Question

Sin120-sin60+cos45+cos135..urgent !

Answer 1

sin120=sin(60+60)=sin60cos60+sin60cos60=√3/2•1/2+√3/2•1/2=√3/2
sin60=√3/2
cos45=√2/2
cos135=cos(180-45)= cos180cos45+sin180sin45=-1•√2/2+0•√2/2=-√2/2
√3/2-√3/2+√2/2-√2/2=0

Answer 2

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