兩條A.math !!!(急問)(10分)

1. cos Acos 3A=cos 7Acos 5A
[cos 4A+cos(-2A)]/2=(cos 12A +cos 2A)/2
cos 4A+cos 2A=cos 12A+cos 2A <== cos(-2A)=cos2A
cos 4A=cos 12A
4A=2n pi + 12A or 4A=2n pi - 12A
A= -n pi /4 or A= n pi /8

2.16 cos^4 x=1
cos^4 x= 1/16
cos^2 x=1/4 or cos^2 x=-1/4 (捨去)
cos x = 1/2 or cos x=-1/2
x= n pi + pi/3 or x=n pi - pi/3 (由於四個象限都有解....可以將ANS合併)

2007-12-08 22:55:00 補充：
在計算通解時cos 4A=cos 12A己經可以用通解拆....不用將佢變番積.....咁會令條數錯的機會增加

1. cosAcos3A - cos7Acos5A = 0
(cos(-2A) - cos4A)/2 - (cos2A - cos12A)/2 = 0 as cosxcosy = (cos(x-y) - cos(x+y))/2
cos2A - cos4A - cos2A + cos12A = 0 as cos(-2A) = cos2A
cos12A - cos4A = 0
2cos8Acos4A = 0 as cosx - cosy = 2cos((x+y)/2)cos((x-y)/2)
cos4A = 0 or cos 8A = 0
cos4A=0 A = n*pi/4 n = any integer
cos8A = 0 A = n*pi/8 n = any integer

2. (2cosx)^4 =1 (2cosx)^2 = +1 or -1(rejected as cos x must be real)
2cosx = m+1 0r -1
x = n*pi +30 or n*pi -30 n= any integer

用
cos(A)cos(B)=(cos(A+B)+cos(A-B))/2
和
cos(A)-cos(B)=2sin((A+B)/2)sin((A-B)/2)


cos(A)cos(3A) = cos(7A)cos(5A)
(cos(4A)+cos(2A))/2 = (cos(12A)+cos(2A))/2
cos(4A)=cos(12A)
cos(12A)-cos(4A)=0
2sin(8A)sin(4A)=0
sin(8A)=0 or sin(4A)=0

如果 sin(8A)=0,
A=n*pi 或 A=n*pi/8

如果 sin(4A)=0,
A=n*pi 或 A=n*pi/4

答案: A=n*pi, n*pi/4 或 n*pi/8

2007-12-08 22:18:35 補充：
2.16 cos^4 x=1cos^4 x=1/16cosx=+/- 1/2x=(n+1/3)pi 或 (n+2/3)pi