In this problem you have to maximize the distance between two particles of positions #s_1# and #s_2#. This means that you have to maximize the following quantity (that is time-dependent)
#d(t)=|s_1(t)-s_2(t)|=|cos(t)-cos(t+pi/4)|#
To maximize this quantity, let's compute the derivative
#d'(t)=(cos(t)-cos(t+pi/4))/|cos(t)-cos(t+pi/4)| * (sin(t+pi/4)-sin(t))#
and study when it's positive:
#(cos(t)-cos(t+pi/4))*(sin(t+pi/4)-sin(t))>0#
The product is positive if and only if the two factors are both positive or both negative. So we can study their sign separately.
#cos(t)-cos(t+pi/4)>0 iff cos(t)>cos(t+pi/4)#
The equality occurs when #t=-pi/8 + k pi# for all #k in ZZ#, i.e. the inequality is satisfied for #t \in (-pi/8+2kpi,7/8pi+2 k pi)# for all #k in ZZ#. So for these values the first factor is positive, and for the complementary values in #RR# it's not.
#sin(t+pi/4)-sin(t)>0 iff sin(t+pi/4)>sin(t)#
The equality occurs when #t=3/8 pi + k pi# for all #k in ZZ#, i.e. the inequality is satisfied for #t in (-5/8pi+2k pi,3/8 pi+2k pi)# for all #k in ZZ#.
If we finally consider the product, we get that the derivative #d'(t)# is positive if and only if #t in (-pi/8+k pi,3/8 pi + k pi)# for all #k in ZZ# and it's negative if and only if #t \in (3/8 pi + k pi,7/8 pi + k pi)# for all #k in ZZ#.
So #3/8 pi+ kpi# for all #k in ZZ# are the point that maximize the distance #d(t)#.
