exact value of sin(10 degrees)

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blackpenredpen

blackpenredpen

Күн бұрын

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@matonphare
@matonphare Жыл бұрын
13:00 for that we can use the polar form of each numbers. For the first, we get e^(2*i*pi/3), then we divide by 3 the exponent (because of the cube root). So at the end we get 1/2(e^(2i*pi/9)+e^(-2*i*pi/9)) Then we can use the trig form and we see that the imaginary parts cancels: (cos(2pi/9)+isin(2pi/9)+cos(2pi/9)-isin(2pi/9))=2cos(2pi/9) Then we multiply by 1/2 and we get cos(2pi/9)≈0.766 In degrees, it gives us cos(40°) which is indeed sin(50°) using trig identity, like in the video
@blackpenredpen
@blackpenredpen Жыл бұрын
Ah! Thank you for the answer and it’s very nice. Could we write that as a radical expression without the i tho?
@adityagiri3600
@adityagiri3600 Жыл бұрын
​@@blackpenredpen great video, I think there is a trade off. one would either have a closed form in trigonometric functions or as roots of complex numbers, both related to rotations. I'm just speculating, not sure if its possible to have a real closed form under radicals, very fun video nevertheless!
@mhm6421
@mhm6421 Жыл бұрын
But the second one gives 10 and 80 degrees?
@deinauge7894
@deinauge7894 Жыл бұрын
you can use the Real part. because the second part is just the conjugate of the first. sin(x/3) = Re[cuberoot(-sinx + i cosx)] and then check all branches of the cuberoot
@matonphare
@matonphare Жыл бұрын
@@mhm6421 With the same logic, using the polar from of W1=-1/2+i*sqrt(3)/2=e^(2i*pi/3) We get : 1/2(e^(2i*pi/9)*e^(2i*pi/3)+e^(-2*i*pi/9)*e^(-2i*pi/3)) which reduce to -cos(pi/9) using euler’s formula (the imaginary part cancels) which is -cos(20°) and this is equal to -sin(70°) using trig identities. Using the same process with W2, we get sin(pi/18) which is sin(10°)≈0.174
@paradoxicallyexcellent5138
@paradoxicallyexcellent5138 Жыл бұрын
Man, the people who discovered this stuff without the use of calculators to sanity check what they were doing... hats off.
@neutronenstern.
@neutronenstern. Жыл бұрын
you can always draw a triangle and measure the sidelength
@mathisnotforthefaintofheart
@mathisnotforthefaintofheart Жыл бұрын
No social media in those days, so they had plenty of time...:)
@cxpKSip
@cxpKSip Жыл бұрын
​@@neutronenstern. It is impossible to trisect an angle with just a ruler and compass.
@neutronenstern.
@neutronenstern. Жыл бұрын
@@cxpKSip ok, but you can measure the length of a segment of a circle, which is then the angle in radiants. (you could use a rope to measure the length). So to get 10° which is π/18 in radiants, you can draw a circle with radius 1, and then take a rope of length of π/18 and lay it on the circle. Its endpoint is where 10° are. And you had rulers for ages, and π/18 can be calculated easily on paper.
@neutronenstern.
@neutronenstern. Жыл бұрын
@@cxpKSip also there are other ways, like the taylor series. Also they had tables with precalculated values.
@DDroog-eq7tw
@DDroog-eq7tw Жыл бұрын
Finally, we can write out the value in roots of sin(1°), which we could then use to get the sine of any integer (despite lengthy expansions.
@renhaiyoutube
@renhaiyoutube Жыл бұрын
I wonder if sine of (any integer in degrees) is always an algebraic number.
@gagadaddy8713
@gagadaddy8713 Жыл бұрын
Yes! Don't spill the bean, bprp is going to release another more general formula for sin(x/n) ..... 😗
@artemetra3262
@artemetra3262 Жыл бұрын
i am doing my research into this! and yes, i got the value of sin(1°) but it's WAY too big to write down here. i will copy paste it here soon
@artemetra3262
@artemetra3262 Жыл бұрын
​@@renhaiyoutube sin(p/q * π) for any p, q from Z (rational multiples of π) is an algebraic number, since there are n-angle formulas that express sin(nx) as a polynomial in terms of sin(x)
@paradoxicallyexcellent5138
@paradoxicallyexcellent5138 Жыл бұрын
I thought all of our known angle values have a multiple of 5 degrees in them? Don't we need the root of a 5th degree polynomial to get down to a trig value at 1 degree?
@micronalpha
@micronalpha Жыл бұрын
I really enjoy when you keep things real and show the struggles you had to reach the result and not only the cleaned-up result. As a math teacher I relate and get a better appreciation of your video. 谢谢
@xozu
@xozu Жыл бұрын
he didn't make this complex...
@bartekabuz855
@bartekabuz855 Жыл бұрын
In Cardano's formula if complex number show up it is not possible to get an answer than involves only reals and radicals.
@mathisnotforthefaintofheart
@mathisnotforthefaintofheart Жыл бұрын
Why not? This is factually untrue. Tartaglia used a cubic equation with only REAL RATIONAL solutions to test the formula and through the use of complex numbers only then he was able to verify the real solutions (after algebra ultimately the imaginary terms would cancel) It is precisely THIS reason why complex numbers ultimately had to be accepted as a new branch of math, as Tartaglia's formula from an algebraic standpoint was correct
@moisesbravopalma9288
@moisesbravopalma9288 6 ай бұрын
​@@mathisnotforthefaintofheart la duda verdadera, radica en cómo sacar la raiz cúbica de manera general a los números complejos. Y saber a cuanto equivale la parte real de la compleja.
