Points of contacts of tangents of the curve $y=\sin x$

Question

Prove that the points of contacts of tangents of the curve $y={\sin x}$ drawn from origin lie on the curve $\frac{1}{x^2} - \frac{1}{y^2} = -1$

Answer 1

Points on the curve $y=\sin x$ that have tangent lines through the origin satisfy:

$$\text{slope of curve at $(x,y)$ is } \dfrac{y}{x} = \dfrac{\sin x}{x}.$$

By simple differentiation, the slope of the curve at $x$ is $\cos x$, so our points satisfy

\begin{align} \cos x &= \dfrac{\sin x}{x} \\ \therefore\quad x &= \tan x \\ \text{or}\quad \tan^{-1} x &= x. \end{align}

So we have

\begin{align} y = \sin x &= \sin\left(\tan^{-1} x\right) = \dfrac{x}{\sqrt{x^2+1}} \\ \therefore\quad y^2 &= \dfrac{x^2}{x^2+1} = \dfrac{1}{1+1/x^2} \\ \text{Re-arranging gives, as required,}\qquad & \dfrac{1}{x^2} - \dfrac{1}{y^2} = -1. \end{align}