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Sagot :
Bonjour Anella771
Démontrer que :
a) sin (PI/8) - sin (3PI/8) + sin (5PI/8) - sin (7PI/8) = 0
[tex]\sin(\dfrac{\pi}{8})-\sin(\dfrac{3\pi}{8})+\sin(\dfrac{5\pi}{8})-\sin(\dfrac{7\pi}{8})\\\\=\sin(\dfrac{\pi}{8})-\sin(\dfrac{3\pi}{8})+\sin(\pi-\dfrac{5\pi}{8})-\sin(\pi-\dfrac{7\pi}{8})\\\\=\sin(\dfrac{\pi}{8})-\sin(\dfrac{3\pi}{8})+\sin(\dfrac{3\pi}{8})-\sin(\dfrac{\pi}{8})\\\\=(\sin(\dfrac{\pi}{8})-\sin(\dfrac{\pi}{8}))+(\sin(\dfrac{3\pi}{8})-\sin(\dfrac{3\pi}{8}))\\\\=0[/tex]
Par conséquent,
[tex]\boxed{\sin(\dfrac{\pi}{8})-\sin(\dfrac{3\pi}{8})+\sin(\dfrac{5\pi}{8})-\sin(\dfrac{7\pi}{8})=0}[/tex]
b) cos^2 (PI/8) + cos^2 (3PI/8) + cos^2 (5PI/8) + cos^2 (7PI/8) = 2
[tex]\cos^2(\dfrac{\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\dfrac{5\pi}{8})+\cos^2(\dfrac{7\pi}{8})\\\\=\cos^2(\dfrac{\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\pi-\dfrac{5\pi}{8})+\cos^2(\pi-\dfrac{7\pi}{8})\\\\=\cos^2(\dfrac{\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\dfrac{\pi}{8})\\\\=2\cos^2(\dfrac{\pi}{8})+2\cos^2(\dfrac{3\pi}{8})\\\\=1+\cos(\dfrac{\pi}{4})+1+\cos(\dfrac{3\pi}{4})\\\\=2+\cos(\dfrac{\pi}{4})-\cos(\pi-\dfrac{3\pi}{4})[/tex]
[tex]\\\\=2+\cos(\dfrac{\pi}{4})-\cos(\dfrac{\pi}{4})\\\\=2[/tex]
Par conséquent,
[tex]\boxed{\cos^2(\dfrac{\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\dfrac{5\pi}{8})+\cos^2(\dfrac{7\pi}{8})=2}[/tex]
c) cos (3PI/8) sin (PI/8) + cos (25PI/8) sin (11PI/8) = 1
[tex]\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})+\cos(\dfrac{25\pi}{8})\times\sin(\dfrac{11\pi}{8})\\\\=\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})-\cos(3\pi-\dfrac{25\pi}{8})\times\sin(\pi-\dfrac{11\pi}{8})\\\\=\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})-\cos(-\dfrac{\pi}{8})\times\sin(-\dfrac{3\pi}{8})\\\\=\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})-\cos(\dfrac{\pi}{8})\times(-\sin(\dfrac{3\pi}{8}))[/tex]
[tex]\\\\=\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})+\cos(\dfrac{\pi}{8})\times\sin(\dfrac{3\pi}{8})\\\\=\sin(\dfrac{\pi}{8}+\dfrac{3\pi}{8})\\\\=\sin(\dfrac{4\pi}{8})\\\\=\sin(\dfrac{\pi}{2})\\\\=1[/tex]
Par conséquent,
[tex]\boxed{\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})+\cos(\dfrac{25\pi}{8})\times\sin(\dfrac{11\pi}{8})=1}[/tex]
Résoudre les équations de type cos (u (x)) = cos (v (x)) suivantes :
a) cos (x) = cos (4x) dans /R
[tex]\cos(x)=\cos(4x)\\\\x=4x+2k\pi\ ou\ x=-4x+2k\pi\ (k\in\mathbb{Z})\\\\x-4x=2k\pi\ ou\ x+4x=2k\pi\ (k\in\mathbb{Z})\\\\-3x=2k\pi\ ou\ 5x=2k\pi\ (k\in\mathbb{Z})\\\\\boxed{x=\dfrac{2k\pi}{3}\ ou\ x=\dfrac{2k\pi}{5}\ (k\in\mathbb{Z})}[/tex]
b) cos (2x) = - ((racine de 2) / 2) dans /R
[tex]\cos(2x)=-\dfrac{\sqrt{2}}{2}\\\\\cos(2x)=\cos(\dfrac{3\pi}{4})\\\\2x=\dfrac{3\pi}{4}+2k\pi\ ou\ 2x=-\dfrac{3\pi}{4}+2k\pi\ \ (k\in\mathhbb{Z})\\\\\boxed{x=\dfrac{3\pi}{8}+k\pi\ ou\ x=-\dfrac{3\pi}{8}+k\pi\ \ (k\in\mathhbb{Z})}[/tex]
c) cos (3x) = 1/2 dans ]-PI ; PI ] puis dans ] 0 ; 3PI ].
