With an appropiate substitution of the Taylor polynomial around $0$ give an estimation of $sin 1$ with the accuracy of $10^{-4}$. I know that the reminder is $R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ and that the Taylor polynom of the sinus is $x-\dfrac {x^{3}} {3!}+\dfrac {x^{5}} {5!}-\ldots +\left( -1\right) ^{n-1}\dfrac {x^{2n-1}} {(2n-1)!}+R_{2n+1}$
$f'(x)= cos (x)$         $f'(0)=1$
$f''(x) = -sin (x)$      $f''(x) = -sin (0)=0$
$f'''(x)=-cos(x)$ $f'''(x)=-cos(0)=-1$
$f^{iv}(x)=sin(x)$ $f^{iv}(x)=sin(0)=0$
$f^{v}(x)=cos(x)=f'(x)$
$T_{5}(x)=0 + x + \frac{0}{2!}x^2+-\frac{1}{3!}x^3+\frac{0}{4!}x^4+\frac{1}{5!}x^5$
how can I get the value of $sin 1$?

Let $A=1-1/3!+1/5!-1/7!.$ $$\text {Then }\quad A-\sin 1=1/9!-(1/10!-1/11!)-(1/12!-1/13!)-...<1/9!$$ $$ \text {and }\quad A-\sin 1=(1/9!-1/10!)+(1/11!-1/12!)+... >0.$$ So $0<A-\sin 1<1/9!=1/362,880.$
Or we could say that with $x=1,$ $a=0$ and $f(x)=\sin x,$ we have $x-a=1, $ so for some $c\in (a,x)=(0,1)$ we have $$A-\sin 1=R_7=\frac {f^{(8)}(c)}{8!} =\frac {\sin c}{8!}.$$  And since $0<\sin c<1$ for $c\in (0,1)$ we have $0<R_7<1/8!=1/40,320.$
Remark: From the first part we see that the $c$ in the second part satisfies $\sin c<1/9.$  And we have $R_7=R_8$ because $f^{(8)}(a) =f^{(8)}(0)=\sin 0=0.$
