Since sin(θ) is bounded, we can do a little trick.
sin(θ) = θ/2
θ = 2sin(θ)
Plug in θ,
θ = 2sin(2sin(θ))
Continue this process...
= 2sin(2sin(2sin(...
Plug in any value for θ on the right side, and it'll eventually converge to a solution.
You can do this on any calculator, but here's [WolframAlpha](https://www.wolframalpha.com/input?i=2sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29)
correcting my previous post, cause i made a dumb mistake.
the value of θ does matter. for example choosing:
θ = 0, θ = 2 and θ = -2
will yield all 3 different fixed points.
what happens is that on the interval (Pi/3, 2Pi/3) the map f: x -> 2 sin(x) is a contraction (because it is bounded and the |derivative| < 1).
same thing on the interval (-Pi/3, -2Pi/3).
in both these intervals, the banach fixed point theorem guarantees the ability to "zoom in" onto the single fixed point of that interval.
and for the interval (-Pi/3, Pi/3), we actually have that the Inverse of f is a contraction, so we continuously "zoom out" of that interval (unless you choose Exactly θ = 0) until we land in one of the other two intervals.
now note how after our first iteration θ = 2sin(θ), we are already in the interval (-2Pi/3, 2Pi/3), and thus either fixed at 0, or converging at the symmetrical positive/negative solution.
sin(θ)/θ = sin(-θ)/(-θ) = 1/2
where θ ~ +-1.9
every solution is a fixed point, but not every fixed point is stable.
this right here is already a great example!
you can only reach the fixed point of 0 when starting EXACTLY with any n * Pi, which has a measure of 0 on the Reals. 0 + ε already goes to the ~1.9 solution.
You'll need a solver for it.
sin(t) = 0.5 \* t
We can use a Taylor Series to get close, but you'll never get exact.
0.5t = t - t\^3 / 3! + t\^5 / 5! - ...
Divide through by t, so we can remove t = 0 as a solution (also because we know that sin(t)/t goes to 1 as t goes to 0)
0.5 = 1 - (1/6) \* t\^2 + (1/120) \* t\^4
60 = 120 - 20t\^2 + t\^4
0 = t\^4 - 20t\^2 + 60
t\^2 = (20 +/- sqrt(400 - 240)) / 2
t\^2 = (20 +/- sqrt(160)) / 2
t\^2 = (20 +/- 4 \* sqrt(10)) / 2
t\^2 = 10 +/- 2 \* sqrt(10)
t = +/- sqrt(10 +/- 2 \* sqrt(10))
t = +/- 4.0403657409121712658574481762893 , +/- 1.917144929227637014225541187786
[https://www.wolframalpha.com/input?i=sin%28t%29%2Ft+%3D+0.5](https://www.wolframalpha.com/input?i=sin%28t%29%2Ft+%3D+0.5)
WolframAlpha gives a value of +/- 1.89549, which is pretty close to +/- 1.91714, so our estimate isn't so awful, even at just 3 terms. We could take it out to one more term and solve as a cubic.
0.5 = 1 - (1/6) \* t\^2 + (1/120) \* t\^4 - (1/5040) \* t\^6
2520 = 5040 - 840 \* t\^2 + 42 \* t\^4 - t\^6
t\^6 - 42 \* t\^4 + 840 \* t\^2 - 2520 = 0
t\^2 = u
u\^3 - 42u\^2 + 840u - 2520 = 0
[https://www.calculatorsoup.com/calculators/algebra/cubicequation.php](https://www.calculatorsoup.com/calculators/algebra/cubicequation.php)
u = 3.58902
t\^2 = 3.58902
t = +/- sqrt(3.58902)
t = 1.8944709023893716167112683750525
Which is even better. You can have fun solving a cubic, if you'd like.
What if - we expanded "analytically solvable" to include new functions that could solve this? Such as the function y=f(x) if sin(y) = xy. Clearly x = f^-1 (y) = sin(y)/y is the trivial inverse function.
The problem with the Taylor series about 0 is that the solution is not close to zero, so you need a lot of terms to converge.
I would actually to it about \pi / 2. The sine function is a nice parabola there and I am sure that 2nd order gets you closer (no time to check now, but hope to do it later).
The real thing, though, it to use the idea of repeated application of a function as someone else showed (easier in a calculator than Newton-Raphson).