[tex]\cos(3x)=\dfrac{1}{2}\\\\\cos(3x)=\cos(\dfrac{\pi}{3})\\\\3x=\dfrac{\pi}{3}+2k\pi\ ou\ 3x=-\dfrac{\pi}{3}+2k\pi\ \ (k\in\mathbb{Z})\\\\x=\dfrac{\pi}{9}+\dfrac{2k\pi}{3}\ ou\ x=-\dfrac{\pi}{9}+\dfrac{2k\pi}{3}\ \ (k\in\mathbb{Z})\\\\x=\dfrac{\pi}{9}+\dfrac{6k\pi}{9}\ ou\ x=-\dfrac{\pi}{9}+\dfrac{6k\pi}{9}\ \ (k\in\mathbb{Z})\\\\Dans\ ]-\pi\ ;\ \pi],\\ \boxed{x=\pm\dfrac{\pi}{9}\ \ ou\ \ x=\pm\dfrac{5\pi}{9}\ \ ou\ \ x=\pm\dfrac{7\pi}{9}}[/tex]
[tex]\\\\Dans\ [0\ ;\ 3\pi],\\ \boxed{x=\dfrac{\pi}{9}\ ou\ \dfrac{7\pi}{9}\ ou\ \dfrac{13\pi}{9}\ ou\ \dfrac{20\pi}{9}\ ou\ \dfrac{26\pi}{9}\ ou\ \dfrac{5\pi}{9}\ ou\ \dfrac{11\pi}{9}\ ou\ \dfrac{17\pi}{9}\ ou\ \dfrac{23\pi}{9}}[/tex]
Résoudre les équations de type sin (u (x)) = sin (v (x)) suivantes :
a) sin (x) = sin (3x) dans /R
[tex]\sin(x)=\sin(3x)\\\\x=3x+2k\pi\ ou\ x=\pi-3x+2k\pi\ \ (k\in\mathbb{Z})\\\\x-3x=2k\pi\ ou\ x+3x=\pi+2k\pi\ \ (k\in\mathbb{Z})\\\\-2x=2k\pi\ ou\ 4x=\pi+2k\pi\ \ (k\in\mathbb{Z})\\\\\boxed{x=k\pi\ ou\ x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\ \ (k\in\mathbb{Z})}[/tex]
b) sin (x) = sin (2x + PI/4) dans /R
[tex]\sin(x)=\sin(2x+\dfrac{\pi}{4})\\\\x=2x+\dfrac{\pi}{4}+2k\pi\ \ ou\ \ x=\pi-(2x+\dfrac{\pi}{4})+2k\pi\ (k\in\mathbb{Z})\\\\x-2x=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ x=\pi-2x-\dfrac{\pi}{4}+2k\pi\ (k\in\mathbb{Z})\\\\-x=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ 3x=\dfrac{3\pi}{4}+2k\pi\ (k\in\mathbb{Z})\\\\\boxed{x=-\dfrac{\pi}{4}+2k\pi\ \ ou\ \ x=\dfrac{\pi}{4}+\dfrac{2k\pi}{3}\ (k\in\mathbb{Z})}[/tex]
c) sin (2x) = ((racine de 2) / 2) dans ] -PI ; PI ] puis dans [ 0 ; 3PI ]
