Hopeooooo wrote: In the coordinate plane,the graps of functions $y=sin x$ and $y=tan x$ are drawn, along with the coordinate axes. Using compass and ruler, construct a line tangent to the graph of sine at a point above the axis, $Ox$, as well at a point below that axis (the line can also meet the graph at several other points) Tangent to $\sin x$ at $t$ is $y=x\cos t +\sin t-t\cos t$ Double tangency not horizontal means $\cos t_1=\cos t_2\ne 0$ and $\sin t_1-t_1\cos t_1=\sin t_2-t_2\cos t_2$ for some $t_1\ne t_2$ Since $\cos t_1=\cos t_2$, we have $\sin t_1=-\sin t_2$ else we'd have $t_1\cos t_1=t_2\cos t_2$ impossible with $t_1\ne t_2$ and $\cos\ne 0$ So $t_2=2k\pi-t_1$ and second equality becomes $\sin t_1-t_1\cos t_1=-\sin t_1-(2k\pi-t_1)\cos t_1$ And so $t_1-k\pi=\tan t_1$ Hence the solution : Mark tha point $M(0,2\pi)$ as the second nonzero intersection of $\sin x$ with $0x$ With compass centered at $(0,0)$ and using $M$, mark the point $N(0,-2\pi)$ With ruler, draw line $MN$ (equation $y=x-2\pi$) Mark intersection $P$ of this line with graph of $\tan x$ between $\frac{\pi}2$ and $\pi$ With compass and ruler, build the symetric $P'$ of $P$ relative to Ox With compass and ruler, draw the line $x=2\pi$ With compass and ruler, build the symetric $Q$ of $P$ relative to the line $x=2\pi$ With compass and ruler, build the symetric $Q'$ of $P'$ relative to the line $x=2\pi$ Line $PP'$ cuts graph of $\sin x$ at point $A$ Line $QQ'$ cuts graph of $\sin x$ at point $B$ Line $AB$ is one suitable required double tangent.

Interesting... Just consider the first intersection $P$ of the curve $y= \tan(x)$ with the line $y=x$ above the $x$ axis and to the right of the $y$ axis. Draw the perpendicular to $x$ axis until it intersect the $y=sin(x)$ curve at point $Q$. By calculus, the line joining the origin and $Q$ works.