Sa se calculeze sin si cos de 2pi/3. 13pi/6. 4pi/3.

[tex] \frac{2 \pi }{3} = \frac{2*180}{3} = \frac{360}{3} = 120[/tex]
[tex] \frac{13 \pi }{6} = \frac{13*180}{6} = \frac{2340}{6} =390 [/tex]
[tex] \frac{4 \pi }{3} = \frac{4*180}{3} = \frac{720}{3} = 240[/tex]

Exemplu de calcul:
sin120*= sin(90*+30*) = sin90cos30+sin30cos90= 1*[tex] \sqrt{3} /2[/tex] +1/2*0= [tex] \sqrt{3} /2[/tex] 
cos120*=cos(90*+30*) = cos90cos30-sin90sin30= 0*[tex] \sqrt{3}/2 [/tex]-1*1/2=-1/2

Formule
sinx*=sin([tex] \alpha + \beta [/tex])= [tex]sin \alpha*cos \beta+sin \beta*cos \alpha [/tex]
sinx*=sin([tex] \alpha - \beta [/tex])= [tex]sin \alpha*cos \beta-sin \beta*cos \alpha [/tex]

cosx=cos([tex] \alpha + \beta [/tex])= [tex]cos \alpha*cos \beta-sin \beta*sin \alpha [/tex]
cosx=cos([tex] \alpha - \beta [/tex])= [tex]cos \alpha*cos \beta+sin \beta*sin \alpha [/tex]

Mult succes in elucidarea problemelor.

sin (2π/3)=sin (π–2π/3)=sin (π/3)=√3/2
cos (2π/3)=–cos(π–2π/3)=–cos(π/3)=–1/2

sin(13π/6)=sin(2π+π/6)=sin(π/6)=1/2
cos(13π/6)=cos(2π+π/6)=cos(π/6)=√3/2

sin(4π/3)=sin(2π–2π/3)=sin(–2π/3)=–sin(2π/3)=–sin(π–2π/3)=–sin(π/3)=–√3/2
cos(4π/3)=cos(2π–2π/3)=cos(–2π/3)=cos(2π/3)=cos(π–2π/3)=cos(π/3)=1/2