[tex]\sin(2x)=\dfrac{\sqrt{2}}{2}\\\\\sin(2x)=\sin(\dfrac{\pi}{4})\\\\2x=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ 2x=\pi-\dfrac{\pi}{4}+2k\pi\ \ (k\in\mathbb{Z})\\\\2x=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ 2x=\dfrac{3\pi}{4}+2k\pi\ \ (k\in\mathbb{Z})\\\\x=\dfrac{\pi}{8}+k\pi\ \ ou\ \ x=\dfrac{3\pi}{8}+k\pi\ \ (k\in\mathbb{Z})[/tex]
[tex]Dans\ ]-\pi\ ;\ \pi]\\\boxed{x=-\dfrac{7\pi}{8}\ \ ou\ \ -\dfrac{5\pi}{8}\ \ ou\ \ \dfrac{\pi}{8}\ \ ou\ \ \dfrac{3\pi}{8}}\\\\Dans\ [0\ ;\ 3\pi]\\\boxed{x=\dfrac{\pi}{8}\ \ ou\ \ \dfrac{3\pi}{8}\ \ ou\ \ \dfrac{9\pi}{8}\ \ ou\ \ \dfrac{11\pi}{8}\ \ ou\ \ \dfrac{17\pi}{8}\ \ ou\ \ \dfrac{19\pi}{8}}[/tex]
Démontrer que :
a) sin (PI/8) - sin (3PI/8) + sin (5PI/8) - sin (7PI/8) = 0
[tex]\sin(\dfrac{\pi}{8})-\sin(\dfrac{3\pi}{8})+\sin(\dfrac{5\pi}{8})-\sin(\dfrac{7\pi}{8})\\\\=\sin(\dfrac{\pi}{8})-\sin(\dfrac{3\pi}{8})+\sin(\pi-\dfrac{5\pi}{8})-\sin(\pi-\dfrac{7\pi}{8})\\\\=\sin(\dfrac{\pi}{8})-\sin(\dfrac{3\pi}{8})+\sin(\dfrac{3\pi}{8})-\sin(\dfrac{\pi}{8})\\\\=(\sin(\dfrac{\pi}{8})-\sin(\dfrac{\pi}{8}))+(\sin(\dfrac{3\pi}{8})-\sin(\dfrac{3\pi}{8}))\\\\=0[/tex]
Par conséquent,
[tex]\boxed{\sin(\dfrac{\pi}{8})-\sin(\dfrac{3\pi}{8})+\sin(\dfrac{5\pi}{8})-\sin(\dfrac{7\pi}{8})=0}[/tex]
b) cos^2 (PI/8) + cos^2 (3PI/8) + cos^2 (5PI/8) + cos^2 (7PI/8) = 2
[tex]\cos^2(\dfrac{\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\dfrac{5\pi}{8})+\cos^2(\dfrac{7\pi}{8})\\\\=\cos^2(\dfrac{\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\pi-\dfrac{5\pi}{8})+\cos^2(\pi-\dfrac{7\pi}{8})\\\\=\cos^2(\dfrac{\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\dfrac{\pi}{8})\\\\=2\cos^2(\dfrac{\pi}{8})+2\cos^2(\dfrac{3\pi}{8})\\\\=1+\cos(\dfrac{\pi}{4})+1+\cos(\dfrac{3\pi}{4})\\\\=2+\cos(\dfrac{\pi}{4})-\cos(\pi-\dfrac{3\pi}{4})[/tex]