@youssefayman78
@youssefayman78 Жыл бұрын
You can reach a formula that always works this way: You can begin using Euler’s formula and say that: Sin(x/3)=(1/2i)(e^(ix/3)- e^(-ix/3)) Next, you can just say that e^(ix/3)= cuberoot( cosx + isinx), and the same can be done to e^(-ix/3). This gives us the formula: Sin(x/3)=(1/2i)(cuberoot( cosx + isinx)- cuberoot( cosx - isinx)) This formula only works for x less than pi but greater than 0. If you want a formula for x less than 2pi but greater than pi, you should multiply the first term in the formula by w1, and multiply the second term by w2, where w1, and w2, are the complex cubic roots of unity. So we get the formula: Sin(x/3)=(1/2i)((-1/2+√3/2 i)cuberoot( cosx + isinx)- (-1/2-√3/2 i)cuberoot( cosx - isinx)) If you want me to explain why these formulas always work, feel free to reach out to me as it would be hard to explain here in the comment section. It has something to do with the rotations in the complex plane.
@youssefayman78
@youssefayman78 Жыл бұрын
If you put x=30 degrees in the first formula you will get sin(10 degrees). If you put x=150 degrees, you get sin(50 degrees). Finally, if you put x=210 degrees in the second formula you get sin(70 degrees). I used wolfram alpha to calculate the results.
@harryliang9263
@harryliang9263 Жыл бұрын
It give me the feel, we don't solve the problem. We just hide the complexity into calculting the cube root of complex number. cuberoot(cosx - isinx)) is almost the same thing as caculating sin(x/3).
@ViguLiviu
@ViguLiviu Жыл бұрын
Too complicated. sin(x/3)=sin(x/3)+1/2icos(x/3)-1/2icos(x/3)= croot(sinx+icosx)+croot(sinx-icosx)
@Gunslinger-us1ek
@Gunslinger-us1ek 6 ай бұрын
yo i just did that and i thought at least someone must've noticed something so obvious
@djttv
@djttv Жыл бұрын
A basic difficulty I have about this is: y=x^3+px+q has constant p and q. But here we have q=q(x). Typically we're trying to find the roots of a fixed polynomial. We vary the x and get the y value along a specific curve, and we're looking for where y=0. Here, as we vary x, the whole curve changes because q=q(x). Or in terms of 2nd order poly.: y=Ax^2+Bx+C(x). We could just use the quadratic formula, and get a result, but thinking about what exactly we are doing is a bit confusing. Could even consider 1st order: y=ax+b. Root formula would be x=-b/a. Now we try it with y=ax+b(x), and use the same formula, so x= -b(x)/a. I'll have to think about this a bit more. Any comments? Thanks
@djttv
@djttv Жыл бұрын
After writing all that, I figured it out! Theta is not varying. It is fixed at 10deg. Then sin(3*10) is a known value. Sin(10) is not known. The domain axis could be labeled sin(10), or to avoid my confusion, I would rename it u. Then we are solving 0=u^3-(3/4)u+(1/8). The u that satisfies this equation is the value of sin(10).
@jb31842
@jb31842 Жыл бұрын
@@djttv That non-constant q bothered me too. I think I half understand your followup. I do think this would be a great topic for BPRP to elaborate on in a future video.
@TheDannyAwesome
@TheDannyAwesome Жыл бұрын
I've always had troubles with the cubic formula, and I believe it's all because weird stuff goes on with branch cuts when you have to take a sum of two nonreal cube roots. You'd have to go back to the derivation of the formula, and make sure everything is defined over a branch cut C^3 -> C so that you make sure you equate the compatible cube roots. It all feels very messy and fiddly and I don't want to be the one who has to delve into it.
@renhaiyoutube
@renhaiyoutube Жыл бұрын
Yes, and that's why the cubic formula is painful and almost nobody uses it in applications and instead just plugs the solution into WolframAlpha. Sad but it has to be this way.
@kamrujjaman1585
@kamrujjaman1585 Жыл бұрын
I have memorised the cubic formula and I can prove the formula
@jomariraphaellmangahas1991
@jomariraphaellmangahas1991 Жыл бұрын
Well, I love this I think your next video is to finally get the exact value of sin1° that is mixed with pythagorean theorem, your special special right triangle of 15-75-90, 18-72-90, 3-87-90 Then you will prove cardano's cubic formula, the trigonometric identities and more
@blackpenredpen
@blackpenredpen Жыл бұрын
I don’t think my board is big enough for sin(1 deg) 😆
@Static_MK3Focus
@Static_MK3Focus Жыл бұрын
I’m finally far enough in my trig that, I was actually able to jump ahead of blackpen and derive the equation myself and then check my solution, just needed a push in the right direction. Well done Blackpen🤝
@atzuras
@atzuras Жыл бұрын
Since the roots of a polynomial are not sorted and are algebraicaly indistinguishable, I claim there is no way to know what is the correct root for a given angle in advance. But you can have a set of solutions called red, blue, and black, which is kind of better.
