[tex] \displaystyle\\
\sin120^o-\sin60^o+\cos45^o+\cos135^o = ?\\\\
\sin 120^o = \sin(180^o-60^o)=\sin60^o~~~\text{Sinusul e pozitiv in cadranul 2.}\\\\
\cos135^o=\cos(180-45^o)=-\cos45^o~~~\text{Cosinusul e negativ in cadranul 2.}\\\\
\text{Rezulta:}\\\\
\sin120^o-\sin60^o+\cos45^o+\cos135^o =\\\\
=\sin60^o-\sin60^o+\cos45^o-\cos45^o = 0+0=\boxed{\bf 0} [/tex]
