Thank you for the shout out! 😺

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

11:03 *thought download begins* 12:07 *thought download complete*

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

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

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

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%

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

Really shows you even an expert has troubled moments

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...

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

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.

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)

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

11:02 Me during an exam

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.

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

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.

Completing that rectangle was lovely. Well done.

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 : )

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

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!

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

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.

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

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

No one: Minecraft Villager: 11:03

Congrats on 300K!!!

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

Congrats for gaining 300k subscribers 👏

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

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

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

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

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

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

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

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

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.

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

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

0.3M subscribers. Congrats!!!

Thank you for all efforts God bless you

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

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.

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

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

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

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

3:05 "One is equal to two"

Wow this was awesome!

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

Very very nice method.. . u are best teacher

This is awesome! Thanks for this solution.

Check out Toby's (Tibees) Channel: kzfaq.info/get/bejne/Y6epeJSI1c3TY4U.html Purchase your Klein Bottle Kitty t-shirt here crowdmade.com/collections/tibees/products/tibees-toby-unisex-t-shirt Check out Chester's Channel: kzfaq.info/get/bejne/r7CZo7doz9y-gas.html Thanks! 100/(1-x)

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

This calculation was so much fun. Thx

Very nice to use Euler's formula and geometry 💕

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.

This is just wonderful.

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 ).

Will you do multivariable calc vids ? Or pde?

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 !

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

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

Very good, I like a lot. A hard work

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

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

@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!!!

You know it’s serious when he becomes blackpenredpenbluepen

An elegant way to combine euler formula and trigonometry 👍

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

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

Nicely done!

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

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

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.

That's so nice!!!

I am from India. Your explanation is really awesome. It's very nice. I haven't words for appreciation. Awesome awesome awesome.........................

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

Wow thanks for sharing this

Thank you

Great video ! ❤️

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.

That was so much fun!

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.

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!

thank you

Thanxx it's helpful

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)

It’s funny the toby(tibees) looks like a new function

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.

also out of interest, what degrees/education things (phd, a level, etc.) do you have in maths? you seem really good at it