"Strengthening" the mean value theorem for the sine function The present discussion arises from this MO question. Below, $e$ stands for Euler's number and let
$$\tau:=\arccos\left(\frac{\sin e-\sin 1}{e-1}\right)\approx 1.82\cdots.$$
An application of the Mean Value Theorem (for derivatives) to the function $f(t)=\sin t$ leads to
$$\frac{\sin(e\,t)-\sin(t)}{e\,t-t}=\cos(\xi_tt) \qquad \text{for some $1\leq\xi_t\leq e$}. \tag1$$

QUESTION. Is it true that for each $t>0$, one can always find some $\xi_t\geq\tau$ such that (1) holds? Example: $\xi_1=\tau$.

 A: In view of the answer by Carlo Beenakker and the comment by Alexandre Eremenko, it appears that what you actually had in mind is the following question: 

By the mean value theorem, for each $t\in(0,1]$ there is some $\xi_t\in(1,e)$ such that 
  \begin{equation*}
 r(t):=\frac{\sin et-\sin t}{(e-1)t}=\cos(\xi_t t). \tag{2}
\end{equation*}
  (Since $\cos u$ is strictly decreasing in $u\in[0,e]$, the value of $\xi_t$ is unique for each $t\in(0,1]$.)
  Is it true that $\xi_t\ge\tau$ for all $t\in(0,1]$? 

The answer to this question is yes. Indeed, for $t\in(0,1)$ we have $\xi_t t\in(0,e)\subset[0,\pi]$ and $\tau t\in(0,\tau]\subset[0,\pi]$. Therefore, in view of (2) and because $\cos$ is strictly decreasing on $[0,\pi]$, we see that 
\begin{equation*}
 \xi_t>\tau\iff d(t):=\cos\tau t-r(t)>0;  \tag{3}
\end{equation*}
here and in what follows, $t\in(0,1)$. 
Next, 
\begin{equation*}
d_1(t):=(e-1) d(t)/t^2=\sum_{j=1}^\infty(-1)^jb_j t^{2j-2}-\sum_{j=1}^\infty(-1)^ja_j t^{2j-2},
\end{equation*}
where 
\begin{equation*}
 a_j:=\frac{e^{2 j+1}-1}{(2 j+1)!},\quad b_j:=\frac{(e-1) \tau^{2 j}}{(2 j)!}. 
\end{equation*}
It is easy to see that $0<a_j<a_{j+1}$ and $0<b_j<b_{j+1}$ for all natural $j$. So, 
\begin{equation*}
 d_1(t)>-b_1+b_2t^2-b_3t^4+a_1-a_2t^2>0 \quad\text{if}\quad 0<t\le4/5. 
\end{equation*}
It remains to prove that 
\begin{equation}
 d_2(t):=(e-1)t d(t)>0\quad\text{if}\quad 4/5<t<1.
\end{equation}
Since $d_2(1)=0$, it suffices to show that 
\begin{equation}
 d_2'(t)=\cos t-e \cos et+(e-1) \cos \tau t-(e-1) \tau t \sin \tau t<0
\end{equation}
for $t\in(4/5,1)$. 
Since $\cos t,\cos et,\cos \tau t$ are decreasing in $t\in(4/5,1)$ and $\sin \tau t$ is concave in $t\in(4/5,1)$, the desired result follows because for $t\in[t_j,t_{j+1}]$ and $j=0,\dots,n-1$
\begin{equation}
 d_2'(t)\le\cos t_j-e \cos et_{j+1}+(e-1) \cos \tau t_j
 -(e-1) \tau t_j \min(\sin \tau t_j,\sin \tau t_{j+1})<0, 
\end{equation}
where $n:=20$ and $t_j:=4/5+j/(5n)$. 

To illustrate the above proof, here is the graph $\{(t,d(t))\colon0<t<1\}$:

A: this is a plot of $\frac{\sin e\,t-\sin t}{e\,t-t}-\cos\xi \,t$ as a function of $\xi$ for $t=1.1$; the curve does not cross zero in the interval $[\tau,e]$, so I would conclude that (1) does not hold.

