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A friend of mine asked how to lower and raise a constant line on the x-y axis as a function of sines and cosines. That is where I found that $$\sin(x) + \sin(x+y) = y $$ would do the trick if I found a function $y$ that works. My first instinct was to solve the equation but it seems like a rather difficult problem and I was unable to do so (although maybe I am wrong) My second thought was that, given that I already have a function for $y$, maybe approximating it with a Taylor series would work.
So far, working out the implicit derivatives and then solving for $y'$ has been a huge problem efficiency-wise. I was wondering if there is a quicker method to solve this problem or in case there isn't, is there an implicit derivative calculator online that will help me do this faster?

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    $\begingroup$ it sounds like representing parallel lines with a changing coefficients using sine,cosine $\endgroup$ – Archis Welankar Jan 7 '16 at 16:56
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    $\begingroup$ maybe you could elaborate a little on what exactly your original problem is. $\endgroup$ – user159517 Jan 7 '16 at 17:05
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    $\begingroup$ @user159517 I am trying to isolate $y$, I simply want to express $y$ in terms of $x$, the original problem was to find a function that used only sines and cosines to produce a constant line with slope 0, but I think that it is much more interesting to simply isolate $y$ in terms of $x$, even if it is an infinite series. $\endgroup$ – Guacho Perez Jan 7 '16 at 17:12
  • $\begingroup$ You could never find an exact function in terms of x or y however the interesting thing is that all points on the implcit function can be expressed in a closed form. If you check wolfram alpha and type "sin(x)+sin(x+a)=a" you will find four general solutions to $x$. $\endgroup$ – Arbuja Jan 8 '16 at 3:13
  • $\begingroup$ @GuachoPerez Are you allowed to use arcsin for raising a constant line? $\endgroup$ – Arbuja Jan 8 '16 at 13:09
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I am not sure what you mean in the first question by "raising a constant line" in terms of $\sin$ or $\cos$. You could just use $a$ in $y=\arcsin{\left(\sin(a)\right)}$ so as $a$ increases, the line raises at a constant rate.

If you want to take only $x$ and $y$-values then shifting a constant line is impossible. Only a constant number in the form of $a$ can shift $x$ and $y$ values; not $x$ and $y$ values themselves. If you take $x^2$ there is no way you shift its vertex from $(0,0)$ to $(a,0)$ without changing $x^2$ to $(x-a)^2$.

For the second part of your question, which is now unrelated to the first part, you cannot to solve $\sin(x+y)+\sin(x)=y$ since one of $y$ variables is free from the sine brackets. However you can find solutions in terms of $x$ but it very difficult to find by hand. On computer I replaced $y$ with $a$ so wolfram alpha could register a solution as shown here (look under "solutions to $x$").

I was also able to find a taylor series as I have found using mathematica. I chose $(\pi,0)$ because the curve was less steep in that region so the rate of convergence can be quick. I found the radius of convergence was near $({\pi/4,3\pi/4})$.

$$[\sin(x+y)+\sin(y)=y]\approx-(x-\pi)+\frac{\left(x-\pi\right)^{3}}{(3!)(2)}-\frac{\left(x-\pi\right)^{5}}{(5!)(2)}+\frac{(x-{\pi})^7}{2(7!)}+\frac{17(x-{\pi})^9}{(9!)}$$

More coefficients of the taylor series can be found here. To find even more coefficients you should get mathematica and for simply graphing functions you can use desmos.

The most exciting thing I found is that $\sin(x+y)+\sin(x)=y$ cannot be solved explcitly for $x$ or $y$ as one equation but has all points that can be described as a closed form.

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We add an answer which is still an implicit function but in another way . $$\sin x +\sin (x+y)=(1+\cos y)\sin x+\sin y\cos x=y$$ Put $A=\sqrt{(1+\cos y)^2+(\sin y)^2}=\sqrt{2(1+\cos y)}$ so $A$ is the hypotenusa of a triangle whose legs are $\sin y$ and $1+\cos y$; dividing by $A$ it follows $$\frac{1+\cos y}{A}\sin x+\frac{\sin y}{A}\cos x=\frac{y}{A}$$ and puting $\sin \theta=\frac {\sin y}{A}$ we have $$\sin x \cos \theta+\cos x \sin \theta=\sin (x+\theta)=\frac{y}{A}$$

Thus $$\color{red}{y=\sqrt{2(1+\cos y)}\sin (x+\theta)}$$ where $$ \theta=\arcsin \left(\frac{\sin y}{\sqrt{2(1+\cos y)}}\right)$$


HINT.-The equation being a trascendental one, it seems doesn't matter of a closed form or getting $y=f(x)$. An approximation could be get solving several times $$\sin x +\sin (x+a_i)=a_i$$ (which is not a nice gift) and interpolating some manner, particularly polynomials, the resulting points $(x_i,a_i)$.

On the other hand some points of the function can be easily get, for example, when you make $$\sin x +\sin (x -\frac{\pi}{3})=-\frac{\pi}{3}\qquad (*)$$ you have the function $3\sin x-\sqrt 3\cos x$ at the LHS and this function has its minimun equal to $-3.464$ which is less than $-\frac{\pi}{3}$ hence the equation $(*)$ has solution; making $\tan \frac x2=t$ you can obtain an exact solution $(x,y)$. But this exactly calculated point is not important for the viewpoint of interpolation, here suggested

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  • $\begingroup$ I checked your exact solution and the arcsin is undefined. If you graph the implicit function there is no value of $-\frac{\pi}{2}$. $\endgroup$ – Arbuja Jan 8 '16 at 2:58
  • $\begingroup$ You are right: the expression is equal to $\sin x-\cos x$ so is never equal to $-\frac {\pi}{2}$; this was an oversight on my part. Anyway, it could be other values for $a_i$ with exact solution, which is purely "anecdotic". Which stay valid although embarrassing, I think is a possible interpolation. Thank you very much for your remark. $\endgroup$ – Piquito Jan 8 '16 at 16:36

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