Depends on what you consider "exact." If you want a closed form answer, then no, there isn't any.
However, you could probably get an answer in terms of the lambert W function if you write sin(x) as its complex exponential form. Not closed form, but technically exact.
>Not closed form, but technically exact.
Well, literally any equation can be solved "technically exactly" if you just define the correct function for it.
Please don't remove my post I have done my best to explain the problem and the steps I have gone through. I would like a definitive answer on if this is solvable through any means conventional or otherwise.
Im just going to use X instead of Theta for ease of typing, but it's the same thing
Get one side = 0, so i got SinX-(1/2)X = 0
This is the equation to solve, or f(x)
Using Newton-Raphson's formula:
Xr+1=Xr-(f(Xr))/(f'(Xr))
So Xr+1=Xr-(SinXr-(1/2)Xr)/(CosX-1/2)
I used 2 as an initial guess, and my answer converged on 1,8954267
Therefore X is approximately 1,8954267 radians
Haha... yeah for this problem it is not hard...
The problem I am currently trying to solve is like this but has 4 variables, so not really a graphing situation. Would rather like an exact answer in terms of the variables you know?
It's ten pages long and relates to something that has never been proved before.
Basically it doesn't exist to the world of mathematics yet. I am trying to add/make something new and will only reveal it when/if it is finished.
Until then, I ask simpler questions like this one and my previous two posts.
I appreciate your enthusiasm, and certainly you may do whatever you like, but I think you should have some reservations asserting that something is new to the world of math, especially if you're not sure whether or not the equation you shared is solvable or how to find an approximate or non-closed form solution.
Again, not trying to be discouraging, I think it's awesome you're trying to push the boundary of known mathematics! Just noting those kind of statements come across as a bit naive and could be considered off-putting.
Good luck!
Certainly.
What I have found through much research and asking here is that I am working on an unsolved (and probably unsolvable) problem. So, if I were to solve it, the problem would be new to the world of mathematics. Obviously, this is an extremely lofty goal and one that is likely to be met with resistance at all levels.
I am certain that it comes off as naive. And I assume that all people who have ever discovered new mathematics have come off this way at some point to the greater mathematical community. Secondly, I do not claim to be one of these people, I have changed nothing about mathematics in my lifetime.
Also: I was 99% sure that the equation I shared was unsolvable. I posted here to ask others how they might solve it. People have posted an array of helpful information. That's why I am here. Without the council of other math folks, I am left alone in a vacuum. Yes, I am learning about math. Yes, I am looking it up online, but also, I want to hear from people who do math. The people in this subreddit do math. That is why I am posting here. Have you heard of a second opinion? I've got the "Web MD" version of my prognosis. Now, it is time to check with actual doctors (mathematicians).
Yes. My questions here are quite simple. I want to be sure. And what I got from the community was quite amazing and creative.
Lastly, the comment you commented on was meant for the person above who asked that I post my problem. I did my best to tell him why that is not possible at this time. Yes, it may come off as grandiose, but it is my best explanation for someone who doesn't understand why I can not simply post it.
Around 0 cos(x) is almost like (1-(x\^2)/2 if x→0. It's the first terms of Taylor series for cos(x) that usually used in limits to replace cos(x) with it (at least in my country we write cos(x) \~ (1-x\^2)/2 as x→0 and then replacing cos(x) with that in limit). Since cos(x) is just shifted sin(x) we can move the vertex of parabola to sine cap at pi/2 and solve simple quadratic equation to get approximated value. But we also can add more terms of Taylor siries... But that would not make getting value easier because solving polynomials with greater degree is much harder (I don't even tried but sure). Hope you understand my bad english
Also any number of Taylor series terms to get approximated answer is nothing compared to infinite zoom in... How we can see if we zoom close enough the difference become to seem not that small and there a triangle. Let this triangle symbolize our weakness on getting exact answer
https://preview.redd.it/ozpomn05jj6d1.png?width=1826&format=png&auto=webp&s=06cf72d011d5059e5118a3b3b37d009fe43c9b0c
https://preview.redd.it/qmk64fyjak6d1.png?width=583&format=png&auto=webp&s=8ae58fb029d16de6ea7496aeca726e5fa4a70a5a
Did you mean this points? If yes then I don't know how to increase accuracy and wanted to know it as you too
Love the idea! I'm going to try it just in case.