@CDChester
@CDChester Жыл бұрын
always knew a video like this was coming
@yoav613
@yoav613 Жыл бұрын
This is nice,but at the end what you get is cos80=sin10 .the formula you get at the end after simlifying is :0.5(e^(4pi/9)i+e^-(4pi/9)i) which is cos(4pi/9)=cos80=sin10,so there is nothing special here,but the way you got it was nice
@laurentwatteau8835
@laurentwatteau8835 Жыл бұрын
Excellent. Thanks very much ! 🙂
@Ninja20704
@Ninja20704 Жыл бұрын
I’m not sure how correct I am, but i read on a wikipedia page that unfortunately according to Galois theory, for cubic equations with all 3 roots being real irrational numbers, the roots cannot be expressed with a finite radical expression using only real numbers. So unfortunately, we can’t write sin(10 degree) in a finite radical expression without the i. If someone could fact check me, that would bemuch appreciated
@blackpenredpen
@blackpenredpen Жыл бұрын
Oooh! That could be the reason! I will check on wiki later too! Thanks.
@jorgegarcia_dragon
@jorgegarcia_dragon Жыл бұрын
It Is possible, to express sin 10°, but you have to use infinite nested radicals
@jorgegarcia_dragon
@jorgegarcia_dragon Жыл бұрын
Another approach is to use hypergeometric function to calculate the real root of the Cubic equation 😊
@theskinegg9168
@theskinegg9168 Жыл бұрын
11:16 you could prob put the terms inside the radicals in terms of cis to simplify the radicals
@galvanaxchampion2751
@galvanaxchampion2751 Жыл бұрын
this is a lot of formulae . It's better if you use euler's formula and some rearangements ; like expressing sine of x|3 as -i ×half of e^ix|3 - e^ix|3 then writing i as e^pi|2, -i as e^-pi|2. This reduces to your obtained answer.
@jsmsj
@jsmsj Жыл бұрын
We could have just used euler form and demoivre's theorem
@khanhtran-gw3pm
@khanhtran-gw3pm Жыл бұрын
If we did that to the formula I think a sin(x/3) germ would pop out (try for yourself), which ruins the point
@DubioserKerl
@DubioserKerl Жыл бұрын
Do you go from sin(50°) to sin(-70°) and then to sin(-190°)=sin(10°) by utilizing the cube roots of 1 because those cube roots are 120° apart?
@WindsorMason
@WindsorMason Жыл бұрын
Yes! 360/3 = 120, so the three values that cube to the same number are 120° from each other.
@kahan1938
@kahan1938 Жыл бұрын
13:00 you can set the expression in parenthesis to x and then ''cube both sides'' similiar to this vid: kzbin.info/www/bejne/a6HEYnSort6fpq8 You can then substitute x to get a cubic equation with real coefficients. However it seems like it will have 3 roots, probably because we cubed both sides. Which root is the correct answer? I dunno
@Sphinxycatto
@Sphinxycatto Жыл бұрын
He just weaves in all conpcept in so easily It seems every sentence is just a revision of a topic
@ancryne
@ancryne Жыл бұрын
I love math.Thank you sir... ❤❤
@liamdonegan9042
@liamdonegan9042 Жыл бұрын
You love math. I love meth. We are not the same.
@koendos3
@koendos3 Жыл бұрын
Lamps love moths, or the otherway around
@jomariraphaellmangahas1991
@jomariraphaellmangahas1991 Жыл бұрын
I literally plug sin(x/3) in the calculator earlier This help me a lot
@denelson83
@denelson83 Жыл бұрын
Those two complex numbers you wrote in your expression for sin 10° have magnitude 1, so you might try writing them in polar form and dividing their arguments by 3.
@farrattalex
@farrattalex Жыл бұрын
Very interesting. I think one problem could be that at around 9:40 you say sqrt(-cos^2 x) is i*cos x. That's not correct; you are missing abs(...). sqrt(-cos^2 x) is +-i*abs(cos x). In the situation you are calculating that it makes no difference because there is +- in front. However, later at 10:38 it makes a difference.
@tunafllsh
@tunafllsh Жыл бұрын
Yea there were a lot of things missing. When working with complex numbers all roots give multiple answers.
@tanhrs8711
@tanhrs8711 11 ай бұрын
14:00 it is because you forgot the 1/64 perviously. + to check it the number is real simply check if it is equals to its cojuge aka replacement every "i" with "-i"
@lychenus
@lychenus Жыл бұрын
This is very similar to a Hong Kong 2007 grade 13 A-level mathematics problem. I regret I had arithmetic panic that time and I didn't score this. The whole paper is damn easy
@BilalAhmad-u6w2u
@BilalAhmad-u6w2u Жыл бұрын
Good writing sir❤❤❤
@michael161
@michael161 Жыл бұрын
Math is so beautiful and helpful in our life.❤️❤️❤️
@masterclash9959
@masterclash9959 Жыл бұрын
If you want to take the cube root of those imaginary numbers, you’re going to have to use DeMoivre’s Theorem. It states the following: For any number a+bi, it can be represented in the form r(cos(a) + isin(a)) or rcis(a). Additionally, for any complex number in this form that has a power applied to it, (r*cis(a))^n = r^n * cis(n*a). For example, if we are to take the fourth root of 1, we know that 1 equals 1(cos(0) + i*sin(0)), or cis(0). Keeping in mind that we can rotate this angle 360 degrees, we can represent 1 as cis(0), cis(360), cis(720), and cis(1080) (you’ll see why we do this in a second). Because we take 1 to the one fourth, or 0.25th power, that means: 1^0.25 = (cis(0))^0.25 = cis(0*0.25) = cis(0) = 1 We can also repeat this for the other reference angles: (cis(360))^0.25 = cis(360*0.25) = cis(90) = i For cis(720), the same process results in cis(180), or -1, and for cis(1080), this results in cis(270), or -i. As such, we can get that the fourth roots of one are 1, i, -1, and -i. Hope this helps!
