thang1704

con gà bài 2 nhé cưng ;v

$A=(\cos 2x + \cos 2y )/2 + \cos ^2z$

$A=\cos (x+y).\cos (x-y)+\cos ^2(n.\pi -z)$

$A=\cos n\(pi -z).\cos (x-y)+\cos ^2(n\pi -z)$

$A=\cos (n\pi -z).(\cos (x-y)+\cos (n.\pi -z))$

$A=2.\cos (n.\pi -z).\cos \frac{x-y+n.\pi -z}{2}.\cos \frac{x-y-n.\pi -z}{2}$$A=2.\cos (n.\pi -z).\cos \frac{2.x+n\pi -x-y-z}{2}.\cos \frac{x+y+z-n.\pi -2.y}{2}$

$A=2.\cos (n.\pi -z).\cos \frac{n.\pi -n\pi +2x}{2}.\cos \frac{n.\pi -n.\pi -2y}{2}$

$A=2.\cos (n.\pi -z).\cos x.\cos y$

$A=(-1)^{n}.2.\cos x.\cos y.\cos z$

ĐIỀU PHẢI CHỨNG MINH

$B.\sin \frac{3\pi }{7}=\sin \frac{3\pi }{7}.\cos \frac{\pi }{7}- \sin \frac{3\pi }{7}.\cos \frac{2\pi }{7}+\sin \frac{3\pi }{7}.\cos \frac{3\pi }{7}$

$=\frac{1}{2}.(\sin \frac{4\pi }{7}+\sin \frac{2\pi }{7}-\sin \frac{5\pi }{7}-\sin \frac{\pi }{7}+\sin \frac{6\pi }{7})$

$=\frac{1}{2}.(\sin \frac{4\pi }{7}+ (\sin \frac{2\pi }{7}-\sin \frac{5\pi }{7})+(\sin \frac{6\pi }{7}-\sin \frac{\pi }{7}))$

$=\frac{1}{2}.(\sin \frac{4\pi }{7}+(\sin \frac{2\pi }{7}-\sin \frac{2\pi }{7})+(\sin \frac{6\pi }{7}-\sin \frac{\pi }{7}))$$B.\sin \frac{3\pi }{7}=\frac{1}{2}.\sin \frac{3\pi }{7}$

$B=\frac{1}{2}$

khỷ khỷ khỷ 

Bài 1: Tính 

1, B = $cos \frac{2\pi }{7} + cos \frac{4\pi }{7} + cos \frac{8\pi }{7}$

2, C = $cos \frac{\pi }{7} - cos \frac{2\pi }{7} + cos \frac{3\pi }{7}$

 

Bài 2: Cho x,y,z thỏa mãn x + y + z = $n\pi$ ( n thuộc Z) 

Chứng minh rằng: $cos^{2}x + cos^{2}y + cos^{2}z - 1 = (-1)^{n}.2.cosx.cosy.cosz$

con gà bài 2 nhé cưng ;v

$A=(\cos 2x + \cos 2y )/2 + \cos ^2z$

$A=\cos (x+y).\cos (x-y)+\cos ^2(n.\pi -z)$

$A=\cos n\(pi -z).\cos (x-y)+\cos ^2(n\pi -z)$

$A=\cos (n\pi -z).(\cos (x-y)+\cos (n.\pi -z))$

$A=2.\cos (n.\pi -z).\cos \frac{x-y+n.\pi -z}{2}.\cos \frac{x-y-n.\pi -z}{2}$$A=2.\cos (n.\pi -z).\cos \frac{2.x+n\pi -x-y-z}{2}.\cos \frac{x+y+z-n.\pi -2.y}{2}$

$A=2.\cos (n.\pi -z).\cos \frac{n.\pi -n\pi +2x}{2}.\cos \frac{n.\pi -n.\pi -2y}{2}$

$A=2.\cos (n.\pi -z).\cos x.\cos y$

$A=(-1)^{n}.2.\cos x.\cos y.\cos z$

ĐIỀU PHẢI CHỨNG MINH