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let $0<x<1$, prove or dispove $$arcos{\left(\dfrac{\sin{1}-\sin{x}}{1-x}\right)}<\dfrac{x+1}{2}$$ or $$\dfrac{\sin{1}-\sin{x}}{1-x}>\cos{\left( \dfrac{x+1}{2}\right)}.\tag{1}$$
if $(1)$ is right.then I find a method solve inequality
I try $$\dfrac{\sin{1}-\sin{x}}{1-x}=\cos{\xi},x<\xi<1$$ it suffuce to show $$\cos{\xi}>\cos{\dfrac{x+1}{2}},x<\xi<1$$
Method 1. Let $a=(1+x)/2$ and $b=(1-x)/2.$ Then $a+b=1$ and $a-b=x . $ So $$\sin 1-\sin x= \sin (a+b)-\sin (a-b)=$$ $$=(\sin a \cos b+\sin b \cos a)-(\sin a \cos b-\sin b \cos a)=$$ $$=2\sin b \cos a.$$ And $(1-x)^{-1} =(2b)^{-1}.$ So the inequality is equivalent to $$(2b)^{-1}(2 \sin b \cos a)>\cos (1+x)/2=\cos a.$$ Since $x\in (0,1)\implies a=(1+x)/2\in (0,1) \implies \cos a>0,$ we may divide out the term $\cos a.$ So the inequality is equivalent $$(2b)^{-1} (2\sin b)>1.$$ Since $b>0,$ this is equivalent to $\sin b>b,$ which is FALSE.
Method 2. With $a=(1+x)/2$ and with $f(y)=\sin y$ for all $y,$ the inequality is equivalent to $f(1)-f(x)-(1-x)f'(a)>0 .$ To show this is false, we have $$(i).\quad f(1)-f(a)=(1-a)f'(a) +\frac {1}{2}(1-a)^2f''(c) \text { for some } c\in (a,1).$$ $$(ii).\quad f(x)-f(a)=(x-a)f'(a)+\frac {1}{2}(x-a)^2f''(d) \text { for some } d\in (x,a).$$ Subtracting (ii) from (i), and using $(1-a)^2=(x-a)^2$ we have $$(iii).\quad f(1)-f(a)=(1-x)f'(a)+\frac {1}{2}(1-a)^2 (f''(c)-f''(d)).$$ Now $0<x<d<a<c<1$ so $0<d<c<1.$ And $f''(y)=-\sin y$ is strictly decreasing for $y\in (0,1), $ so $$f''(c)-f''(d)<0.$$ Applying this to $(iii),$ considering that $(1-a)^2=((1-x)/2)^2>0,$ we have $$(iv).\quad f(1)-f(a)-(1-x)f'(a)=\frac {1}{2}(1-a)^2(f''(c)-f''(d))<0.$$
Disproof: Let $x=\frac13$, then $$ \frac{\sin{1}-\sin{x}}{1-x}-\cos{\left( \frac{x+1}{2}\right)} < -0.01 < 0 $$
