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Ok, I scoured the internet for more than a few months for this one (whenever I had the time). But just because of this question I've created my first Stack Exchange account.

I have thrown this at all Computer Algebra System software I was accessible to. (not Maple or Mathematica though)

I also tried complex transformation, no luck there (not too much experience there)

I know this could be easily numerically solved because it is guaranteed that there is exactly one solution for every interval when: $a \sin(bx + c) \le 1$. In my case, the general solution is crucial because I'm applying this to an optimization where I can't afford to have an additional dimension to be optimized.

This problem can be visualized as the intersection of two sine waves.

A simplified version of the problem (when a = 1) could be easily solved with the sum-to-product identity:

$\sin(bx + c) = \sin(x)$

$0 = \sin(bx + c) - \sin(x)$

$0 = 2 \cos(\frac{x(b + 1) + c}{2})\sin(\frac{x(b - 1) + c}{2})$

$0 = \cos(\frac{x(b + 1) + c}{2})$ and $0 = \sin(\frac{x(b - 1) + c}{2})$ etc...

But for the time being, the only valid assumption in my case is that b is rational

Andrew Karasev
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  • I'm quite curious about the mentioned optimization problem that leads to this particular equation. Would you mind giving a reference? –  Jan 19 '21 at 22:55
  • @mrsamy I don't have like a tangible reference, its just my personal project. I am doing this for an optimization problem that generates epicycloid geometry. This is just the last bit that has to do with the constraints of the problem. – Andrew Karasev Jan 19 '21 at 23:07
  • That makes sense. Epicycloids are fun. Thanks for the info and good luck to your project :-) –  Jan 19 '21 at 23:12
  • So for "solving", do you expect a solution $x$ written in terms of $a,b,c$? –  Jan 19 '21 at 23:13
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    in the $a=1$ example, I think it should be.. $$\sin(bx+c)-\sin(x) = 2 \cos \bigg(\frac{(b+1)x + c}{2} \bigg)\sin\bigg(\frac{(b -1)x+ c}{2}\bigg)$$ – WW1 Jan 19 '21 at 23:14
  • @mrsamy Yes, since the answer is going to have an infinite amount of answers, it will also have an "n" and probably $\pi$ in it – Andrew Karasev Jan 19 '21 at 23:16
  • @WW1 thank you, I typed this out by memory. edited now. My notes do say the same. – Andrew Karasev Jan 19 '21 at 23:20
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    I guess you must go to a "numerical side" analysis. – Felix Marin Jan 20 '21 at 03:17

2 Answers2

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COMMENT.-I doubt that there is a general closed solution as a function of $a, b, c$. However, I offer you here an equation of which approximate solutions could be found.

We have $$a\sin(bx+c)=a(\sin(c)\cos(bx)+\cos(c)\sin(bx))$$ from which we get the equation $$a\sin(c)\sum_{k=0}^{\infty}\frac{(-1)^k(bx)^{2k}}{(2k)!}+a\cos(c)\sum_{k=0}^{\infty}\frac{(-1)^k(bx)^{2k+1}}{(2k+1)!}=\sum_{k=0}^{\infty}\frac{(-1)^k(x)^{2k+1}}{(2k+1)!}$$ This is a really difficult problem and what you want is probably impossible, I think.

Piquito
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This is a supplement to the answer of @Piquito. Recalling the identity theorem for power series, we make a coefficient comparison of the series expansion \begin{align*} a\sin(c)\sum_{k=0}^\infty (-1)^k\frac{\left(bx\right)^{2k}}{(2k)!} +a\cos(c)\sum_{k=0}^\infty (-1)^k\frac{\left(bx\right)^{2k+1}}{(2k+1)!} =\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!} \end{align*} In the following we denote with $[x^n]$ the coefficient of $x^n$ of a series.

We consider even and odd coefficients separately and obtain for $n\geq 0$: \begin{align*} &[x^{2n}]:\quad &a\sin(c)\frac{(-1)^nb^{2n}}{(2n)!}&=0&\\ &&a\sin(c)b^{2n}&=0&\tag{1}\\ \\ &[x^{2n+1}]:\quad &a\cos(c)\frac{(-1)^nb^{2n+1}}{(2n+1)!}&=\frac{(-1 )^n}{(2n+1)!}&\\ &&a\cos(c)b^{2n+1}&=1&\tag{2}\\ \end{align*}

From (1) we conclude that at least one factor has to be equal to zero. Looking at (2) we see that $a,b\ne 0$ and it follows \begin{align*} \sin(c)=0\qquad\rightarrow\qquad c=j\pi\qquad j\in\mathbb{Z}\tag{3} \end{align*} Putting $c=j\pi$ in (2) we obtain \begin{align*} a(-1)^jb^{2n+1}=1\qquad\qquad n\geq 0,\ j\in\mathbb{Z}\tag{4} \end{align*} From (4) we see that $j\in2\mathbb{Z}$. Since (4) has to be valid for all $n\geq 0$ we conclude that $b=1$ and finally that $a=1$.

Conclusion: The equation \begin{align*} a\sin(bx+c)=\sin(x)\qquad\qquad a,b,c\in\mathbb{C} \end{align*} has a solution if and only if $a=b=1$ and $b\in 2j\pi, j\in\mathbb{Z}$. In this case we have the identity \begin{align*} \sin(2j\pi x)=\sin(x)\qquad\qquad x\in\mathbb{C} \end{align*}

Markus Scheuer
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