# two points with same tangent line

curve $y=\sin(x)−x/2+10$.
The slopes of the tangent lines of this function are the same for any two points that are separated by a distance of $2\pi$. Find the two points $(x_0,f(x_0))$ and $(x_0+2\pi,f(x_0+2\pi))$ with $0< x_0<2$ whose tangent lines are the same line.

as much as i can calculated I got, that the slope of tangent line is equal to the derivative of $y$, therefore $= \cos x -1/2 = -1/2$
how should i find the points? $y(x_0)=\pi/2-9.5$

$y(x_0+2\pi)=(\pi+19)/2\pi$ or $19/2\pi$?

i was confused here.

could anyone help me please, any help would be appreciated

• The two contact points have the same $y$, so the common tangent is horizontal and $\cos(x)=1/2$. – Yves Daoust Jun 19 '15 at 16:54
• Sorry, I am wrong. – Yves Daoust Jun 19 '15 at 17:25

Since we have $y'=\cos x-\frac 12$, the tangent line at $(x_0,f(x_0))$ is $$y-f(x_0)=\left(\cos x_0-\frac 12\right)(x-x_0),$$i.e.$$y=\left(\cos x_0-\frac 12\right)x+f(x_0)-x_0\left(\cos x_0-\frac 12\right)$$ and the tangent line at $(x_0+2\pi,f(x_0+2\pi))$ is $$y-f(x_0+2\pi)=\left(\cos(x_0+2\pi)-\frac 12\right)(x-(x_0+2\pi)),$$i.e.$$y=\left(\cos x_0-\frac 12\right)x+f(x_0+2\pi)-(x_0+2\pi)\left(\cos x_0-\frac 12\right).$$ Since these two lines are the same, we have $$f(x_0)-x_0\left(\cos x_0-\frac 12\right)=f(x_0+2\pi)-(x_0+2\pi)\left(\cos x_0-\frac 12\right),$$ i.e. $$\cos x_0=0.$$ So, $x_0=\frac{\pi}{2}+k\pi$ for $k\in\mathbb Z$. From $0\lt x_0\lt 2$, we have $k=0$, so $x_0=\frac{\pi}{2}$.
• $\cos x = 0 , x = (2 k -1) \pi/2$ – Narasimham Jun 19 '15 at 17:03
• @Sarah: Yes, but note that $x_0=\pi/2$, not $x$. What we want to know is $(x_0,f(x_0))$ and $(x_0+2\pi,f(x_0+2\pi))$. So, all we need to know is $x_0$ to know both $f(x_0)$ and $f(x_0+2\pi)$. – mathlove Jun 19 '15 at 18:04