A line tangent to the graph of sin

Question

We have the function $f(x)=\sin x$, and a line tangent to the function at 2 different points.
The slop of the line is $\alpha$.
I need to show that $\tan\alpha=\alpha$.
Here's what I have tried:
Let $A(x_1;\sin x_1)$ be a point on $f(x)$ where the line with slope $\alpha$ tangent to the functiom.
$f'(x)=\cos x$, And we get the tangent line to the graph at $A$ is the following:
$y=\cos(x_1)\cdot x+\sin x_1-x_1\cdot \cos(x_1)$.
Now let $A'(x_0,\sin(x_0))$ be the other point.
Thus:
$$\cos(x_0)=\cos(x_1)=\alpha$$ $$\sin(x_0)-x_0\cdot \cos(x_0) = \sin(x_1)-x_1\cdot \cos(x_1)$$ From the first equation we get $x_0=x_1+2\pi k$ or $x_0=-x_1+2\pi k$.
From the first case, if we plug into the second equation we get: $$\cos(x_0)=0$$ So $\alpha=0$ and $\tan\alpha=\alpha$ I could not get any progress with the second case. Does anyone have another idea to solve this problem or can help me with the second case?
Thanks!

Answer 1

It seems that as of the time I last visited this question, there is something wrong about it. Inspection of the sine function indicates that if the slope of the line is not zero, the horizontal distance between tangent points is greater than $2\pi,$ while the vertical distance is less than $2.$ So $\lvert \alpha \rvert < \frac 1\pi,$ and the only possible solution of $\tan\alpha=\alpha$ under these constraints is $\alpha=0.$

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Let’s consider the case $\alpha\neq 0$ with tangent points at $(x_0,y_0)$ and $(x_1,y_1).$ As you found, this implies that $x_0 = 2\pi k - x_1.$ Therefore $\frac12(x_0 + x_1) = k \pi.$

Let $x_m = k\pi$ and let $g(x) = \sin(x - x_m).$ Then $g(x) = \sin x$ or $g(x) = -\sin x,$ depending on whether $k$ is even or odd. Without loss of generality, let $x_0 = x_m - \xi$ and let $x_1= x_m +\xi,$ where $\xi > 0.$ Then the line $y=\alpha x$ is parallel to the original tangent line specified in the problem statement, and $y=\alpha x$ is tangent to $y=g(x)$ at $(-\xi,y_0)$ and at $(\xi,y_1).$

For even $k,$ we have \begin{align} \sin \xi &= g(x) = \alpha \xi,\\ \cos \xi &= g’(x) = \alpha, \end{align} and therefore $$ \tan \xi = \frac{ \alpha \xi }{ \alpha } = \xi.$$ For odd $k,$ the signs of both $g(x)$ and $g’(x)$ are reversed, and we again have $ \tan \xi = \xi.$

That is as close as I was able to get to any formula like “$\tan\alpha = \alpha.$”