A) On sait que Cos²x+Sin²x=1 quel que soit x
Sin²(π/5)=1-Cos²(π/5)
Cos²(π/5)=[tex](\frac{1+\sqrt{5}}{4})^{2}=\frac{1+2\sqrt{5}+5}{16}=\frac{3+\sqrt{5}}{8} [/tex]
Donc
Sin²(π/5)=[tex]1-\frac{3+\sqrt{5}}{8}=\frac{5-\sqrt{5}}{8} [/tex]
Donc Sin(π/5)=[tex]\sqrt{\frac{5-\sqrt{5}}{8}} [/tex]
B) 4π/5=π-π/5
Cos(π-π/5)=-cos(π/5)
Donc Cos(4π/5)=-cos(π/5)
Sin(π-π/5)=Sin(π/5)
Donc Sin(4π/5)=Sin(π/5)
9π/5=π+4π/5
Donc Cos(π+4π/5)=-Cos(4π/5)=Cos(π/5)
Cos(9π/5)=Cos(π/5)
Sin(π+4π/5)=-Sin(4π/5)=-Sin(π/5)
Sin(9π/5)=-Sin(π/5)
