exact value of sin(10 degrees)

0:00 what's a formula for sin(x/3), i.e. 1/3 angle formula for sine
0:23 deriving sin(3x) by using double-angle formula
4:34 using the cubic formula (the depressed version)
11:11 attempting to get sin(10 degrees) but we ran into some issues
19:11 finally got sin(10 degrees)
but I wonder if we can write sin(10 deg) as a radical expression without using the imaginary unit i
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Thank you all!

[matonphare]

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
Then we can use the trig form and we see that the imaginary parts cancels:
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

[paradoxicallyexcellent]

Man, the people who discovered this stuff without the use of calculators to sanity check what they were doing... hats off.

[DDroog-eqtw]

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.

[micronalpha]

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

[youssefayman]

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.

[djttv]

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

[TheDannyAwesome]

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.

[Static_MKFocus]

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🤝

[bartekabuz]

In Cardano's formula if complex number show up it is not possible to get an answer than involves only reals and radicals.

[jomariraphaellmangahas]

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

[DubioserKerl]

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?

[galvanaxchampion]

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.

[yoav]

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

[kahan]

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

[atzuras]

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]

always knew a video like this was coming

[farrattalex]

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.

[Ninja]

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

[mathmanic]

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.

[Sphinxycatto]

He just weaves in all conpcept in so easily
It seems every sentence is just a revision of a topic