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 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!
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
@@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
@@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.
@@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)
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?
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
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.
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.
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).
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.
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.
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
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
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).
@@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.
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
@@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.
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🤝
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
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
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.
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
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.
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)
"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).
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
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.
I guess Σx symbolizes the sum of all the weights of the packages, so you just add up all the weights. On the 2nd part, you do the same, but subtracting 4 to each weight, because this time you have x-4 instead of x inside of the sum
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)
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.
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"
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
-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.
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.
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)
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).
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 - ).
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!
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
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
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.
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}
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.
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
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.
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°
3 sin 10 - 4 sin^3 10 = sin (3x10 = 30) = 1/2, so sin 10 is one of the 3 roots of cubic equation 8 x^3 - 6 x + 1 = 0. What about the other 2 roots ? sin (-210) = sin (150-360) = sin (150) = sin (180-30) = sin (30) = 1/2, So the other 2 roots are sin (150/3 = 50) and sin (-210/3) = - sin(70). If T = 10, 50, -70 degrees, then sin 3T = sin 30 = sin (180 - 30 = 150) = sin (150 - 360 = -210) = 1/2 = 3 sin T - 4 sin^3 T = 3x - 4x^3 where x = sin T are the 3 roots of the cubic equation. The 3 roots of cubic equation (x^3 + 3mx = 2n) are x = (A + B, Aw + B/w, A/w + Bw) where 1, w and w^2=1/w=w* are the 3 (real and complex conjugate) cube roots of unity=1 and A^3 = n+d and B^3 = n-d where d^2 = discriminant D = n^2 + m^3. So, 8x^3 - 6x + 1 = 0 can be re-written as x^3 + 3 (m = -3/4) x = 2 (n = -1/2), therefore d^2 = discriminant D = n^2 + m^3 = (-1/2)^2 + (-3/4)^3 = 1/4 - 27/64 = -11/64, So d = ± i √11/8 A^3 = n + d = -1/2 + i √11/8 = (-4 + i √11)/8 and B^3 = -1/2 - i √11/8 = (-4 - i √11)/8 and complex conjugate cube roots of unity = w,w* = cos(2π/3) ± i sin (2π/3) = (-1 ± i√3)/2 So, sin(10) = Aw + B/w (after substituting the values of A^3, B^3 and w above).
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
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) )
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.
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 😅
Hay have found a more general solution. I you put the ecuation in terms of e, you will obtain (1/2)*(e^(i*(pi*(1+4n)/6 + x/3)) + e^(i*(pi*(1+4n)/6 - x/3))). For a real solution, the condition is sin(pi*(1+4n)/6 + x/3) + sin(pi*(1+4m)/6 - x/3) = 0, that complies with pi*(1+4n)/6 + x/3 = 2*pi*p and (pi*(1+4m)/6 - x/3) = 2*pi*q, (n, m, p, q)€(N,N,N,N), for example. I haven't go further, but I think it might be a interesting way.
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!
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.
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.
13:00 you can set the expression in parenthesis to x and then ''cube both sides'' similiar to this vid: ua-cam.com/video/5pa1AryylpM/v-deo.html 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
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
..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) )
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....)
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).
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
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!
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
.. 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.
This whole approach seems to have gotten me nowhere. I have exact values for sine and cosine of 60° and 18° and used them to get exact values for sine and cosine of 42°. Then I wanted to get exact values for sine and cosine of 14° = 42°/3. Using the methods described here I can solve two depressed cubic equations for cos14° and sin14°, but in the process I must find cube roots of complex numbers, which requires computing approximate values for cos14° and sin14° to get those cube roots. I don't want to use some canned program to evaluate the cube roots, but do it myself. The only way I know to evaluate (a + bi)^(1/3) is the r^(1/3)[cos(θ/3) + i sin(θ/3)] approach where r = √(a^2 + b^2) and θ = arctan(b/a), but that requires approximating the values that I want to solve for. In general it appears there are no other ways to evaluate those roots without using approximate trig functions, which defeats the whole purpose in the first place.