@XJWill1
@XJWill1 Жыл бұрын
It is impossible to write an expression for a non-multiple of 3 degrees for sine without having a cube root of a complex number (or worse). By the way, here are some expressions for sine of non-multiple of 3 degrees that I find interesting. Note that you can copy and paste them into Wolfram Alpha in order to visualize them (and verify that they work). You have to be careful when taking the cube root of a complex number below. Choose the "principal value" (since there are 3 possibilities, but only one fits the conventional definition for principal value). I avoid writing "i" to make the syntax more universal for computer entry. Also, sind() is the sine degrees function. First the simplified version of what was derived in the video: sind(10)= sqrt(2-(1/2+sqrt(-3)/2)^(1/3)-(1/2-sqrt(-3)/2)^(1/3))/2 Now here is an alternative form that only has one cube-root in it: sind(10)= 1/sqrt(1-(1-4/(2-( 4+sqrt(-48))^(1/3)))^2) And finally, the holy grail, for 1 degree: sind( 1)= 1/sqrt(1-(1-16/(8-(4+sqrt(-48))^(1/3)*(sqrt(-1)-sqrt(-5)+sqrt(10+sqrt(20)))))^2)
@XJWill1
@XJWill1 Жыл бұрын
By the way, since computing a numerical value of the cube root of a complex number implicitly involves trigonometric functions, these expressions do not really provide a way to compute a number for the sine of 10 degrees using only arithmetic (and not trigonometric approximations).
@paradeeza
@paradeeza Жыл бұрын
Yes ! Your solution for sin(1°) works ! It seems that our man redpenblackpen overlooked some terms in the expression on the whiteboard. It lacks sqrt(2 - ).
@paradeeza
@paradeeza Жыл бұрын
Hence, should be sqrt(2 - )/2.
@Zoyfad
@Zoyfad Жыл бұрын
w0 works for the range 90-180, w2 for the range 0-180. I have a code in R plotting all the options. I will post the code if there is interest.
@Alex-vH
@Alex-vH Жыл бұрын
Exact expression for sin(10): Use: cos(80) = sin(10) Use: half angle formula for cos: cos(A/2) = +/- sqrt((1 + cos(A))/2). Set x = 2/A, multiply by 2, assume 0
@muskyoxes
@muskyoxes Жыл бұрын
I can just use the power series of sin to get a decimal approximation. I was hoping this would end with some sort of real closed-form radical expression. (knowing my luck, if i tried to manipulate the imaginaries in the radicals with polar forms and such, i'd end up with sin (10) = sin (10) )
@parimalpandya9645
@parimalpandya9645 Жыл бұрын
What is the alternative solution of the strend problem solved by ramanujan?
@mathmanic1438
@mathmanic1438 Жыл бұрын
Alternate Method: Let θ = 10° sin θ = sin 10° sin 3θ = sin 30° 3 sin θ - 4 sin³θ = (1/2) 6 sin θ - 8 sin³θ = 1 8 sin³θ - 6 sin θ + 1 = 0 Let sin³θ = t 8t³ - 6t + 1 = 0 Solve the above cubic equation to get 3 solutions. The solution which is closest to 0.174 will be the value of sin 10 degrees.
@wes9627
@wes9627 Жыл бұрын
"Can we tell which solution will give sin(10°)?" I may have an answer to your question. When solving for cos(θ/3) we have x_j = [cos(2πj/3 + θ/3) + i sin(2πj/3 + θ/3) + cos(-2πj/3 - θ/3) + i sin(-2πj/3 - θ/3)]/2, j = 0, 1, 2 When j = 0 we get x_0 = [cos(θ/3) + cos(-θ/3)]/2 = cos(θ/3). So, when solving for cos(θ/3) use the first root corresponding to j = 0. But, when solving for sin(θ/3) things are a bit more complicated... x_j = [cos(2πj/3 + γ/3) + i sin(2πj/3 + γ/3) + cos(-2πj/3 - γ/3) + i sin(-2πj/3 - γ/3)]/2, j = 0, 1, 2 where γ = θ + π/2. Then, after combining terms, substitute θ + π/2 back for γ and set j = 2 to get x_2 = [cos(3π/2 + θ/3) + cos(-3π/2 - θ/3)]/2 = sin(θ/3)! In summary when solving for cos(θ/3) set j = 0 and when solving for sin(θ/3) initially replace θ with γ - π/2 and set j = 2. Then replace γ back with θ + π/2 and simplify. So, similar to what you observed, solving for cos(θ/3) requires the first root (j = 0) and solving for sin(θ/3) requires the third root (j = 2).
@MathOrient
@MathOrient Жыл бұрын
Interesting results :)
@sacielo
@sacielo Жыл бұрын
once you have the sinx+icosx under roots, can you use euler identity to simplfy it further?
@carultch
@carultch Жыл бұрын
Yes. It's called DeMoivre's theorem.
@victortesla2935
@victortesla2935 Жыл бұрын
8x³-6x+1=0 Where x=sin(10) This equation also have the same three roots We can also get approximate value by (10×π)÷180
@cruzcostilla9796
@cruzcostilla9796 Жыл бұрын
Hi Blackpenredpen. Make a video demonstrating why π is irrational. I am subscribed to your channel, very good videos. Greetings from Mexico
@stevemaskal5933
@stevemaskal5933 Жыл бұрын
You should have pointed out that sin 50 = sin 130 and -sin 70 = sin 250. The three values you calculated are for angles all 120 degrees apart. 10, 130 and 250 degrees.
