Official solution
y = sinx.sin(x+2) - sin2(x+1)
\[ \Rightarrow {1 \over 2}\left\{ {2\sin \left( {x + 2} \right)\sin x - 2{{\sin }^2}(x + 1)} \right\}\]
\[ \Rightarrow {1 \over 2}\left\{ {\cos 2 - \cos (2x + 2) + cos(2x + 2) - 1} \right\} = - {\sin ^2}1 < 0\]
Hence the line passes through III and IV quadrant