Remainder for Maclaurin series is R5(x)=sin(x)-S5(x). When x=1:
R5(1)=sin(1)-305353/362880,
R5(1)=∑(-1)n/(2n+1)!-(1-1/3!+1/5!-1/7!+1/9!) where summation limits are 0≤n<∞.
R5(1)=∑(-1)n/(2n+1)! where summation limits are 5≤n<∞. We can take S5 to be equivalent to S5(1) and R5 to be equivalent to R5(1). The first term of this reduced series is -1/11! and continues +1/13!-1/15!+1/17!-... Because of alternating signs, the Maclaurin series for sin(1) oscillates between being a little below sin(1) and being a little above sin(1). The discrepancy (absolute distance between true sin(1) and the series summation up to a given n), represented by |S-Sn|, decreases with progressive terms. R5<0 because the lead term is negative and the next term (positive 1/13!) has a lower magnitude so the overall sum is negative. The same applies to the next pair of terms, implying that the overall sum is negative: -1/11!+1/13!=-1/11!+1/(13×12×11!)=(1/11!)(-1+1/156)=-155/13!. We need to compare magnitudes using |S-S5| and |R5|. S as an approximation to sin(1) to which it converges, will always be slightly above or below the true value. Similarly R5 will always be slightly above or below the true difference. -R5≤S-S5≤R5 is another way to express the relationship.
I hope this helps.