@fishstab5406
@fishstab5406 Жыл бұрын
Congrats on 1000000ln3 subs🎉🎉🎉
@takyc7883
@takyc7883 Жыл бұрын
You can use de movies theorem to find the formula for sin(x/3) much faster, and for the final bit you can use polar exponent form. I know you didn’t want to go into complex too much because that restricts the viewers who understand it and you’d have to teach them an entire new subject which is fair.
@Firefly256
@Firefly256 Жыл бұрын
9:57 why is the absolute value not needed?
@jonathanchen1834
@jonathanchen1834 9 ай бұрын
since theres +/-, absolute value will do nothing
@danielkovacs6809
@danielkovacs6809 Жыл бұрын
Hi! I made up a function that is almost perfectly equal to e^(-x^2), hoping for that we'll be able to come up with an elementary integral function for it. The function is 9/(7+2,8^(2,8x)+2,8^(-2,8x)), I created it using the reciprocal of cosh(x). How should I go on to make it more (or perfectly) equal?
@nathansweetman5101
@nathansweetman5101 Жыл бұрын
Try it
@Real_Cotton
@Real_Cotton Жыл бұрын
I was wondering if you could do this problem sometime. This is a very interesting problem, and I can't get the proof anywhere. Prove that a^x = log (base a) x (a
@outofthebots3122
@outofthebots3122 Жыл бұрын
being a cube root it has 3 solutions for each cubic where 2 of the solutions are sin(10) = 1/2 [ cos(80) + i sin(80) + cos(80) + i sin(-80) ] which simplifies to cos(80) and cos (-80) so it is a long way of proving sin(10) = cos(90 - 10) and sin(10) = cos(10 - 90) surprising the other solutions that produce real solution are cos (40), cos(-40), cos(160), cos(-160)
@0xDolve
@0xDolve Жыл бұрын
Keep it up bro, What's the best books to learn advanced mathematics after graduating high school?
@eckhardtdom
@eckhardtdom Жыл бұрын
This is me every other class but math 😂
@aditya-u7s9v
@aditya-u7s9v Жыл бұрын
Another method to evaluate the complex sued is by assuming it to be u Then u^3 = -1 + 3(cbrt(-1/2 + iroot 3/2)(cbrt(-1/2 -root3/2)u U^3 = -1 + 3(1)u So this becomes a polynomial that is p(x) = u^3 -3u +1 Then again applying cardanos formula for depressed cubics you can get an answer @blackpenredpen I just checked with wolfram alpha I got an 3 real roots and when you divide one of the answers namely 0.3470 by 2 in the original formula you get 0.1735 which is very close to sin 10 degrees
@ravitripathy2479
@ravitripathy2479 Жыл бұрын
Very informative and nice idea
@RabiSankarTripathy
@RabiSankarTripathy Жыл бұрын
Was thinking along the same lines
@theMathSinger
@theMathSinger Жыл бұрын
Interesting and very well done. Thank you for this!
@blackpenredpen
@blackpenredpen Жыл бұрын
Thank you! : )
@Humanly-whatever
@Humanly-whatever Жыл бұрын
Blackpenredpen: I don't know what the exact value That Euler's formula: Hold my beer
@Porkfu
@Porkfu Жыл бұрын
for all my fellow irresponsible engineers out there, the answer is about pi/18, basically 1/6 😊
@varun3282
@varun3282 Жыл бұрын
I was just about to tell you to use cube root of unity, but then you figured it out yourself
@Theraot
@Theraot Жыл бұрын
This works from 0 to π: sqrt(1 - (1/2 (1/(cos(x) + sqrt(-1 + cos^2(x)))^(1/3) + (cos(x) + sqrt(-1 + cos^2(x)))^(1/3)))^2) And this works from π/2 to 2π + π/2: 1/(2 (-sin(x) + sqrt(-cos^2(x)))^(1/3)) + 1/2 (-sin(x) + sqrt(-cos^2(x)))^(1/3)
@sriprasadjoshi3036
@sriprasadjoshi3036 Жыл бұрын
This man is mathing which we cant even math about mathematics
@SummerFrost23
@SummerFrost23 Жыл бұрын
The final solutions can be changed into hyperbolic functions sinh (ix) = i sin (x), cosh (ix) = cos (x) to verify your answer , probably without the use of calculator.
@SummerFrost23
@SummerFrost23 Жыл бұрын
Reversely, You can reach the same answer directly using hyperbolic functions.
@joshdanziger8367
@joshdanziger8367 Жыл бұрын
There's a nice way of getting there by looking at the properties of the 80:80:20deg isosceles triangle: if the short (unequal side) has unit length, then the two equal sides have length 3-x^2, satisfying the equation 3-x^2=1/x. sin 10 is then given as x/2. Unfortunately you still get a cubic equation to solve to find x, and it's still not immediately obvious from the solutions that x is real. I wanted to post a picture to show this but I don't know how to; so I'll have to describe the construction: if you bear with me, once you've done the construction it's very nice ... Construct triangle ABC with angle 20 at A and the other two angles 80; choose BC to be unit length. All construction lines are inside ABC: first draw a line B at 20 deg to BC; this meets AC at D; let x be the length BC; now draw a line from D at 60 deg to BD; this meets AB at E; now draw a line from E at 100 deg from BE meeting AC at F; finally draw two lines from F, one at 60 deg to EF meeting AB at G, the other at 80 degrees to EF meeting AB at H. You should now be able to satisfy yourself that triangle BDE is equilateral and triangles ABC, FGH, BCD, AGF and EFH are similar isosceles triangles, with the last three congruent. Since additionally DEF is isosceles, BD, BE, DE, EF, AF, AG, EF and EH are the same unit length as BC, and because BCD, AFD and EFH are congruent, FG and FH are the same length as BC, namely x, and the similarity of FGH to this triangle allows us to deduce GH is x^2. Finally the similarity between BCD and ABC gives the result 3-x^2=1/x ==> x^3 - 3x +1 = 0, with sin 10 = x/2
@joshdanziger8367
@joshdanziger8367 Жыл бұрын
Correction: BCD, *AFG* (not AFD) and EFH are congruent
@joshdanziger8367
@joshdanziger8367 Жыл бұрын
.. and the first construction line runs *from* B at 20 deg to BC; and *CD* is the length x, not BC. Playing around with this technique I can also find sin (90/7), ie, sin(12.857 ....) from the irreducible cubic x^3 - x^2 - 2x + 1=0: sin(90/7) = x/2.
