中學數學一問(三角函數)

1. sinx=1=cosy=sinz,
(x,y,z)=(π/2 + 2nπ, 2m, π/2 + 2kπ)
so, cos(x+y+z)=cos(π)= -1
2. z=π/2 -(x+y)
cosxsinycosz
=(1/2)[sin(x+y)-sin(x-y)]*sin(x+y)
又 x+y= π/2 - z <= 5π/12
x>=y >= π/12, 如圖

圖片參考：http://imgcld.yimg.com/8/n/AE03435620/o/101103210202213869528110.jpg


(取右端點x+y最大, x-y 亦最大)
故取 x+y=5π/12, x-y= 3π/12時,可得最小值
=(1/2)sin(5π/12)[sin(5π/12)-sin(π/4)] = 1/8
(此時(x,y,z)=(π/3, π/12, π/12)


2011-03-21 14:42:04 補充：
註: x+y固定, 取y=π/12時, x-y最大, 故欲得最小值 y=π/12代回cosxsinycosz得 cosx *(1/4)
x越大時,cos(x)/4越小, 故(x,y,z)=(π/3, π/12, π/12)時可得最小值

2011-03-21 15:55:48 補充：
Q1: (x,y,z)=(π/2 + 2nπ, 2mπ, π/2 + 2kπ), m,n,k為整數
Q2: (x,y,z)=(5π/24, 5π/24, π/12)時,可得最大值

1.x,y,z沒有限定
也可能是+1吧!?