Substitute cubic root sum as *A* sin 10° = (1/2) • *A* then count: A''' = -1 + 3A• *1* ➡ A = (A''' +1) / 3 Substitute back ⬆ we get: sin 10° = (1 + A''') / 6 --- --- --- Original Formula: sin 3x = sin x •( 3 - 4.sin"x) sin 30° = sin 10°•( 3 - 4.sin"10) 1/2 = 3.sin 10° - 4.sin'''10° 1 = 6.sin 10° - 8.sin''' 10° 1 + 8.sin'''10° = 6.sin 10° ↔ sin 10° = [ 1 + (2.sin10°)'''] / 6 --- --- So, A = 2.sin 10° !?!? 🔄 Round and round 😵 --- I don't like imaginary number. It Only exist in dreams world. √(-1) = not 1, not -1 = Error! 😜
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?
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?
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!!!)
It's not possible to trisect an angle with ruler and compass, but takes only a few trig tricks to do it algebraically. What steps here are illegal in ruler and compass?
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 ❤😊😊
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
Ah! Thank you for the answer and it’s very nice. Could we write that as a radical expression without the i tho?
@@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!
But the second one gives 10 and 80 degrees?
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
@@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
Man, the people who discovered this stuff without the use of calculators to sanity check what they were doing... hats off.
you can always draw a triangle and measure the sidelength
No social media in those days, so they had plenty of time...:)
@@neutronenstern. It is impossible to trisect an angle with just a ruler and compass.
@@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.
@@cxpKSip also there are other ways, like the taylor series.
Also they had tables with precalculated values.
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.
I wonder if sine of (any integer in degrees) is always an algebraic number.
Yes! Don't spill the bean, bprp is going to release another more general formula for sin(x/n) ..... 😗
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
@@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)
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?
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
he didn't make this complex...
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.
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.
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).
Too complicated. sin(x/3)=sin(x/3)+1/2icos(x/3)-1/2icos(x/3)= croot(sinx+icosx)+croot(sinx-icosx)
yo i just did that and i thought at least someone must've noticed something so obvious
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.
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.
I have memorised the cubic formula and I can prove the formula
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
I don’t think my board is big enough for sin(1 deg) 😆
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
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).
@@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.
In Cardano's formula if complex number show up it is not possible to get an answer than involves only reals and radicals.
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
@@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.
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🤝
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
Oooh! That could be the reason! I will check on wiki later too! Thanks.
It Is possible, to express sin 10°, but you have to use infinite nested radicals
Another approach is to use hypergeometric function to calculate the real root of the Cubic equation 😊
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
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.
We could have just used euler form and demoivre's theorem
If we did that to the formula I think a sin(x/3) germ would pop out (try for yourself), which ruins the point
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()
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
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.
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)
"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).
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
Very informative and nice idea
Was thinking along the same lines
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
I literally plug sin(x/3) in the calculator earlier
This help me a lot
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.
Excuse me! Does anyone understands this one?
16. The weights of seven packages mailed by a company are 7Kg, 9Kg, 6Kg, 12Kg, 10Kg, 9Kg and 8Kg. Find
a) ∑x
b) ∑(x − 4)
🙏🙏
I guess Σx symbolizes the sum of all the weights of the packages, so you just add up all the weights. On the 2nd part, you do the same, but subtracting 4 to each weight, because this time you have x-4 instead of x inside of the sum
@@octane9309 Or, for part 2, simply subtract 7*4=28 from the first part,as you are subtracting 4 seven times
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)
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.
Yea there were a lot of things missing. When working with complex numbers all roots give multiple answers.
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?
Yes! 360/3 = 120, so the three values that cube to the same number are 120° from each other.
I love math.Thank you sir... ❤❤
You love math. I love meth. We are not the same.
