Exercice 30
1) on sait que sin²(π/5)+cos²(π/5)=1
donc sin²(π/5)= 1-cos²(π/5)
= 1- [(√5+1)/4]²
= 1-(√5+1)²/16
= 1-(5+2√5+1)/16
= 1-(6+2√5)/16
= (10-2√5)/16
or π/5∈[0;π/2] alors sin( π/5)>0
Donc sin(π/5)= √(10-2√5)/4
2)
a) sin(4π/5)=sin(π-π/5)=sin(π/5)=√(10-2√5)/4
cos(4π/5)= cos(π-π/5)=-cos(π/5)=-(√5+1)/4
b) sin(-π/5)=-sin(π/5)=- √(10-2√5)/4
cos(-π/5)=cos(π/5)=(√5+1)/4
c) cos(6π/5)=cos(π+π/5)=-cos(π/5)=-(√5+1)/4
sin(6π/5) = sin(π+π/5)=-sin(π/5)=- √(10-2√5)/4
d) cos(3π/10)=cos(π/2-π/5)=sin(π/5)=√(10-2√5)/4
sin(3π/10)=sin(π/2-π/5)=cos(π/5)=(√5+1)/4
Bon courage !