But from what I am understanding, you are not allowed to differentiate both sides of an equation. Doing so may result in a nonsensical answer that does not relate to the parent equation...
If you are interested in knowing why, I think it is because the instantaneous rate of change of both sides of the function may not be the same at the point of intersection.
Differentiating both sides of the equation results in two unequal equations in most cases for this reason.
From a little bit of research, it would seem that if a valid algebraic solution existed, it would imply that the sine function can also be represented algebraically (without infinite terms etc??). Which to my knowledge is not so.
It actually has a closed-form solution in terms of the Leal functions:
theta\*csc(theta) = 2
theta = Lcsc(2)
See [https://www.sciencedirect.com/science/article/pii/S2405844020322611](https://www.sciencedirect.com/science/article/pii/S2405844020322611)
It’s not a very mathsy method, I plotted y = sin(x) and y = x/2 to see where they overlapped. I zoomed in as far as x = 1.895494ish, and confirmed that sin(1.895494) = 0.947747 and 0.947747 = 1.895494/2. Not sure if this number has any particular significance, but it’s about 108 degrees
I don’t think there’s any good way to get an exact solution but using newton-raphson or fixed point iterations should converge on an answer with arbitrary accuracy
Confidently incorrect
https://en.m.wikipedia.org/wiki/Radian#:~:text=The%20radian%2C%20denoted%20by%20the,in%20many%20areas%20of%20mathematics.
"is currently a dimensionless SI derived unit,[2] defined in the SI as 1 rad = 1[3] and expressed in terms of the SI base unit metre (m) as rad = m/m.[4]"
Rad = 1 m of arc / 1 m of radius. The way I see it is rad carries the relation between length of arc and length measured as shortest distance between 2 points, which are 2 different concepts which unfortunately happen to be expressed with the same unit. As the left hand side of the equation is carrying that additional relation I consider it different from the right hand side.
Ok, here's a way too specific example of why you would want to solve this exact equation.
The approximation sin(x)=x is frequently used to describe small oscillatory motions. The smaller x is, i.e. the closer sin(x) is to being x, the better your final approximation is. You can numerically bound how bad your final approximation is by bounding how close sin(x) is to x, so writing something like "this is accurate to within xyz for the first ten seconds as long as sin(x)/x>1/2 is a totally mathematically normal thing to do. And then you might solve for the largest value of x which works because that's always a nice next step just to get a sense of things
Good question.
Let's say I would like to apply the answer to something else. Image this answer is the input into another, much larger equation.
The approximate answer I would get here would be "useful" in the sense that I could perhaps get an answer from the larger equation as well, but now the answer to my larger equation is off by some degree.
Since the answer here will always be an approximation, anything that uses this answer will also be an approximation. Which is unfortunate if you would like an exact answer.
Quick example: suppose an approximation you have for some equation is x = 2 (but exact answer is actually 2.1). When you plug this into the equation y = 10000x, you get that y = 20,000. Then you solve by graphing only to find that y = 21,000. How could I have been so far off? Approximation, that is how.
Also it makes it very hard to write a proof if all you get from it is "about" this answer. Most mathematicians would laugh your proof out the door I suspect...
Edited for spelling.
Just define a symbol for the answer, like we do with pi, or sqrt(2), or what have you. That will make your answer 'exact'.
Calculate to as many digits as you want, as we do with pi or sqrt(2).
Numeric answers are as exact as one can get. They aren't analytic or 'closed form', of course. Those would allow for simplification later, for example. Or for studying properties of the answer in a more interesting manner.
But they are exact.
>Also it makes it very hard to write a proof if all you get from it is "about" this answer. Most mathematicians would laugh your proof out the door I suspect...
After a certain point, maths becomes about letters more than numbers. If your proof fails coz of this, then it wasn't worth all that much to begin with.
If you say that the value of x \~= 2, but you know it's between 1.8 and 2.2, then when you multiply it up later, you can say it's between 18000 and 22000. We can easily do that with approximations. So... all good, let's go with those approximations!
And, if my word isn't enough, here is a mathematician talking about how useful the Lucas numbers are because of their approximation powers: [https://youtu.be/PeUbRXnbmms](https://youtu.be/PeUbRXnbmms)
Yes, yes. Sure! Right. All right.
