Question 1158099
I believe there was a mistake copying the formula in the question.
As written in the question, it says
{{{sin(-theta)sin(90theta+theta)-cos(180theta-theta)cos(270theta+theta)}}} ,
which could be written more simply as {{{sin(-theta)sin(91theta)-cos(179theta)cos(271theta)}}}
I do not know of a way to further simplify that, and it does not sound like a typical high school math question.
 
I suspect that the expression to simplify was meant to be
{{{sin(-theta)*sin(90^o+theta)-cos(180^o-theta)*cos(270^o+theta)}}}
That could  be simplified using the relations know between the trigonometric functions sine and cosine of angles that are reflected or turned by {{{90^o}}} and {{{180^o}}} .
You could find those relations in any list of "trigonometric identities",
but no need to memorize them, because you can visualize them for any angle in the unit circle.
 
For example, if we change the sign of the angle , sine changes sign, but cosine stays the same:
{{{highlight(sin(-theta)=-sin(theta))}}} , but {{{cos(-theta)=cos(theta))}}}
{{{drawing(200,300,-0.1,1.4,-0.9,1.35,
arrow(-0.1,0,1.35,0),arrow(0,0,0,1.3),arrow(0,0,0,-1.3),
circle(0,0,1),triangle(0,0,0.92,0.39,0.92,-0.39),
rectangle(0.92,-0.05,0.87,0.05),arc(0,0,0.6,0.6,0,22.9),
arc(0,0,0.7,0.7,-22.9,0),locate(0.35,-0.03,theta),
locate(1.37,0,x),locate(0.05,1.35,y),green(line(0.92,-0.02,0.92,-0.39)),
red(line(0,0,0.92,0)),green(line(0.92,0.02,0.92,0.39)),
locate(0.92,0.5,P),locate(0.92,-0.39,Q)
)}}} {{{P(red(cos(theta)),green(sin(theta)))}}}  {{{Q(red(cos(theta)),green(sin(-theta)))}}}

 
Adding {{{90^o}}} to an angle is turning it 1/4 of a circle, and turns sine into cosine and cosine into -sine.
{{{drawing(300,200,-1.1,1.15,-0.1,1.4,
arrow(-0.1,0,1.15,0),arrow(0,0,0,1.3),arrow(0,0,-1.3,0),
circle(0,0,1),triangle(0,0,0.92,0.39,0.92,-0),
rectangle(0.92,0,0.87,0.05),arc(0,0,0.6,0.6,-112.9,-90),
arc(0,0,0.7,0.7,-22.9,0),locate(0.35,0.15,theta),
locate(1.13,0,x),locate(0.05,1.35,y),
red(line(0,0,0.92,0)),green(line(0.92,0.02,0.92,0.39)),
locate(0.92,0.5,P),
triangle(0,0,-0.39,0.92,0,0.92),rectangle(0,0.92,-0.05,0.87),
green(line(-0.39,0.92,-0.02,0.92)),red(line(0,0,0,0.92)),
locate(-0.39,1.05,Q)
)}}} {{{P(red(cos(theta)),green(sin(theta)))}}}  {{{Q(green(cos(90^o+theta)),red(sin(90^o+theta)))}}}
So, {{{highlight(sin(90^o+theta)=cos(theta))}}} and {{{cos(90^o+theta)=-sin(theta)}}} .
 
Using the two highlighted relations above we can start simplifying:
{{{sin(-theta)*sin(90^o+theta)-cos(180^o-theta)*cos(270^o+theta)}}}
{{{"="}}}{{{-sin(theta)*cos(theta)-cos(180^o-theta)*cos(270^o+theta)}}}}
 
Also, adding {{{180^o}}} to an angle (turning it half a circle), causes the sign to change for the sine and cosine functions.
So, 
{{{cos(180^o-theta)=-cos(-theta)}}} , but we know that {{{cos(-theta)=cos(theta)}}}, so {{{highlight(cos(180^o-theta)=-cos(theta))}}}
Similarly
{{{cos(270^o+theta)=cos(180^o+(90^o+theta))=-cos(90^o+theta)}}} , but we know that {{{cos(90^o+theta)=-sin(theta)}}} ,so
{{{highlight(cos(270^o+theta)=sin(theta))}}}
 
Using the highlighted relations above we can continue simplifying:
{{{sin(-theta)*sin(90^o+theta)-cos(180^o-theta)*cos(270^o+theta)}}}
{{{"="}}}{{{-sin(theta)*cos(theta)-cos(180^o-theta)*cos(270^o+theta)}}}}
{{{"="}}}{{{-sin(theta)*cos(theta)-(-cos(90^o+theta))*(sin(theta))}}}
{{{"="}}}{{{-sin(theta)*cos(theta)+sin(theta)*cos(theta)=highlight(0)}}}