15题的第二问求详细过程谢谢了
2个回答
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(I)
f(x)=2(cosx)^2 +2√3sinxcosx -1
= cos2x +√3sin2x
=2sin(2x+π/3)
最小正周期=π
(II)
x在[0,π/2)
max f(x) =2
min f(x) =2sin(π+π/3)
= -√3
f(x)=2(cosx)^2 +2√3sinxcosx -1
= cos2x +√3sin2x
=2sin(2x+π/3)
最小正周期=π
(II)
x在[0,π/2)
max f(x) =2
min f(x) =2sin(π+π/3)
= -√3
追问
最大值最小值怎么算得？
追答
f(x)=2sin(2x+π/3)
max sin(2x+π/3) = 1
at
2x+π/3 = π/2
x = π/12
x在[0,π/2]
min sin(2x+π/3)
at
2x+π/3 = 4π/3
x = π/2
min 2sin(2x+π/3) = 2sin( π + π/3)
=-2sin(π/3)
=-√3
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