Question 923952
you have {{{y=sin(2x-pi)}}}

{{{y = a sin(bx + c)}}}=> in your case  {{{a=1}}},{{{b=2}}}, and {{{c=-pi}}}

Both {{{b}}} and {{{c}}} in these graphs affect the phase shift (or displacement), given by:

    {{{Phase_ shift=(-c)/b}}}  => in your case {{{(-(-pi)/2)=pi/2}}}



you can write {{{pi)}}} as {{{4pi/4)}}} or {{{2pi/2}}}, and that was used because {{{x}}} values

The choice of the values of {{{x}}} in the table correspond  to {{{x}}}  and {{{y}}} intercepts, maxima  and minima points.

{{{x}}}|{{{y}}}
{{{2pi/4}}}|{{{ 0}}}.........{{{y=sin(2(2pi/4)-4pi/4)=sin(pi-pi)=sin(0)=0}}}

{{{3pi/4}}}|{{{1}}}.......{{{y=sin(2(3pi/4)-4pi/4)=sin(3pi/2-2pi/2)=sin(pi/2)=1}}}

{{{4pi/4}}}| {{{0}}}........{{{y=sin(2(4pi/4)-4pi/4)=sin(2pi-pi)=sin(pi)=0}}}

{{{5pi/4}}}|{{{-1}}}........{{{y=sin(2(5pi/4)-4pi/4)=sin(5pi/2-2pi/2)=sin(3pi/2)=-1}}}

{{{6pi/4}}}|{{{0}}}........{{{y=sin(2(6pi/4)-4pi/4)=sin(3pi-pi)=sin(2pi)=0}}}

{{{drawing( 600,600, -5, 5, -5, 5,circle(-pi,0,.1),locate(-3.2,-0.2,(-pi)),circle(pi,0,.1),locate(3,-0.3,(pi)),circle(pi/2,0,.1),locate(pi/2,0,(pi/2)),circle(-pi/2,0,.1),locate(-2,0,(-pi/2)), graph( 600,600, -5, 5, -5, 5, sin(2x-pi))) }}}


domain: all real numbers
range:
{ {{{y}}} element {{{R}}} : {{{-1<=y<=1}}} }