@tarkeshwarsingh9051
@tarkeshwarsingh9051 Жыл бұрын
By Ramanujan theorem : Sinθ=θ-(θ³/3!)+(θ⁵/5!)+(θ⁷/7!) Note:- Theta's value(input) will be in radians but output will it give in degrees {make sure that you putted value in rad}
@programmer4047
@programmer4047 Жыл бұрын
You can only approximate the value of sin with that formula. It can't tell you the exact value
@AdityaRaj-vd6lu
@AdityaRaj-vd6lu Жыл бұрын
His name should be written Guinness world record for the fastest pen changing
@mokouf3
@mokouf3 Жыл бұрын
Sorry for asking an unrelated question. There is a statement "For complex numbers, square roots can only accept principal values as output." Is that statement true? Example: √-1 = i, but not -i, while i² = (-i)² = -1 If yes, some of your previous video may have wrong answers!
@romanvolotov
@romanvolotov Жыл бұрын
bro really cube rooted the cube root of unity at some point
@EliteCubingAlliance
@EliteCubingAlliance Жыл бұрын
I have been watching your videos since I was about 13. Now I am 16 and I can gladly say that I can understand almost all of your videos. (Some of them still confuse me, but just wait another year!!!)
@busycow8334
@busycow8334 7 ай бұрын
While trying to simplify the complex solution. I got cos(2pi/9) in the way and it is interesting that it is equal to exactly sin50 degrees
@wes9627
@wes9627 Жыл бұрын
This cubic equation has three roots, which would be [e^(i2π/9) + e^(-i2π/9)]/2, [e^(i8π/9) + e^(-i8π/9)]/2, and [e^(i14π/9) + e^(-i14π/9)]/2. The third of these roots gives cos(280°) = sin(10°) Before taking the cube roots we have e^[±i(2π/3 + 2πj)], j = 0, 1, 2. The three roots above come from e^[±i(2π/3 + 2πj)/3], j = 0, 1, 2.
@jeremyc6054
@jeremyc6054 Жыл бұрын
my approach was to look at those two radicals in polar form, and realize that each has three roots in the complex plane, all equal in magnitude but splayed out 120 degrees from one another. the magnitude of the quantity in each radical is 1, so you just have to determine the three phase angles. for the first radical, the possible phase angles are 2pi/9, 8pi/9, and 14pi/9. (roughly 40, 160, and 280 degrees if you prefer thinking in degrees) for the second radical, the possible phase angles are 4pi/9, 10pi/9, and 16pi/9. (roughly 80, 200, and 320 degrees) If you draw these in the complex plane, you'll see that each triad is splayed out 120 degrees (2pi/3) about the origin. Then, you need to add these two values. Consider all possible combinations of the three, and figure out which ones have wholly real-valued sums. This is easy, just pick off the ones in "group 1" and "group 2" which have equal and opposite imaginary values. There are three possible additions which result in a real-valued result: (1) 2pi/9 from the first radical with 16pi/9 from the second radical. (2) 14pi/9 from the first radical with 4pi/9 from the second radical. (3) 8pi/9 from the first radical with 10pi/9 from the second radical. Each of these pairs has equal and opposite imaginary components, so all you have to do is add the real components. It turns out that in each pair, the real values are identical, so you pick one and double it. And then as the last step you multiply by 1/2 to complete the expression from the video. combination (1) gives you sin(50). (coming from sin(150) in the 1/3-angle formula) combination (2) gives you sin(10). (coming from sin(30) in the 1/3-angle formula) combination (3) gives you sin(70). (coming from sin(210) in the 1/3-angle formula) The problem with Wolfram is that it arbitrarily chooses one of the three roots and discards the others. Apparently, the roots chosen correspond to combination (1), which is why you got the result for 50 degrees. In this case, combination (2) is the one you want. (my final quandary was that there are FOUR angles which have sines of +-1/2: 30, 150, 210, and 330 degrees. The first three are represented in the combinations above, but I couldn't find a combination corresponding to 330 degrees. That, in turn, would give us sin(110) when put through the 1/3-angle formula. However, sin(110) = sin(70) so that simply collapses into combination (3) above). This continues: sin(390) degrees is also equal to 1/2, and plugging that into the 1/3-angle formula would give you sin(130), but that's the same as sin(50), so it collapses into combination (1). And sin(360+150) = sin(510) = 1/2, which would give sin(170) after dividing the angle by 3, but that's exactly the same as sin(10) which corresponds to combination (2). This continues forever....)
