = (5/12)²

@bobbyheffley4955

10 ай бұрын

You can use the half-angle formula to obtain values for quarter angles.

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.

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.

@SatyaVenugopal

5 жыл бұрын

That is kind of what he did. Just... geometrically

@christianalbina6217

5 жыл бұрын

Are we not able to to 45 degrees divided by 15 degrees or is that not allowed?

@darkseid856

5 жыл бұрын

@@christianalbina6217 boi thats not how it works ! (As much as I know it doesn't )

@ashtonsmith1730

3 жыл бұрын

if you want to do it with algebra you can, he did it with geometry

@joshmcdouglas1720

3 жыл бұрын

You could do this to find the values of sin15 and cos15 but you would need to use the angle difference identities again

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°)

11:03 .. A magical moment of thought .. See how your mind works :) Like it

@leonhardeuler6811

4 жыл бұрын

Can someone explain to me why he paused

@aldobernaltvbernal8745

4 жыл бұрын

@@leonhardeuler6811 to think

For cos15 and sin15, couldn't you just use compound formula again...cos(45-30) and sin(45-30)?

@jomama3465

5 жыл бұрын

@Blackpenredpen

@peterchan6082

5 жыл бұрын

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).

@Gold161803

5 жыл бұрын

@@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 :)

@peterchan6082

5 жыл бұрын

@@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)

@Gold161803

5 жыл бұрын

@@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

This is just wonderful.

Noting that (√ 5 - 1) = (√5 + 1 - 2) = 2⋅φ - 2 = 2⋅(φ - 1) = 2⋅√[ φ² - 2φ +1 ] = 2⋅√[ 2 - φ] and √[ 5 + √5 ] = √[ 4 + 2⋅φ ] = √2 ⋅ √[ 2 + φ] your final expression can be reduced to the nice anti-symmetric form [ √[2 - φ] ⋅ (√3 + 1) - √[2 + φ] ⋅ (√3 - 1) ] / 4√2. 😃

Love the way you base this proof on (1) = (2)

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.

@MattMcIrvin

Жыл бұрын

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.)

Those were some badass triangles, no doubt! Great vid, thanks

The Euler formula is much harder to prove than trig identities, bro!

This is awesome! Thanks for this solution.

An alternative method (just a suggestion, but may be somewhat easier). Sine and cosine of 36 deg is as easy as that of 18 deg. Then calculate cos(6 deg) = cos(36 - 30), using the known values for 30 deg angles. Then sin(3 deg) = sqrt ( (1 - cos(6 deg))/2 ).

It’s 4am, I have lessons at 9am, and idk why am I watching this now.

I finally understood one of your math videos, I'm so happy!! :)

Will you do multivariable calc vids ? Or pde?

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.

@artemetra3262

5 жыл бұрын

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)_/¯

@diegomullor8605

5 жыл бұрын

@@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.

@ningchin8476

5 жыл бұрын

@@diegomullor8605 That's why I endorse Aops. Check them out at aops.com They've been a life changer for me!

Wow this was awesome!

Thank you for all efforts God bless you

11:03 Literally me on my test two days ago.

Very nice to use Euler's formula and geometry 💕

@blackpenredpen

5 жыл бұрын

Sergio H yea!!

From the fundamental theorem of engineering, this trivially reduces to π/60 ~= 0.0523

@leonardoventura9641

3 жыл бұрын

No! From the fundamental theorem of engineering π=3 and = = ~=, so sin(30°)=π/60=3/60=1/20

You should do more videos on Feynman’s method bprp! I love all your videos tho.

6:20 thanks for a new way of proving angle difference. It blew my mind.😀🔥

This is evidence that mathematicians really don't mind that long walk for a short drink of water

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👍

after watching this ... I'm ready to consume 3 large pizzas (with each slice's tip at 18 degrees wide)

@blackpenredpen

5 жыл бұрын

nimmira hahaha nice!!

This calculation was so much fun. Thx

Very very nice method.. . u are best teacher

Starting from scratch would mean creating the number system, then algebra, basic principles of geometry then trig and then calculus. Now for the Taylor Series ( and yes I am going to say Taylor and not McLauren series) and then prove Euler's Formula. Man thats a lot of work. 🙃🙃

@blackpenredpen

5 жыл бұрын

Oh man! I don’t know I can do all that in 30 min

@jayapandey2541

5 жыл бұрын

However, doesn't it just amaze you when you look back at the journey of Mathematics?

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)))

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 KZread videos ever! Take my “Like,” Sir! 🤣🤣🤣🤣🤣🤣🤣

You are a nice teacher !!! Good luck

Wow this is really nice thanks for sharing this...

@blackpenredpen

5 жыл бұрын

maths- physicshub thank 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)

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

11:04 - I like my math videos like I like my Jerry Springer videos: Raw and Uncut.

@blackpenredpen

5 жыл бұрын

Edward Sanville I once put “raw footage” in my title but YT demonetized it.

@pianoforte17xx48

3 жыл бұрын

@@blackpenredpen filthy youtube

You tried to sneak in the true proof of the angle addition formula with the boxes, and you thought we wouldn't notice!

@blackpenredpen

5 жыл бұрын

noahtaul hahahaha yea

Gracias. Aprendi, sin querer, a demostrar la identidad cos(a-b) y sen(a-b). En serio, genial!!

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)

every triangle is special for me

Time to grab some popcorn 🍿