大师作文网数学三角函数题,急求过程
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数学三角函数题,急求过程
设函数f(x)=a.b,其中向量a=(2cosx,1),b=(cosx,√3sin2x)
①求函数f(x)的单调减区间
②若x属于[-派/4,0],求函数f(x)的值域
设函数f(x)=a.b,其中向量a=(2cosx,1),b=(cosx,√3sin2x)
①求函数f(x)的单调减区间
②若x属于[-派/4,0],求函数f(x)的值域
f(x)=a*b=(2cosx,1)*(cosx,√3sin2x)=2cos²x+√3sin2x
=1+cos2x+√3sin2x=1+2sin(2x+π/6)
(1)2kπ+π/2≤2x+π/6≤2kπ+3π/2
2kπ+π/3≤2x≤2kπ+4π/3
kπ+π/6≤x≤kπ+2π/3
f(x)的单调减区间:[kπ+π/6,kπ+2π/3]
(2)-π/4≤x≤0 -π/2≤2x≤0
-π/3= -π/2+π/6≤2x+π/6≤π/6
-√3/2≤sin(2x+π/6)≤1/2
1-√3≤1+2sin(2x+π/6)≤2*1/2+1=2
函数f(x)的值域:[1-√3,2]
=1+cos2x+√3sin2x=1+2sin(2x+π/6)
(1)2kπ+π/2≤2x+π/6≤2kπ+3π/2
2kπ+π/3≤2x≤2kπ+4π/3
kπ+π/6≤x≤kπ+2π/3
f(x)的单调减区间:[kπ+π/6,kπ+2π/3]
(2)-π/4≤x≤0 -π/2≤2x≤0
-π/3= -π/2+π/6≤2x+π/6≤π/6
-√3/2≤sin(2x+π/6)≤1/2
1-√3≤1+2sin(2x+π/6)≤2*1/2+1=2
函数f(x)的值域:[1-√3,2]