@jeremyc6054
@jeremyc6054 Жыл бұрын
an interesting corollary: 1^(1/3) - 1^(1/3) = ? you might reflexively say 0, but that's only one of the possible answers. the cube root of 1 can be 1, but it can also be (-1/2 + i*sqrt(3)/2), and (-1/2 - i*sqrt(3)/2). Considering all possible combinations of the above, you can get zero, but also answers like (3/2 + i*sqrt(3)/2). All told, there are nine possible combinations with seven unique answers - 0 (coming from three of the combinations), and six complex values forming a hexagon about the origin (all with magnitude sqrt(3, and phase angles of 30, 90, 150, 210, 270, and 330 degrees).
@ManjulaMathew-wb3zn
@ManjulaMathew-wb3zn Жыл бұрын
Just substitute. a fot the first radical and b for the second one. Then the product ab would be the cubic root of the product of conjugated which is real. Then a^3 +b^3 would be the sum of conjugates which is also real. Those two quantities will generate a cubic equation of a+b which has real roots. I suggest using a numerical method to solve as opposed to cubic formula.
@xy-st9dz
@xy-st9dz Жыл бұрын
Interesting that this is an algebraic number. Can we say when this is the case and when sin(x) is transcendental?
@IceFlamesYT
@IceFlamesYT 9 ай бұрын
4:54 “then we know that” me: what kind of ultra specific formula is that?
@binehtesham
@binehtesham Жыл бұрын
Can you teach me all maths that you have ever learned, watching videos and learning is great, but physical tuitions will be next level
@rupertolababwe5973
@rupertolababwe5973 Жыл бұрын
Please could u do a video on why sin(54 degrees) = φ/2 , where φ is the golden ratio? I feel like this is a result not many people know about. Thank you!
@thenateman27
@thenateman27 Жыл бұрын
We know the expression is real without checking because it is of the form z + z* where z* is the complex conjugate of z. z + z* is guaranteed always real. Now about the nice expression involving only radicals, I have no fuckin idea 😅
@RoKa-Fby
@RoKa-Fby Жыл бұрын
To solve the equation 4(sinx)^3-3sinx+sin(3x)=0, substitute sinx=u+v to obtain u^3 = -1/8*sin(3x)±i/8*cos(3x). Before taking the cubic root to obtain u, we must multiply u^3 by e^(2𝜋i*n), n=0,1,2 to obtain all three branches of the cubic root. Then we have u = ( -1/8*sin(3x)±i/8*cos(3x))^(1/3)*e^(2𝜋i*n/3), n=0,1,2. Since sinx = u+v and v=1/(4u) we get sinx = (1/8*sin(3x)+i/8*cos(3x))^(1/3)*e^(2𝜋i*n/3) + (1/8*sin(3x)-i/8*cos(3x))^(1/3)*e^(-2𝜋i*n/3) , n=0,1,2. For 3x=𝜋/6 we get sin(𝜋/18) = ½*e^(2𝜋i/3)^(1/3)* e^(2𝜋i*n/3) + ½*e^(-2𝜋i/3)^(1/3)* e^(-2𝜋i*n/3) =cos(2𝜋/9*(1+3n)) = cos(40°)≈0.7660.. , -cos(20°)≈-0.9396.. , sin(10°)≈0.1736.. , for n=0,1,2, respectively. Obviously, we take n=2 to obtain the correct solution. /Roland Karlsson
@wes9627
@wes9627 Жыл бұрын
Here is another challenge. 18° = 60° - 3*14°. Thus, given x = sin14° evaluate sin18° as a function of x. That solution is relatively simple, but let us go the other way around. Given that sin18° = (Φ - 1)/2 = 1/(2Φ) where Φ = (√5 + 1)/2 is the Golden Ratio, evaluate sin14° as a function of Φ, where 14° = (60° - 18°)/3. I'm more impressed by how fast @blackpenredpen switches between pens. I try that and there's more ink on me than the board.
@Szzle77_GD
@Szzle77_GD Жыл бұрын
Revolutionary
@Gg-ij7li
@Gg-ij7li Жыл бұрын
Jesus loves you and wants to free you from any shackles in your life! Call upon His name and He will save you! If you confess with your mouth “Jesus is Lord” and believe that He rose from the dead and abandon your sins, you will be saved! God bless you, have an awesome day ❤😊😊
@guy_with_infinite_power
@guy_with_infinite_power Жыл бұрын
@@Gg-ij7li you got a lot of time dude
@jillerkore
@jillerkore Жыл бұрын
how big is your board?
@Patrik6920
@Patrik6920 11 ай бұрын
..working on it from time to time.. not there yet with a common usable formula ...so faar the function is off by +2.7 degres at 'sin(10)' degrees ..but its a start... ..if someone want to help pitch in this is the startinng formula, x is in rad, and the output is in sin of x +0 to 3 degrees of error, with an error at zero and pi/2 of zero (3 degees of error at sin(45) ) to far off for any use yet Y= (-pi(x)^2 + pi^2x) / ( 2pi^2- (4pi - 2pi/11) )
@Sam_on_YouTube
@Sam_on_YouTube Жыл бұрын
-sinx+icosx=(-i)(cosx+isinx)=-ie^ix -sinx-icosx=(i)(cosx-isinx)=ie^-ix Cube root of i= -i, root(3)/2+/-(i/2) Cube root of -i= i, -root(3)/2+/-(i/2) Wait... I'm going to get back to the polar form expression for sin. This doesn't help you go further... but it does give you an alternate route to prove the same expression. Maybe exploring it will help you understand which form to use.
@anupamamehra6068
@anupamamehra6068 Жыл бұрын
Hi blackpenredpen! A challenge for you from my side - prove the following: The volume of an n-dimensional sphere of radius R is : (Pi to the power (n/2))/(n/2) factorial
@emilioreyes7369
@emilioreyes7369 Жыл бұрын
Could someone explain to me why even though he plugged in x = 30° he still got the sin(150°)? Is there a complex analysis reason for this in understandable terms?
