sin(a+b)=sin(a)cos(b)+cos(a)sin(b);
sin(a-b)=sin(a)cos(b)-cos(a)sin(b).
So, sin(a+b)-sin(a-b)=2cos(a)sin(b).
If a+b=x and a-b=1, then 2a=(x+1), so a=½(x+1) and 2b=(x-1), so b=½(x-1).
Therefore sin(x)-sin(1)=2cos(½(x+1))sin(½(x-1)).
Now, let x-1=h where h is very small.
(sin(x)-sin(1))/(x-1)=2cos(½(h+2))sin(½h)=hcos(½(h+2)) because sin(½h)≈½h when h is small.
h is very small when compared to 2, so hcos(½(h+2))=hcos(1) as h→0.
Therefore (sin(x)-sin(1))/(x-1)=hcos(1)/h=cos(1)=0.5403 (approx) in the limit as h→0, which is when x→1.