amaths

Question

http://i270.photobucket.com/albums/jj88/ucc041087/...

help me to prove the Q25 above .thz!~

Answer 1

最愛解答

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

Answer 2

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?

Answer 3

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

Answer 4

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