Lamps love moths, or the otherway around
always knew a video like this was coming
Excellent. Thanks very much ! 🙂
Math is so beautiful and helpful in our life.❤️❤️❤️
11:16 you could prob put the terms inside the radicals in terms of cis to simplify the radicals
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"
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
-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.
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.
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
He just weaves in all conpcept in so easily
It seems every sentence is just a revision of a topic
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)
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).
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 - ).
Hence, should be sqrt(2 - )/2.
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.
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!
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
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
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.
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}
You can only approximate the value of sin with that formula. It can't tell you the exact value
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.
Hi Blackpenredpen. Make a video demonstrating why π is irrational. I am subscribed to your channel, very good videos. Greetings from Mexico
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
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.
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°
3 sin 10 - 4 sin^3 10 = sin (3x10 = 30) = 1/2, so sin 10 is one of the 3 roots of cubic equation 8 x^3 - 6 x + 1 = 0. What about the other 2 roots ?
sin (-210) = sin (150-360) = sin (150) = sin (180-30) = sin (30) = 1/2, So the other 2 roots are sin (150/3 = 50) and sin (-210/3) = - sin(70).
If T = 10, 50, -70 degrees, then sin 3T = sin 30 = sin (180 - 30 = 150) = sin (150 - 360 = -210) = 1/2 = 3 sin T - 4 sin^3 T = 3x - 4x^3 where x = sin T are the 3 roots of the cubic equation.
The 3 roots of cubic equation (x^3 + 3mx = 2n) are x = (A + B, Aw + B/w, A/w + Bw) where 1, w and w^2=1/w=w* are the 3 (real and complex conjugate) cube roots of unity=1 and A^3 = n+d and B^3 = n-d where d^2 = discriminant D = n^2 + m^3.
So, 8x^3 - 6x + 1 = 0 can be re-written as x^3 + 3 (m = -3/4) x = 2 (n = -1/2), therefore d^2 = discriminant D = n^2 + m^3 = (-1/2)^2 + (-3/4)^3 = 1/4 - 27/64 = -11/64, So d = ± i √11/8
A^3 = n + d = -1/2 + i √11/8 = (-4 + i √11)/8 and B^3 = -1/2 - i √11/8 = (-4 - i √11)/8 and complex conjugate cube roots of unity = w,w* = cos(2π/3) ± i sin (2π/3) = (-1 ± i√3)/2
So, sin(10) = Aw + B/w (after substituting the values of A^3, B^3 and w above).
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
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) )
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.
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 😅
Hay have found a more general solution. I you put the ecuation in terms of e, you will obtain (1/2)*(e^(i*(pi*(1+4n)/6 + x/3)) + e^(i*(pi*(1+4n)/6 - x/3))).
For a real solution, the condition is sin(pi*(1+4n)/6 + x/3) + sin(pi*(1+4m)/6 - x/3) = 0, that complies with pi*(1+4n)/6 + x/3 = 2*pi*p and (pi*(1+4m)/6 - x/3) = 2*pi*q, (n, m, p, q)€(N,N,N,N), for example. I haven't go further, but I think it might be a interesting way.
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!
This is me every other class but math 😂
Blackpenredpen: I don't know what the exact value
That Euler's formula: Hold my beer
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.
Reversely, You can reach the same answer directly using hyperbolic functions.
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.
13:00 you can set the expression in parenthesis to x and then ''cube both sides'' similiar to this vid: ua-cam.com/video/5pa1AryylpM/v-deo.html
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
His name should be written Guinness world record for the fastest pen changing
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
I have a challenge for you, calculate 2^^0.5, 2^^π, -1^^i, 2^^i and i^^i (tetrations)
..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) )
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!
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....)
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).
