amaths

cos(x+y+z) + cos(x+y-z) + cos(x-y+z) + cos(-x+y+z) = 4cosxcosycosz
cos(x+y+z) + cos(x+y-z)
= 2 cos[(x+y+z+x+y-z)/2] cos[(x+y+z-x-y+z)/2]
= 2 cos(x+y) cosz
cos(x-y+z) + cos(-x+y+z)
= 2 cos[(x-y+z-x+y+z)/2] cos[(x-y+z+x-y-z)/2]
= 2 cosz cos(x-y)
L.H.S. = 2 cos(x+y) cosz + 2 cosz cos(x-y)
= 2 cosz [cos(x+y) + cos(x-y)]
= 2 cosz (2) cos[(x+y+x-y)/2] cos[(x+y-x+y)/2]
= 2 cosz (2) cosx cosy
= 4cosxcosycosz
= R.H.S.
只重複用左一條公式
cos M cos N = 2 cos[(M+N)/2] cos[(M-N)/2]
OK? 要補習嘛? 有我, 囉 A, 無難度

2008-02-18 21:32:30 補充：
oh, 高手好多, 搶答問題, 有難度...記得投我一票

都唔係靠實力攞最佳ge! 講咩「OK? 要補習嘛? 有我, 囉 A, 無難度」, 你估你真係補習天王咩?

cos(x+y+z)+cos(x+y-z)+cos(x-y+z)+cos(-x+y+z)=4cosxcosycosz
L.H.S=cos(x+y+z)+cos(x+y-z)+cos(x-y+z)+cos(-x+y+z)
=2cos(x+y+z+x+y-z/2)cos(x+y+z-x-y+z)+2cos(x-y+z-x+y+z)cos(x-y+z+x-y-z)
=2cos(x+y)cos(z)+2cos(z)cos(x-y)
=2cos(z)[cos(x+y)+cos(x-y)]
=2cos(z)[2cos(x+y+x-y)cos(x+y-x+y)]
=2cos(z)[2cos(x)cos(y)]
=4cosxcosycosz

2008-02-18 21:28:31 補充：
sorry.....我所有部份公式都漏左除2...cos(x y z) cos(x y-z) cos(x-y z) cos(-x y z)=4cosxcosycoszL.H.S=cos(x y z) cos(x y-z) cos(x-y z) cos(-x y z)=2cos(x y z x y-z/2)cos(x y z-x-y z) 2cos(x-y z-x y z)cos(x-y z x-y-z)=2cos(x y)cos(z) 2cos(z)cos(x-y)=2cos(z)[cos(x y) cos(x-y)]

2008-02-18 21:28:41 補充：
=2cos(z)[2cos(x y x-y/2)cos(x y-x y/2)]=2cos(z)[2cos(x)cos(y)]=4cosxcosycosz=R.H.S

Prove the identity:

cos(x+y+z)+cos(x+y-z )+cos(x-y+z)+cos(-x+ y+z)=4cosxcosycosz
cosx +cosy = 2 cos((x +y)/2) cos((x –y)/2).
cosx –cosy = -2 sin((x +y)/2) sin((x –y)/2).
Consider
cos(x+y+z)+cos(x+y-z )
=2cos(x+y)cos(z )
cos(x-y+z)+cos(-x+ y+z)
=2cos(z)cos(x- y)
So
cos(x+y+z)+cos(x+y-z )+cos(x-y+z)+cos(-x+ y+z)
=2cos(x+y)cos(z )+2cos(z)cos(x- y)
=2[cos(x+y)cos(z )+cos(z)cos(x- y)]
=2cos(z)[cos(x+y) +cos(x- y)]
=2cos(z)[2cos(x)cos(y)]
=4cosxcosycosz