Question 476923
{{{cos(90^o + x) = -sinx}}}
{{{cos(180^o - x) = -cosx}}}

{{{sin(180^o + x) = -sinx}}}
{{{sin(90^o + x) = cosx}}}

==>sin (90° + x) sin (180° + x) + cos (90° + x) cos (180° – x) = 
cosx*-sinx + -sinx*-cosx = -sinx*cosx + sinx*cosx = 0.