(2010•杭州一模)已知函数f(x)=2sinxcosx+2cos2x(x∈R).
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(2010•杭州一模)已知函数f(x)=2sinxcosx+2cos2x(x∈R).
(Ⅰ)求f(x)的最小正周期,并求f(x)的最小值.
(Ⅱ)令g(x)=f(x+
)−1
(Ⅰ)求f(x)的最小正周期,并求f(x)的最小值.
(Ⅱ)令g(x)=f(x+
π |
8 |
(Ⅰ)∵f(x)=2sinxcosx+2cos2x=sin2x+cos2x+1=
2sin(2x+
π
4)+1,
∴f(x)的最小正周期 T=
2π
2=π.由于-1≤sin(2x+
π
4)≤1,∴1-
2≤f(x)≤
2+1,
故f(x)的最小值是 1-
2.
(Ⅱ)由题意可得 g(x)=f(x+
π
8)−1=
2sin[2(x+
π
8)+
π
4]+1-1=
2cos2x,
∵-
π
6≤x≤
π
3,∴-
π
2sin(2x+
π
4)+1,
∴f(x)的最小正周期 T=
2π
2=π.由于-1≤sin(2x+
π
4)≤1,∴1-
2≤f(x)≤
2+1,
故f(x)的最小值是 1-
2.
(Ⅱ)由题意可得 g(x)=f(x+
π
8)−1=
2sin[2(x+
π
8)+
π
4]+1-1=
2cos2x,
∵-
π
6≤x≤
π
3,∴-
π
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