On utilise les formules d'Euler
cos(5x)=1/2(e^(5ix)+e^(-5ix))
de plus e^(5ix)=cos(5x)+i.sin(5x)
et e^(5ix)=(e^(ix))^5=(cos x+i. sin x)^5
donc cos(5x)=Re [ (cos x+i.sin x)^5 ]
or (a+b)^5=a^5+5a^4.b+10a^3.b^2+10a^2.b^3+5a.b^4+b^5
donc (cos x+i.sin x)^5
=16 i sin^5(x)-20 i sin^3(x)+5 i sin(x)+cos(x)+16 sin^4(x) cos(x)-12 sin^2(x) cos(x)
donc cos(5x)
=Re [ (cos x+i.sin x)^5 ]
=cos^5(x)-10 sin^2(x).cos^3(x)+5 sin^4(x).cos(x)