1.....cos(a)=4/7
on sait que cos²a+sin²a=1
donc sin²a=1-cos²a=1-(4/7)²=1-16/49=49/49-16/49=33/49
alors sin a=√33/7 ou sin a=-√33/7 a ∈[-5π;-4π] donc sin a<0 ⇒sin a=-√33/7
2....a... sin (13π/12)=sin (π+π/12)=-sin(π/12)=(-√6+√2)/4
cos(π+π/12)=-cos(π/12)=(-√6-√2)/4
b....sin (11π/12)=sin(π-π/12)=sin (π12)
cos(11π/12)=cos(π-π/12)=-cos(π/12)
c....sin(-π/12)=-sin(π/12)
cos(-π/12)=cos(π/12)
3.. ca demande bcp d'explication