[tex]\\\\=2+\cos(\dfrac{\pi}{4})-\cos(\dfrac{\pi}{4})\\\\=2[/tex]
Par conséquent,
[tex]\boxed{\cos^2(\dfrac{\pi}{8})+\cos^2(\dfrac{3\pi}{8})+\cos^2(\dfrac{5\pi}{8})+\cos^2(\dfrac{7\pi}{8})=2}[/tex]
c) cos (3PI/8) sin (PI/8) + cos (25PI/8) sin (11PI/8) = 1
[tex]\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})+\cos(\dfrac{25\pi}{8})\times\sin(\dfrac{11\pi}{8})\\\\=\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})-\cos(3\pi-\dfrac{25\pi}{8})\times\sin(\pi-\dfrac{11\pi}{8})\\\\=\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})-\cos(-\dfrac{\pi}{8})\times\sin(-\dfrac{3\pi}{8})\\\\=\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})-\cos(\dfrac{\pi}{8})\times(-\sin(\dfrac{3\pi}{8}))[/tex]
[tex]\\\\=\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})+\cos(\dfrac{\pi}{8})\times\sin(\dfrac{3\pi}{8})\\\\=\sin(\dfrac{\pi}{8}+\dfrac{3\pi}{8})\\\\=\sin(\dfrac{4\pi}{8})\\\\=\sin(\dfrac{\pi}{2})\\\\=1[/tex]
Par conséquent,
[tex]\boxed{\cos(\dfrac{3\pi}{8})\times\sin(\dfrac{\pi}{8})+\cos(\dfrac{25\pi}{8})\times\sin(\dfrac{11\pi}{8})=1}[/tex]
Résoudre les équations de type cos (u (x)) = cos (v (x)) suivantes :
a) cos (x) = cos (4x) dans /R
[tex]\cos(x)=\cos(4x)\\\\x=4x+2k\pi\ ou\ x=-4x+2k\pi\ (k\in\mathbb{Z})\\\\x-4x=2k\pi\ ou\ x+4x=2k\pi\ (k\in\mathbb{Z})\\\\-3x=2k\pi\ ou\ 5x=2k\pi\ (k\in\mathbb{Z})\\\\\boxed{x=\dfrac{2k\pi}{3}\ ou\ x=\dfrac{2k\pi}{5}\ (k\in\mathbb{Z})}[/tex]
b) cos (2x) = - ((racine de 2) / 2) dans /R
[tex]\cos(2x)=-\dfrac{\sqrt{2}}{2}\\\\\cos(2x)=\cos(\dfrac{3\pi}{4})\\\\2x=\dfrac{3\pi}{4}+2k\pi\ ou\ 2x=-\dfrac{3\pi}{4}+2k\pi\ \ (k\in\mathhbb{Z})\\\\\boxed{x=\dfrac{3\pi}{8}+k\pi\ ou\ x=-\dfrac{3\pi}{8}+k\pi\ \ (k\in\mathhbb{Z})}[/tex]
c) cos (3x) = 1/2 dans ]-PI ; PI ] puis dans ] 0 ; 3PI ].
