[tex]\sin (120^o)=\sin (90+30)[/tex][tex]\sin (a+b)=\sin a\cos b+\cos a\sin b[/tex][tex]\begin{gathered} \sin (90+30)=(\sin 90)(\cos 30)+(\cos 90)(\sin 30) \\ \sin (90+30)=1(\frac{\sqrt[]{3}}{2})+0(\frac{1}{2})_{} \\ \sin (90+30)=\frac{\sqrt[]{3}}{2} \\ \sin (120)=\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]
