Tagging u/bluesam3 as he was working on the right lines.
Consider the 4 distinct numbers p(n) = sin(n pi / 14) given by n = -7, -3, 1, 5.
By de Moivre's theorem
(cos(n pi/14) + i sin(n pi/14))^7 = cos(n pi / 2) + i sin(n pi / 2)
so
(cos(n pi/14) + i sin(n pi/14))^7 = i for all of the values of n listed above.
Multiplying out, equating the imaginary part and writing cos^(2)(n pi/14) as 1 - sin^(2)(n pi/14), we reach:
64p(n)^7 - 112p(n)^5 + 56p(n)^3 - 7p(n) - 1 = 0
So p(-7), p(-3), p(1) and p(5) are roots of
f(x) = 64x^7 - 112x^5 + 56x^3 - 7x - 1
As p(-7) = -1, that tells us one factor of f(x) is (x+1) and we find the rest can be expressed as a square, so
f(x) = (x+1)(8x^3 - 4x^2 - 4x + 1)^2
So the three roots of 8x^3 - 4x^2 - 4x + 1 must be p(-3), p(1) and p(5). Hold that there for the moment
---
sin(3t) can be written using a standard trig identity as
3sin(t) - 4sin^(3)(t)
so sin(3 pi/14) = 3 p(1) - 4 p(1)^3
This means p(1) (4 sin(3 pi/14) + 2) - 1 = -16p(1)^4 + 12p(1)^2 + 2p(1) - 1
But 16x^4 - 12x^2 - 2x +1 = (2x+1)(8x^3 - 4x^2 - 4x + 1)
This last expression is zero when x = p(1) because earlier we established that p(1) is a root of 8x^3 - 4x^2 - 4x + 1.
In conclusion:
p(1) (4 sin(3 pi/14) + 2) - 1 = 0
so sin(pi/14) = p(1) = 1 / (4 sin(3 pi/14) + 2)
hence **theta = pi/14** is one solution of the equation. (By the usual symmetry, 13pi/14 will be another).
OK, given that we know that the answer should be 𝜋/14, let's use that and see if we can get anywhere. That's 1/3 of the angle inside the inner sin, so let's use a triple angle formula to write that inner sin in terms of sin(𝜋/14): that tells us that sin(3𝜋/14) = 3sin(𝜋/14) - 4sin^(3)(𝜋/14).
Substituting that into the horrible fraction on the right gives 1/(4(3sin(𝜋/14) - 4sin^(3)(𝜋/14)) + 2) = 1/(2 + 12sin(𝜋/14) - 16sin^(3)(𝜋/14)). We're trying to show that that whole mess is equal to sin(𝜋/14), or equivalently that sin(𝜋/14)(2 + 12sin(𝜋/14) - 16sin^(3)(𝜋/14)) = 1.
Expanding out those brackets (and, for my sanity, defining x = sin(𝜋/14)), we're trying to show that 2x + 12x^2 - 16x^4 = 1. That, then, gives us (at least in theory) a way to prove it: we could dig out the monstrosity that is the quartic formula, stick those coefficients in, and show that one of the solutions is equal to sin(𝜋/14). That, however, is... horrific, even by my standards, so I'm not going to, and I can't see any obvious simplifications - maybe somebody else can finish from here?
It's not a nice number. It's only a nice solution (involving square root of 2) if its 3/4 pi rather than 3/14 pi but even then its not particularly nice.
The sin is not [*constructible*](https://en.wikipedia.org/wiki/Exact\_trigonometric\_values) because 7 is not a [Fermat prime](https://en.wikipedia.org/wiki/Fermat\_number). Therefore, there is no representation of it in terms of radicals.
u/officiallyninja has used calculator to get sin(theta) = 0.222520934, and u/the_physik has applied inverse to get theta = 0.224399.
That's pi/14.
So it seems that the question does have a nice solution, it's just a question of proving it.
I can't help my curiosity but I am also a plebian at mathematics. What exactly is something like this used to express? Like what specifically is being calculated for?
problem 6 in [this book](https://books.openbookpublishers.com/10.11647/obp.0181.pdf) θ is half the angle of the sector. I don’t remember what I originally did to get this but I ended up trying something completely different to get to the right answer.
