已知函数f(x)=2sin(x/4)cos(x/4)-2√3sin²(x/4)+√3,且g(x)=f(x+π/
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已知函数f(x)=2sin(x/4)cos(x/4)-2√3sin²(x/4)+√3,且g(x)=f(x+π/3)
(1)判断g(x)的奇偶性 (2)求g(x)的单调递增区间
(1)判断g(x)的奇偶性 (2)求g(x)的单调递增区间
f(x)=sin(x/2) -√3[1-cos(x/2)]+√3
=2[(1/2)sin(x/2) +(√3/2)cos(x/2)]
=2sin(x/2+π/3)
(1) g(x)=f(x+π/3)=2sin[(x+π/3)/2 +π/3)]=2sin(x/2 +π/2)=2cos(x/2)
所以 g(x)是偶函数.
(2)令 -π+2kπ≤x/2≤2kπ,得
-2π+4kπ≤x≤4kπ,
所以 增区间为
[-2π+4kπ,4kπ],k是整数.
=2[(1/2)sin(x/2) +(√3/2)cos(x/2)]
=2sin(x/2+π/3)
(1) g(x)=f(x+π/3)=2sin[(x+π/3)/2 +π/3)]=2sin(x/2 +π/2)=2cos(x/2)
所以 g(x)是偶函数.
(2)令 -π+2kπ≤x/2≤2kπ,得
-2π+4kπ≤x≤4kπ,
所以 增区间为
[-2π+4kπ,4kπ],k是整数.
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