Răspuns :
[tex] \displaystyle\\
\sin120^o + \cos150^o = \\\\
\sin(180^o-60^o) + \cos(180^o-30) = \\\\
\sin60^o + \Big(- \cos30^o\Big)=\\\\
\sin60^o-\cos30^o = \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=\frac{\sqrt{3}-\sqrt{3}}{2}=\frac{0}{2}=\boxed{\bf 0}
[/tex]
[tex] \it sin120^o+cos150^o=sin(180^o-120^o) -cos(180^o-150^o)=
\\ \\
= sin60^o-cos30^o=\dfrac{\sqrt3}{2} -\dfrac{\sqrt3}{2} = 0 [/tex]