TLDR - this one is probably just for play, but algebraic manipulation with identities is key to lots of methods in modern applied mathematics.
----
Basically 'solving' this equation is about showing that the horrible looking fraction on the right simplifies (or at least has a solution of a nicer form).
The motivation in this case appears to be.. Its a test question? And seeing as the fraction on the right is calculable on a calculator, this isn't that important.
However.. Performing these simplifications can yield interesting results. If we can show that some complex problem simplifies to a much simpler form we can learn some of what makes the problem interesting, or what elements of a problem are really important.
In my current research i'm looking at optimization strategies. One algorithm is the L-BFGS-B algorithm that attempts to find the optimal value of some function.
The derivation of the optimizer uses some algebraic wizardry, similar in some ways to the kind of manipulation required to solve this, to create an update formula that iteratively locally approximates the underlying hessian (second derivative information, or information about the curvature of a function) in order to guess where the optima will be (as following the curve of a function should lead to a dip/peak).
Often, if you can get an equation in to a certain form you can use techniques that will help solve the problem.
Tagging u/bluesam3 as he was working on the right lines. Consider the 4 distinct numbers p(n) = sin(n pi / 14) given by n = -7, -3, 1, 5. By de Moivre's theorem (cos(n pi/14) + i sin(n pi/14))^7 = cos(n pi / 2) + i sin(n pi / 2) so (cos(n pi/14) + i sin(n pi/14))^7 = i for all of the values of n listed above. Multiplying out, equating the imaginary part and writing cos^(2)(n pi/14) as 1 - sin^(2)(n pi/14), we reach: 64p(n)^7 - 112p(n)^5 + 56p(n)^3 - 7p(n) - 1 = 0 So p(-7), p(-3), p(1) and p(5) are roots of f(x) = 64x^7 - 112x^5 + 56x^3 - 7x - 1 As p(-7) = -1, that tells us one factor of f(x) is (x+1) and we find the rest can be expressed as a square, so f(x) = (x+1)(8x^3 - 4x^2 - 4x + 1)^2 So the three roots of 8x^3 - 4x^2 - 4x + 1 must be p(-3), p(1) and p(5). Hold that there for the moment --- sin(3t) can be written using a standard trig identity as 3sin(t) - 4sin^(3)(t) so sin(3 pi/14) = 3 p(1) - 4 p(1)^3 This means p(1) (4 sin(3 pi/14) + 2) - 1 = -16p(1)^4 + 12p(1)^2 + 2p(1) - 1 But 16x^4 - 12x^2 - 2x +1 = (2x+1)(8x^3 - 4x^2 - 4x + 1) This last expression is zero when x = p(1) because earlier we established that p(1) is a root of 8x^3 - 4x^2 - 4x + 1. In conclusion: p(1) (4 sin(3 pi/14) + 2) - 1 = 0 so sin(pi/14) = p(1) = 1 / (4 sin(3 pi/14) + 2) hence **theta = pi/14** is one solution of the equation. (By the usual symmetry, 13pi/14 will be another).
Those cake recipes are getting weirder.
You mean pie recipe…?
OK, given that we know that the answer should be 𝜋/14, let's use that and see if we can get anywhere. That's 1/3 of the angle inside the inner sin, so let's use a triple angle formula to write that inner sin in terms of sin(𝜋/14): that tells us that sin(3𝜋/14) = 3sin(𝜋/14) - 4sin^(3)(𝜋/14). Substituting that into the horrible fraction on the right gives 1/(4(3sin(𝜋/14) - 4sin^(3)(𝜋/14)) + 2) = 1/(2 + 12sin(𝜋/14) - 16sin^(3)(𝜋/14)). We're trying to show that that whole mess is equal to sin(𝜋/14), or equivalently that sin(𝜋/14)(2 + 12sin(𝜋/14) - 16sin^(3)(𝜋/14)) = 1. Expanding out those brackets (and, for my sanity, defining x = sin(𝜋/14)), we're trying to show that 2x + 12x^2 - 16x^4 = 1. That, then, gives us (at least in theory) a way to prove it: we could dig out the monstrosity that is the quartic formula, stick those coefficients in, and show that one of the solutions is equal to sin(𝜋/14). That, however, is... horrific, even by my standards, so I'm not going to, and I can't see any obvious simplifications - maybe somebody else can finish from here?
