Question

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

Answer 1

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

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

Formule
sinx*=sin($\alpha + \beta$)= $sin \alpha*cos \beta+sin \beta*cos \alpha$
sinx*=sin($\alpha - \beta$)= $sin \alpha*cos \beta-sin \beta*cos \alpha$

cosx=cos($\alpha + \beta$)= $cos \alpha*cos \beta-sin \beta*sin \alpha$
cosx=cos($\alpha - \beta$)= $cos \alpha*cos \beta+sin \beta*sin \alpha$

Mult succes in elucidarea problemelor.

Answer 2

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