Bonsoir,
Voici une méthode utilisant les nombres complexes et en particulier la formule de Moivre.
[tex]\sin(5x)=Im(\cos(5x)+i\sin(5x))\\\\\sin(5x)=Im(\cos(x)+i\sin(x))^5\\\\(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\\\\\sin(5x)=Im(\cos^5(x)+5\cos^4(x)i\sin(x)+10\cos^3(x)i^2\sin^2(x)\\+10\cos^2(x)i^3\sin^3(x)+5\cos(x)i^4\sin^4(x)+i^5\sin^5(x))\\\\\sin(5x)=Im(\cos^5(x)+5i\cos^4(x)\sin(x)-10\cos^3(x)\sin^2(x)\\-10\cos^2(x)i\sin^3(x)+5\cos(x)\sin^4(x)+i\sin^5(x))\\\\\sin(5x)=\sin^5(x)-10cos^2(x)\sin^3(x)+5\cos^4(x)sin(x)[/tex]
Or sin²(x) + cos²(x) = 1 ==> cos²x = 1- sin²x
et cos^4(x) = (1 - sin²(x))² = 1 - 2sin²(x) + sin^4(x)
D'où
[tex]\sin(5x)=\sin^5(x)-10\cos^2(x)\sin^3(x)+5\cos^4(x)\sin(x)\\\\\sin(5x)=\sin^5(x)-10(1-\sin^2(x))\sin^3(x)\\+5(1-2\sin^2(x)+\sin^4(x))\sin(x)\\\\\sin(5x)=\sin^5(x)-10(\sin^3(x)-\sin^5(x))\\+5(\sin(x)-2\sin^3(x)+\sin^5(x))\\\\\sin(5x)=\sin^5(x)-10\sin^3(x)+10\sin^5(x)\\+5\sin(x)-10\sin^3(x)+5\sin^5(x)\\\\\boxed{\sin(5x)=16\sin^5(x)-20\sin^3(x)+5sin(x)}[/tex]