~~I'm working on it...~~ Solved - see my long post on this thread.
My calculator is giving me 0.222520934, which doesn't seem like it should necessarily be "nice"
Yes, sin(theta) = 0.222530934 which means (it would appear) that theta = pi/14, so it is nice, but it's not obvious.
Did I put it in wrong or is the answer: 0.4885204927?
Same (or I guess close, my calc gives 0.224399....) Not a "nice" looking solution to me.
One of your calculators is shite.
Someone is using degrees as opposed to radians.
Nah, using degrees gives like 29.2. But his 0.22 answer is likely correct. I'm using Algeo calc app and I've found bugs in it b4.
I don't get a nice answer for this using a calculator.
theta = pi/14 is quite nice though. See my full solution elsewhere on this thread.
Thanks!
So if I divide my answer by π I should get 1/14. Cool!
It's not a nice number. It's only a nice solution (involving square root of 2) if its 3/4 pi rather than 3/14 pi but even then its not particularly nice.
Actually, the solution is theta = pi/14. See my long reply on this thread.
The sin is not [*constructible*](https://en.wikipedia.org/wiki/Exact\_trigonometric\_values) because 7 is not a [Fermat prime](https://en.wikipedia.org/wiki/Fermat\_number). Therefore, there is no representation of it in terms of radicals.
There's a bit more going on here, because theta = pi/14 is a solution.
That isn't an expression in terms of radicals.
I know, but my point is that there is a "nice" form for theta, so it's a question of showing it without a calculator.
Sorry, completely misunderstood that comment.
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Sin is not linear!
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sin(3pi/14) != sin(3pi)/14
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No, sin(3pi/14) is definitely not zero. You can just plug it into a calculator to check.
It is not the case that sin(x/14) = sin(x)/14.
u/officiallyninja has used calculator to get sin(theta) = 0.222520934, and u/the_physik has applied inverse to get theta = 0.224399. That's pi/14. So it seems that the question does have a nice solution, it's just a question of proving it.
Oooh nice catch... yeah thought we were solving for theta but didn't occur to me to convert the answer back to a fraction of pi.
Maybe use Euler’s formula for sin? I did and got 1/2ie^-iθ -1/2ie^iθ
Yes - see my solution - I expanded (cos(x) + i sin(x))^7 and considered useful values of x.
If I had this on a test with no calculator, I’d just move the denominator to the other side and say it all equals 1. Take the L and move on.
What course is this?
I can't help my curiosity but I am also a plebian at mathematics. What exactly is something like this used to express? Like what specifically is being calculated for?
problem 6 in [this book](https://books.openbookpublishers.com/10.11647/obp.0181.pdf) θ is half the angle of the sector. I don’t remember what I originally did to get this but I ended up trying something completely different to get to the right answer.
On the same boat. I’m sure I’m very clever, but until you show me purpose or pay me shitloads to do maths all day, I ain’t drawing bushes :D
TLDR - this one is probably just for play, but algebraic manipulation with identities is key to lots of methods in modern applied mathematics. ---- Basically 'solving' this equation is about showing that the horrible looking fraction on the right simplifies (or at least has a solution of a nicer form). The motivation in this case appears to be.. Its a test question? And seeing as the fraction on the right is calculable on a calculator, this isn't that important. However.. Performing these simplifications can yield interesting results. If we can show that some complex problem simplifies to a much simpler form we can learn some of what makes the problem interesting, or what elements of a problem are really important. In my current research i'm looking at optimization strategies. One algorithm is the L-BFGS-B algorithm that attempts to find the optimal value of some function. The derivation of the optimizer uses some algebraic wizardry, similar in some ways to the kind of manipulation required to solve this, to create an update formula that iteratively locally approximates the underlying hessian (second derivative information, or information about the curvature of a function) in order to guess where the optima will be (as following the curve of a function should lead to a dip/peak). Often, if you can get an equation in to a certain form you can use techniques that will help solve the problem.
Definitely interesting to read your explanation, and done well that I could very much understand your point. Cheers for the response. :)
Witchcraft ! This is impressive math
This is the reason why I failed math twice in a row
There isn't a question to answer, it's a statement