Can you teach me all maths that you have ever learned, watching videos and learning is great, but physical tuitions will be next level
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
Even though I am from india❤ ..
your videos are helpful to crack competitive exams(after 10+2)
🎉🎉🎉You are a mathematician sir ...🎉🎉🎉
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@@bobh6728 NO! First find a real spam comment and then report *THAT* , otherwise you're just a liar...
@@erikkonstas they deleted the link to an inappropriate site when they edited their comment.
@@bobh6728 Oh shit...
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!
Interesting and very well done. Thank you for this!
Thank you! : )
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
Correction: BCD, *AFG* (not AFD) and EFH are congruent
.. 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.
bro really cube rooted the cube root of unity at some point
What is the alternative solution of the strend problem solved by ramanujan?
This whole approach seems to have gotten me nowhere. I have exact values for sine and cosine of 60° and 18° and used them to get exact values for sine and cosine of 42°. Then I wanted to get exact values for sine and cosine of 14° = 42°/3. Using the methods described here I can solve two depressed cubic equations for cos14° and sin14°, but in the process I must find cube roots of complex numbers, which requires computing approximate values for cos14° and sin14° to get those cube roots. I don't want to use some canned program to evaluate the cube roots, but do it myself. The only way I know to evaluate (a + bi)^(1/3) is the r^(1/3)[cos(θ/3) + i sin(θ/3)] approach where r = √(a^2 + b^2) and θ = arctan(b/a), but that requires approximating the values that I want to solve for. In general it appears there are no other ways to evaluate those roots without using approximate trig functions, which defeats the whole purpose in the first place.
Substitute cubic root sum as *A*
sin 10° = (1/2) • *A* then count:
A''' = -1 + 3A• *1* ➡ A = (A''' +1) / 3
Substitute back ⬆ we get:
sin 10° = (1 + A''') / 6
--- --- ---
Original Formula:
sin 3x = sin x •( 3 - 4.sin"x)
sin 30° = sin 10°•( 3 - 4.sin"10)
1/2 = 3.sin 10° - 4.sin'''10°
1 = 6.sin 10° - 8.sin''' 10°
1 + 8.sin'''10° = 6.sin 10° ↔
sin 10° = [ 1 + (2.sin10°)'''] / 6
--- ---
So, A = 2.sin 10° !?!? 🔄
Round and round 😵
---
I don't like imaginary number.
It Only exist in dreams world.
√(-1) = not 1, not -1 = Error! 😜
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?
Good writing sir❤❤❤
9:57 why is the absolute value not needed?
since theres +/-, absolute value will do nothing
I was just about to tell you to use cube root of unity, but then you figured it out yourself
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?
Try it
4:54 “then we know that”
me: what kind of ultra specific formula is that?
Keep it up bro, What's the best books to learn advanced mathematics after graduating high school?
Interesting results :)
once you have the sinx+icosx under roots, can you use euler identity to simplfy it further?
Yes. It's called DeMoivre's theorem.
13.26 use polar method to solve it
Interesting that this is an algebraic number. Can we say when this is the case and when sin(x) is transcendental?
PROF U have potential of being Profi Kray teacher's international MATH competition gold medalist)
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!!!)
10 degrees in radians = 0.174533
sin(0.174533) = 0.173648251
Those two numbers are very close to each other :P
Very interesting. That polynomial has one real root and two imaginary roots which are of course conjugates.
Still waiting for that sine of 1 degree.
It's too complex to handle Toooo complex
This man is mathing which we cant even math about mathematics
Sir Can you solve x^y=y^x and told some unknown symbols with value please request from India(south India ,T.N.,TNJ.).
X^y^z=z^y^x proof with derivation .
It's not possible to trisect an angle with ruler and compass, but takes only a few trig tricks to do it algebraically. What steps here are illegal in ruler and compass?
Congrats on 1000000ln3 subs🎉🎉🎉
Revolutionary
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 ❤😊😊
@@Gg-ij7li you got a lot of time dude
easy to hard logs questions for preparation.