[tex]\cos(3x)=\dfrac{1}{2}\\\\\cos(3x)=\cos(\dfrac{\pi}{3})\\\\3x=\dfrac{\pi}{3}+2k\pi\ ou\ 3x=-\dfrac{\pi}{3}+2k\pi\ \ (k\in\mathbb{Z})\\\\x=\dfrac{\pi}{9}+\dfrac{2k\pi}{3}\ ou\ x=-\dfrac{\pi}{9}+\dfrac{2k\pi}{3}\ \ (k\in\mathbb{Z})\\\\x=\dfrac{\pi}{9}+\dfrac{6k\pi}{9}\ ou\ x=-\dfrac{\pi}{9}+\dfrac{6k\pi}{9}\ \ (k\in\mathbb{Z})\\\\Dans\ ]-\pi\ ;\ \pi],\\ \boxed{x=\pm\dfrac{\pi}{9}\ \ ou\ \ x=\pm\dfrac{5\pi}{9}\ \ ou\ \ x=\pm\dfrac{7\pi}{9}}[/tex]
[tex]\\\\Dans\ [0\ ;\ 3\pi],\\ \boxed{x=\dfrac{\pi}{9}\ ou\ \dfrac{7\pi}{9}\ ou\ \dfrac{13\pi}{9}\ ou\ \dfrac{20\pi}{9}\ ou\ \dfrac{26\pi}{9}\ ou\ \dfrac{5\pi}{9}\ ou\ \dfrac{11\pi}{9}\ ou\ \dfrac{17\pi}{9}\ ou\ \dfrac{23\pi}{9}}[/tex]
Résoudre les équations de type sin (u (x)) = sin (v (x)) suivantes :
a) sin (x) = sin (3x) dans /R
[tex]\sin(x)=\sin(3x)\\\\x=3x+2k\pi\ ou\ x=\pi-3x+2k\pi\ \ (k\in\mathbb{Z})\\\\x-3x=2k\pi\ ou\ x+3x=\pi+2k\pi\ \ (k\in\mathbb{Z})\\\\-2x=2k\pi\ ou\ 4x=\pi+2k\pi\ \ (k\in\mathbb{Z})\\\\\boxed{x=k\pi\ ou\ x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\ \ (k\in\mathbb{Z})}[/tex]
b) sin (x) = sin (2x + PI/4) dans /R
[tex]\sin(x)=\sin(2x+\dfrac{\pi}{4})\\\\x=2x+\dfrac{\pi}{4}+2k\pi\ \ ou\ \ x=\pi-(2x+\dfrac{\pi}{4})+2k\pi\ (k\in\mathbb{Z})\\\\x-2x=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ x=\pi-2x-\dfrac{\pi}{4}+2k\pi\ (k\in\mathbb{Z})\\\\-x=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ 3x=\dfrac{3\pi}{4}+2k\pi\ (k\in\mathbb{Z})\\\\\boxed{x=-\dfrac{\pi}{4}+2k\pi\ \ ou\ \ x=\dfrac{\pi}{4}+\dfrac{2k\pi}{3}\ (k\in\mathbb{Z})}[/tex]
c) sin (2x) = ((racine de 2) / 2) dans ] -PI ; PI ] puis dans [ 0 ; 3PI ]
[tex]\sin(2x)=\dfrac{\sqrt{2}}{2}\\\\\sin(2x)=\sin(\dfrac{\pi}{4})\\\\2x=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ 2x=\pi-\dfrac{\pi}{4}+2k\pi\ \ (k\in\mathbb{Z})\\\\2x=\dfrac{\pi}{4}+2k\pi\ \ ou\ \ 2x=\dfrac{3\pi}{4}+2k\pi\ \ (k\in\mathbb{Z})\\\\x=\dfrac{\pi}{8}+k\pi\ \ ou\ \ x=\dfrac{3\pi}{8}+k\pi\ \ (k\in\mathbb{Z})[/tex]
[tex]Dans\ ]-\pi\ ;\ \pi]\\\boxed{x=-\dfrac{7\pi}{8}\ \ ou\ \ -\dfrac{5\pi}{8}\ \ ou\ \ \dfrac{\pi}{8}\ \ ou\ \ \dfrac{3\pi}{8}}\\\\Dans\ [0\ ;\ 3\pi]\\\boxed{x=\dfrac{\pi}{8}\ \ ou\ \ \dfrac{3\pi}{8}\ \ ou\ \ \dfrac{9\pi}{8}\ \ ou\ \ \dfrac{11\pi}{8}\ \ ou\ \ \dfrac{17\pi}{8}\ \ ou\ \ \dfrac{19\pi}{8}}[/tex]
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