@AdvaitBhalerao
@AdvaitBhalerao 3 ай бұрын
I found out the easiest way to obtain sin(1°). For this, you have to know the values of sines and cosines of 37° and 53° each, (sin37°=3/5, cos37°=4/5), which can be used to find sin 16°, by using sin(53°-37°), i.e the compound angle formula. After we get sin 16°, just use sin(16°-15°), for which, we already know the value of sine and cosine of 15°
@mokshithreddy2375
@mokshithreddy2375 Жыл бұрын
Even though I am from india❤ .. your videos are helpful to crack competitive exams(after 10+2) 🎉🎉🎉You are a mathematician sir ...🎉🎉🎉
@bobh6728
@bobh6728 Жыл бұрын
SPAM ALERT Everyone needs to report this.
@TheZudik
@TheZudik Жыл бұрын
​@@bobh6728 bro shut up
@erikkonstas
@erikkonstas Жыл бұрын
​@@bobh6728 NO! First find a real spam comment and then report *THAT* , otherwise you're just a liar...
@bobh6728
@bobh6728 Жыл бұрын
@@erikkonstas they deleted the link to an inappropriate site when they edited their comment.
@erikkonstas
@erikkonstas Жыл бұрын
@@bobh6728 Oh shit...
@armanavagyan1876
@armanavagyan1876 Жыл бұрын
PROF U have potential of being Profi Kray teacher's international MATH competition gold medalist)
@sharadvyas2025
@sharadvyas2025 Жыл бұрын
I'll ask you a question What is -ln(-1) =? A . iπ B . -iπ C. A ans B both D. Can't possible E. None of the above I am not testing you but by watching just your videos i got that question in my mind
@xanderlastname3281
@xanderlastname3281 Жыл бұрын
Can someone explain to me the cube roots of unity? Is it as simple as "assuming you have 1 answer, you can find the other 2?" .....assuming you only have 1 eeal root, if all 3 roots are real i doubt unity would work as youre introducing unessesary "i"s Anyway so like if I had a cubic equation whose first root is... idk 7, could I find the other two roots using omega? Or is there some sort of special requirements for this to work? (For example... idk maybe you require two terms like in the video)
@lilstooni5629
@lilstooni5629 Жыл бұрын
You can easily get two other roots of any cubic polynomial given one of its roots. In case of x^3=1 you know x_1=1. So, divide the x^3-1=0 polynomial by (x-1), and you'll get the simple quadratic polynomial x^2+x+1=0, which gives us omega 1 and omega 2 after applying the quadratic formula.
@gheffz
@gheffz Жыл бұрын
First think I noticed (and apology as well)... Black pen, red and pen and now blue pen!! I am sure I am about the 1000 plus to remind you of that!
@HelenYera
@HelenYera Жыл бұрын
can u answer this : prod n = 2 to 20 (n ^ 3 - 1)/(n ^ 3 + 1) I try but I don't know how
@Drawingjeffy
@Drawingjeffy 2 ай бұрын
2:44 Shouldn't -2sin^3theta x -2sin^3theta = 4sin^3theta not negative.
@dugong369
@dugong369 Жыл бұрын
I think what you are really doing is getting sin(x/3) = Im(e^(xi/3) = [e^(xi/3) - e^(-xi/3)] / 2i . If you substitute for the exponentials using Euler's formula, you will end up with the same 1/2(sum of cube roots) that you got after simplifying. But I don't think this helps to find a closed form using real numbers?
@antoniogarofalo4587
@antoniogarofalo4587 Жыл бұрын
Hey, the values you get in first place and in the first check are sin(60-10°) and sin(60+10°). using sine addition formula for each and subtracting the equation you get sin(10°)=-(-sin(70°)+sin(50°))/(cos(pi/3)*2). Not the answer to why you choose a solution respect to the other but it was funny to note the simmetry around pi/3
@sankardebnath4724
@sankardebnath4724 Жыл бұрын
Sir, would you please find out an accurate way to trisect any angle?
@bradycall1889
@bradycall1889 Жыл бұрын
What is the sine of an imaginary number? Or the cosine of such?
@edwardhudson815
@edwardhudson815 Жыл бұрын
isnt sin(10) an exact value, what makes a bunch of roots more exact than sin(10), given those roots are also just operators like sin()
@oinkityoink
@oinkityoink Жыл бұрын
sin 1deg should be (relatively) easy now! sin and cos 72deg can be found with the golden ratio triangle, and then we could use the half angle formula thrice to get sin and cos 9 deg! then sin 1deg = sin10deg*cos9deg - cos10deg*sin9deg
@arditgaming349
@arditgaming349 Жыл бұрын
Can you show that 1/(2sin10°) + 2cos20°=1 ?
@jwangosho
@jwangosho Жыл бұрын
Very interesting. That polynomial has one real root and two imaginary roots which are of course conjugates.
@lilstooni5629
@lilstooni5629 Жыл бұрын
On this regard I would like to recommend you to check out the book called "One-third Angle in Trigonometry: New method for Cubic Equation" by Bhava Nath Dahal, which is available online. I am to greedy to pay a quid for it, but as the description says, it might have the general formula we all are looking for. If someone does buy this book, please share the answer :)
@k0pstl939
@k0pstl939 Жыл бұрын
Apply the third using 60 and then halve it using that original one?
@dhruvarai1895
@dhruvarai1895 Жыл бұрын
De Moivre's formula?
precalculus law of sine, but be careful
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