When does the tangent line to the sine curve pass through the origin?

Question

I am trying to find values $a$ and $w$ for which the line $y=ax$ is tangent to the curve $y=\sin(x)$ at $x=w$.

One immediate solution is $a=1$ and $w=0$, but I would like $a<0$ and $w\in (\pi,3\pi/2)$, and I am having a hard time finding such a solution. Asking Mathematica doesn't help (Solve, Reduce, Minimize all say This system cannot be solved) and asking Google gives me elementary calculus problems. Here is a picture showing such values of $a$ and $w$ exist (and giving an impression of how the problem changes with $a$ changing):

It is clear that there is some $a\in(-.3,-.2)$ such that $y=ax$ is tangent to $\sin(x)$, and I could find a numerical approximation, but I want an exact solution.

Thought 1: Describe the tangent line to $\sin(x)$ at $(w,\sin(w))$ as $y=\sin(w)+\cos(w)(x-w)$, and find when that intersects the origin. This leads to solving the equation $\tan(w)=w$, seemingly impossible.

Thought 2: Given $a$, I can find a spot on the sine curve with slope $a$. Indeed, such a point is $\cos(w)=a$, or $w=\arccos(a)$. Due to the limited range of $\arccos$ and wanting $w\in (\pi,3\pi/2)$, I need $w=2\pi-\arccos(a)$. Now I have to solve $a(2\pi-\arccos(a))=\sin(2\pi-\arccos(a))$, which can be simplified by the periodicity of $\sin$ and knowing $\sin(\arccos(a))=\sqrt{1-a^2}$, but that doesn't help too much.

My problem is actually slightly more general, I have the curve $y=\sin(cx)$ for some changing $c$. I know there exists exactly one triple $(a,c,w)$, for $a<0$ and $w\in (\pi/c, 3\pi/2c)$ such that $y=ax$ is tangent to $y=\sin(cx)$ at $x=w$, but I am worried this is a problem that cannot be solved. I tried doing this in the (seemingly) simplest case $c=1$ above, but to no avail. I would be very glad to know if there are related problems / solutions / approaches.

Answer 1

As already said in answers and comments, there is no explicit solution to the equation $x=\tan(x)$; this is already tha case for $x=\cos(x)$ and numerical methods are required.

However, you can have very good numerical approximations using Padé approximants, that is to say writing $$f(x)=\tan(x)-x \approx \frac {P_n(x)}{Q_m(x)}$$ where $P_n(x)$ and $Q_m(x)$ are polynomials of preselected degrees $n$ and $m$. When this is done, we just need to solve for $x$ $P_n(x)=0$. For sure, the developments need to be done around $x=\frac{3\pi}2$.

For example, using $n=1$ and $m=1$ would give $$P_1(x)=-\frac{3}{2} \pi \left(x-\frac{3 \pi }{2}\right)-1\implies x=\frac{9 \pi ^2-4}{6 \pi }\approx 4.500182$$

Using $n=2$ and $m=1$ would give $$P_2(x)=-\frac{2}{3} \left(x-\frac{3 \pi }{2}\right)^2-\frac{3}{2} \pi \left(x-\frac{3 \pi }{2}\right)-1\implies x=\frac{6 \pi +\sqrt{324 \pi ^2-384}1}{16} $$ which is $\approx 4.493398$ while the numerical solution would be $\approx 4.493409$.

Using $n=2$ and $m=3$ would give $$P_2(x)=\frac{1}{80} \left(9 \pi ^2-56\right) \left(x-\frac{3 \pi }{2}\right)^2-\frac{57}{40} \pi \left(x-\frac{3 \pi }{2}\right)-1$$ which would lead to $$x=\frac{-216 \pi +108 \pi ^3-\sqrt{254016 \pi ^2-286720}}{72 \pi ^2-448}\approx 4.493409 $$ which is the numerical solution for seven significant figures.

Edit

I you prefer to avoid radicals,we can consider $x_m$ as the solution of $P_1(x)$ for various $m$. The first terms are given below $$\left( \begin{array}{ccc} m & x_m & \approx \\ 1 & \frac{-4+9 \pi ^2}{6 \pi } & 4.500182390 \\ 2 & \frac{3 \pi \left(-20+27 \pi ^2\right)}{2 \left(-8+27 \pi ^2\right)} & 4.500182390 \\ 3 & \frac{32-252 \pi ^2+243 \pi ^4}{6 \pi \left(-16+27 \pi ^2\right)} & 4.493404773 \\ 4 & \frac{81 \pi \left(16-60 \pi ^2+45 \pi ^4\right)}{2 \left(112-1080 \pi ^2+1215 \pi ^4\right)} & 4.493408901 \\ 5 & \frac{-448+7296 \pi ^2-17820 \pi ^4+10935 \pi ^6}{2 \pi \left(992-4320 \pi ^2+3645 \pi ^4\right)} & 4.493409379 \\ 6 & \frac{\pi \left(-34304+244944 \pi ^2-442260 \pi ^4+229635 \pi ^6\right)}{2 \left(-2176+41328 \pi ^2-113400 \pi ^4+76545 \pi ^6\right)} & 4.493409454 \\ 7 & \frac{8704-242112 \pi ^2+1070496 \pi ^4-1530900 \pi ^6+688905 \pi ^8}{2 \pi \left(-25600+205632 \pi ^2-408240 \pi ^4+229635 \pi ^6\right)} & 4.493409458 \end{array} \right)$$

Answer 2

The tangent to a function $f$ a point $x_0$ has the equation $$y=f(x_0)+f'(x_0)(x-x_0)$$ which can be rewritten as $$y=f(x_0)-f'(x_0)+f'(x_0)x\ .$$ Thus, the intercept with the $y$ axis is at $y=f(x_0)-f'(x_0)$. Therefore, finding tangents that pass through the origin is equivalent to finding roots of the function $g(x)=f-xf'$. In your case $f(x)=\sin(x)$ so you need to find a solution to $$\sin(x)-x\cos(x)=0\ ,$$ or equivalently, $$x=\tan(x)\ .$$ This can be easily done in Mathematica:

Answer 3

I guess the equation you need to solve is $$\frac{\sin(x)}{x} = \cos(x).$$ The left hand side is the slope of the line through the points $(x,\sin(x))$ and the right hand side is the slope of the line tangent to the graph of $y=\sin(x)$ at the point $x$. You want these to be equal.

The reason you're getting the complaint that "This system cannot be solved" is because you need numerical, rather than symbolic techniques to solve the system. The logical way to do this with Mathematica is to use the FindRoot command:

FindRoot[Cos[x] == Sin[x]/x, {x, Pi, 3 Pi/2}]
(* Out: {x -> 4.49341} *)

This yields the following graph:

Answer 4

well, as already pointed out, the $x$ values of the intersection points solve $x = \tan x.$ There are infinitely many such points, you are asking about the first one with $x > 0.$ The slope $a = \cos x.$