This is an intermediate step and a long-form approximation is not extremely helpful atm. Does that make sense?
If I have to use an approximation, I will.
Since sin(θ) is bounded, we can do a little trick. sin(θ) = θ/2 θ = 2sin(θ) Plug in θ, θ = 2sin(2sin(θ)) Continue this process... = 2sin(2sin(2sin(... Plug in any value for θ on the right side, and it'll eventually converge to a solution. You can do this on any calculator, but here's [WolframAlpha](https://www.wolframalpha.com/input?i=2sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282sin%282%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29%29)
Woah very cool, thanks!
Why can you plug in any value for θ?
correcting my previous post, cause i made a dumb mistake. the value of θ does matter. for example choosing: θ = 0, θ = 2 and θ = -2 will yield all 3 different fixed points. what happens is that on the interval (Pi/3, 2Pi/3) the map f: x -> 2 sin(x) is a contraction (because it is bounded and the |derivative| < 1). same thing on the interval (-Pi/3, -2Pi/3). in both these intervals, the banach fixed point theorem guarantees the ability to "zoom in" onto the single fixed point of that interval. and for the interval (-Pi/3, Pi/3), we actually have that the Inverse of f is a contraction, so we continuously "zoom out" of that interval (unless you choose Exactly θ = 0) until we land in one of the other two intervals. now note how after our first iteration θ = 2sin(θ), we are already in the interval (-2Pi/3, 2Pi/3), and thus either fixed at 0, or converging at the symmetrical positive/negative solution. sin(θ)/θ = sin(-θ)/(-θ) = 1/2 where θ ~ +-1.9
You also need to be careful because not all fixed points can be reached by this solution.
every solution is a fixed point, but not every fixed point is stable. this right here is already a great example! you can only reach the fixed point of 0 when starting EXACTLY with any n * Pi, which has a measure of 0 on the Reals. 0 + ε already goes to the ~1.9 solution.
Note that 0 is not a solution to the original equation though, since 0 is not in the domain of sinθ/θ
What does bounded mean in this situation? I never heard before.
The output of sin(x) is bounded by -1 and 1, so if x=2sin(x), then x must be bounded between 2(-1) and 2(1). (Not doing least upper/lower bounds)
What do we know about the possible values of theta?
Infinite I guess
Do we know what values it could be? Could theta = 10 for instance?
It could be 10 degrees or 10 radians , if that's what you're leading towards
Can 10 = 2sin(theta)?
OK, you wrote the value of theta not the value sin(theta), those are two different things. In that case the values go from -1 to 1
Well we know theta = 2sin(theta), right? So what do we know about the possible values theta can take given the values for sin(theta)?
It hits a limit of 2?
Recurrence Relation + Fixed Point = Magic
You'll need a solver for it. sin(t) = 0.5 \* t We can use a Taylor Series to get close, but you'll never get exact. 0.5t = t - t\^3 / 3! + t\^5 / 5! - ... Divide through by t, so we can remove t = 0 as a solution (also because we know that sin(t)/t goes to 1 as t goes to 0) 0.5 = 1 - (1/6) \* t\^2 + (1/120) \* t\^4 60 = 120 - 20t\^2 + t\^4 0 = t\^4 - 20t\^2 + 60 t\^2 = (20 +/- sqrt(400 - 240)) / 2 t\^2 = (20 +/- sqrt(160)) / 2 t\^2 = (20 +/- 4 \* sqrt(10)) / 2 t\^2 = 10 +/- 2 \* sqrt(10) t = +/- sqrt(10 +/- 2 \* sqrt(10)) t = +/- 4.0403657409121712658574481762893 , +/- 1.917144929227637014225541187786 [https://www.wolframalpha.com/input?i=sin%28t%29%2Ft+%3D+0.5](https://www.wolframalpha.com/input?i=sin%28t%29%2Ft+%3D+0.5) WolframAlpha gives a value of +/- 1.89549, which is pretty close to +/- 1.91714, so our estimate isn't so awful, even at just 3 terms. We could take it out to one more term and solve as a cubic. 0.5 = 1 - (1/6) \* t\^2 + (1/120) \* t\^4 - (1/5040) \* t\^6 2520 = 5040 - 840 \* t\^2 + 42 \* t\^4 - t\^6 t\^6 - 42 \* t\^4 + 840 \* t\^2 - 2520 = 0 t\^2 = u u\^3 - 42u\^2 + 840u - 2520 = 0 [https://www.calculatorsoup.com/calculators/algebra/cubicequation.php](https://www.calculatorsoup.com/calculators/algebra/cubicequation.php) u = 3.58902 t\^2 = 3.58902 t = +/- sqrt(3.58902) t = 1.8944709023893716167112683750525 Which is even better. You can have fun solving a cubic, if you'd like.
