Valeur exacte de cos 72°

 

Problθme  :

Formulaire utilisι  :

cos (a + b) = cos a  cos b – sin a  sin b

sin (a + b) = sin a  cos b + cos a  sin b

cos 2x = 2 cos2 x – 1

sin 2x = 2 cos x  sin x

sin2 x = 1 – cos2 x

Solution  :

Tout d’abord, posons  X = cos x  et exprimons  cos 5x  en fonction de  X :

cos 5x = cos(3x + 2x) = cos 3x  cos 2x – sin 3x  sin 2x

cos 5x = cos(2x + x) (2 cos2 x – 1) – sin(2x + x) 2 cos x  sin x

cos 5x = (cos 2x  cos x – sin 2x  sin x) (2 X2 – 1) – (sin 2x  cos x + cos 2x  sin x) 2 X sin x

cos 5x = [(2 X2 – 1) X – 2 X sin x sin x] (2 X2 – 1) – [2 X sin x  X + (2 X2 – 1) sin x] 2 X sin x

cos 5x = [2 X3 – X – 2 X sin2 x] (2 X2 – 1) – [2 X2 + 2 X2 – 1] 2 X sin2 x

cos 5x = [2 X3 – X – 2 X (1 – X2)] (2 X2 – 1) – [4 X2 – 1] 2 X (1 – X2)

cos 5x = (2 X3 – X – 2 X + 2 X3) (2 X2 – 1) – (4 X2 – 1) (2 X – 2 X3)

cos 5x = (4 X3 – 3 X ) (2 X2 – 1) – (4 X2 – 1) (2 X – 2 X3)

cos 5x = (8 X5– 4 X3 – 6 X3 + 3 X) – (8 X3 – 8 X5 – 2 X + 2 X3)

cos 5x = 8 X5– 4 X3 – 6 X3 + 3 X – 8 X3 + 8 X5 + 2 X – 2 X3

cos 5x = 16 X5– 20 X3 + 5 X