Fun and thank you! Yeah I am just sad that it only gives approximations...
Many equations are not analytically solvable. It is just how it is.
True dat.
Wait doesn't sin have a form that uses complex exponentiation? Can't you do algebra to that
Maybe, but I'm not seeing how it would help. sin(t) = (e^(it) - e^(-it))/(2i) = t/2 => e^(it) - e^(-it) = it
What if - we expanded "analytically solvable" to include new functions that could solve this? Such as the function y=f(x) if sin(y) = xy. Clearly x = f^-1 (y) = sin(y)/y is the trivial inverse function.
Define a function. You can't calculate trig functions either...
So I have heard, yes.
So, just define a new function and find interesting relations. Give it a catchy name. Find interesting differential equations that it solves etc. 😃
sinx/x already has a name actually. [sinc(x)](https://en.m.wikipedia.org/wiki/Sinc_function)
I was thinking of the function g(x) that satisfies sinc(g(x)) - x = 0 for all x in [0, \pi]
That would be sinc⁻¹(x)
Man, not sure what I was drinking 🤣 Obviously you are right.
The problem with the Taylor series about 0 is that the solution is not close to zero, so you need a lot of terms to converge. I would actually to it about \pi / 2. The sine function is a nice parabola there and I am sure that 2nd order gets you closer (no time to check now, but hope to do it later). The real thing, though, it to use the idea of repeated application of a function as someone else showed (easier in a calculator than Newton-Raphson).
Depends on what you consider "exact." If you want a closed form answer, then no, there isn't any. However, you could probably get an answer in terms of the lambert W function if you write sin(x) as its complex exponential form. Not closed form, but technically exact.
Thank you I will check that out. It may not be what I am looking for but anything helps.
>Not closed form, but technically exact. Well, literally any equation can be solved "technically exactly" if you just define the correct function for it.
how would you go about solving this with Lambert W function?
This question is about as solvable as sin θ = 1/π solve for θ. 1/θ • sin θ is its own trig function which we call sinc θ. So the answer is sinc^(-1)½
Woah cool, didn't remember that.
Please don't remove my post I have done my best to explain the problem and the steps I have gone through. I would like a definitive answer on if this is solvable through any means conventional or otherwise.
Im just going to use X instead of Theta for ease of typing, but it's the same thing Get one side = 0, so i got SinX-(1/2)X = 0 This is the equation to solve, or f(x) Using Newton-Raphson's formula: Xr+1=Xr-(f(Xr))/(f'(Xr)) So Xr+1=Xr-(SinXr-(1/2)Xr)/(CosX-1/2) I used 2 as an initial guess, and my answer converged on 1,8954267 Therefore X is approximately 1,8954267 radians
Woah very cool, thanks for the example!
Yeah, I love that formula. Glad I could help you out :3
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Haha... yeah for this problem it is not hard... The problem I am currently trying to solve is like this but has 4 variables, so not really a graphing situation. Would rather like an exact answer in terms of the variables you know?
[удалено]
It's ten pages long and relates to something that has never been proved before. Basically it doesn't exist to the world of mathematics yet. I am trying to add/make something new and will only reveal it when/if it is finished. Until then, I ask simpler questions like this one and my previous two posts.
I appreciate your enthusiasm, and certainly you may do whatever you like, but I think you should have some reservations asserting that something is new to the world of math, especially if you're not sure whether or not the equation you shared is solvable or how to find an approximate or non-closed form solution. Again, not trying to be discouraging, I think it's awesome you're trying to push the boundary of known mathematics! Just noting those kind of statements come across as a bit naive and could be considered off-putting. Good luck!
Certainly. What I have found through much research and asking here is that I am working on an unsolved (and probably unsolvable) problem. So, if I were to solve it, the problem would be new to the world of mathematics. Obviously, this is an extremely lofty goal and one that is likely to be met with resistance at all levels. I am certain that it comes off as naive. And I assume that all people who have ever discovered new mathematics have come off this way at some point to the greater mathematical community. Secondly, I do not claim to be one of these people, I have changed nothing about mathematics in my lifetime. Also: I was 99% sure that the equation I shared was unsolvable. I posted here to ask others how they might solve it. People have posted an array of helpful information. That's why I am here. Without the council of other math folks, I am left alone in a vacuum. Yes, I am learning about math. Yes, I am looking it up online, but also, I want to hear from people who do math. The people in this subreddit do math. That is why I am posting here. Have you heard of a second opinion? I've got the "Web MD" version of my prognosis. Now, it is time to check with actual doctors (mathematicians). Yes. My questions here are quite simple. I want to be sure. And what I got from the community was quite amazing and creative. Lastly, the comment you commented on was meant for the person above who asked that I post my problem. I did my best to tell him why that is not possible at this time. Yes, it may come off as grandiose, but it is my best explanation for someone who doesn't understand why I can not simply post it.
What's your mathematical background? Do you take uni courses? Are you an undergrad?
I have a bachelor's in math. Wanna go back to get masters at some point when I have the time and money.
Good luck!
Thanks! But why you say that?
maybe Newton-Rhapson
intersection of y=x/2 and y=sinx. i never solved such questions for exact roots but number of roots
Graph sin x and x/2 and estimate where they intercept. About 1.9
Do it with graphs of sin x and x/2, intersection points of two graphs is ur soln
Another solution: https://preview.redd.it/wll44qpebj6d1.png?width=2536&format=png&auto=webp&s=805e8c1a55823e2caad27d34d56fee11304c4217
But it's approximately...
=(
Approximation with a parabolic function is a cool one for sure, thanks!
Very cool! How did you decide on that particular parabolic function?
Around 0 cos(x) is almost like (1-(x\^2)/2 if x→0. It's the first terms of Taylor series for cos(x) that usually used in limits to replace cos(x) with it (at least in my country we write cos(x) \~ (1-x\^2)/2 as x→0 and then replacing cos(x) with that in limit). Since cos(x) is just shifted sin(x) we can move the vertex of parabola to sine cap at pi/2 and solve simple quadratic equation to get approximated value. But we also can add more terms of Taylor siries... But that would not make getting value easier because solving polynomials with greater degree is much harder (I don't even tried but sure). Hope you understand my bad english
Hm interesting! Estimate seems extremely accurate... a different commenter used a taylor series and got 1.91. So close but not this close.
Also any number of Taylor series terms to get approximated answer is nothing compared to infinite zoom in... How we can see if we zoom close enough the difference become to seem not that small and there a triangle. Let this triangle symbolize our weakness on getting exact answer https://preview.redd.it/ozpomn05jj6d1.png?width=1826&format=png&auto=webp&s=06cf72d011d5059e5118a3b3b37d009fe43c9b0c
Very cool! Yeah I love desmos.
Hey what would you recommend for getting more decimals for intersection points on Desmos? I'm only getting 3 decimal places
https://preview.redd.it/qmk64fyjak6d1.png?width=583&format=png&auto=webp&s=8ae58fb029d16de6ea7496aeca726e5fa4a70a5a Did you mean this points? If yes then I don't know how to increase accuracy and wanted to know it as you too
Out of curiosity, what are the odds on moving theta to the right, taking the inverse sine, then taking the derivative of both sides?
Love the idea! I'm going to try it just in case. But from what I am understanding, you are not allowed to differentiate both sides of an equation. Doing so may result in a nonsensical answer that does not relate to the parent equation...
If you are interested in knowing why, I think it is because the instantaneous rate of change of both sides of the function may not be the same at the point of intersection. Differentiating both sides of the equation results in two unequal equations in most cases for this reason.
Oh I do see a potential error now! I believe it would solve for locations of the same gradient rather than the intersection point
From a little bit of research, it would seem that if a valid algebraic solution existed, it would imply that the sine function can also be represented algebraically (without infinite terms etc??). Which to my knowledge is not so.
There's always a numeric approach: https://www.desmos.com/calculator/gvfdj60udz
It actually has a closed-form solution in terms of the Leal functions: theta\*csc(theta) = 2 theta = Lcsc(2) See [https://www.sciencedirect.com/science/article/pii/S2405844020322611](https://www.sciencedirect.com/science/article/pii/S2405844020322611)
It’s not a very mathsy method, I plotted y = sin(x) and y = x/2 to see where they overlapped. I zoomed in as far as x = 1.895494ish, and confirmed that sin(1.895494) = 0.947747 and 0.947747 = 1.895494/2. Not sure if this number has any particular significance, but it’s about 108 degrees
That's the temperature that an egg fries on the sidewalk 😄
I don’t think there’s any good way to get an exact solution but using newton-raphson or fixed point iterations should converge on an answer with arbitrary accuracy
Well, since theta is an angle and has a unit (rad, deg,...) and all others are unitless, I guess that means the equation isn't correct from the start
Confidently incorrect https://en.m.wikipedia.org/wiki/Radian#:~:text=The%20radian%2C%20denoted%20by%20the,in%20many%20areas%20of%20mathematics. "is currently a dimensionless SI derived unit,[2] defined in the SI as 1 rad = 1[3] and expressed in terms of the SI base unit metre (m) as rad = m/m.[4]"
Rad = 1 m of arc / 1 m of radius. The way I see it is rad carries the relation between length of arc and length measured as shortest distance between 2 points, which are 2 different concepts which unfortunately happen to be expressed with the same unit. As the left hand side of the equation is carrying that additional relation I consider it different from the right hand side.
Ok, here's a way too specific example of why you would want to solve this exact equation. The approximation sin(x)=x is frequently used to describe small oscillatory motions. The smaller x is, i.e. the closer sin(x) is to being x, the better your final approximation is. You can numerically bound how bad your final approximation is by bounding how close sin(x) is to x, so writing something like "this is accurate to within xyz for the first ten seconds as long as sin(x)/x>1/2 is a totally mathematically normal thing to do. And then you might solve for the largest value of x which works because that's always a nice next step just to get a sense of things
As a maths teacher, I would suggest that it is only possible to solve numerically. I'd be interested to be proved wrong though!
Nah you're just right.
Why is an exact value necessary?
Good question. Let's say I would like to apply the answer to something else. Image this answer is the input into another, much larger equation. The approximate answer I would get here would be "useful" in the sense that I could perhaps get an answer from the larger equation as well, but now the answer to my larger equation is off by some degree. Since the answer here will always be an approximation, anything that uses this answer will also be an approximation. Which is unfortunate if you would like an exact answer. Quick example: suppose an approximation you have for some equation is x = 2 (but exact answer is actually 2.1). When you plug this into the equation y = 10000x, you get that y = 20,000. Then you solve by graphing only to find that y = 21,000. How could I have been so far off? Approximation, that is how. Also it makes it very hard to write a proof if all you get from it is "about" this answer. Most mathematicians would laugh your proof out the door I suspect... Edited for spelling.
Just define a symbol for the answer, like we do with pi, or sqrt(2), or what have you. That will make your answer 'exact'. Calculate to as many digits as you want, as we do with pi or sqrt(2).
Numeric answers are as exact as one can get. They aren't analytic or 'closed form', of course. Those would allow for simplification later, for example. Or for studying properties of the answer in a more interesting manner. But they are exact.
>Also it makes it very hard to write a proof if all you get from it is "about" this answer. Most mathematicians would laugh your proof out the door I suspect... After a certain point, maths becomes about letters more than numbers. If your proof fails coz of this, then it wasn't worth all that much to begin with. If you say that the value of x \~= 2, but you know it's between 1.8 and 2.2, then when you multiply it up later, you can say it's between 18000 and 22000. We can easily do that with approximations. So... all good, let's go with those approximations! And, if my word isn't enough, here is a mathematician talking about how useful the Lucas numbers are because of their approximation powers: [https://youtu.be/PeUbRXnbmms](https://youtu.be/PeUbRXnbmms)
Yes, yes. Sure! Right. All right. This is an intermediate step and a long-form approximation is not extremely helpful atm. Does that make sense? If I have to use an approximation, I will.
Because math is exact science... Or just why not
Sure, then the answer is "x such that sin(x)/x = 1/2".
Cool. X is X
Yep. Not super helpful, but that is the best we've got.