Question

已知f(x)=sin²x+sinxcosx,x∈[0,π/2]。(1)求f(x)的最小正周期及值域。

2）若f（α)=5/6，求sin2α的值

Answer 1

(1)∵cos2x=cosxcosx-sinxsinx=cos²x-sin²x=(1-sin²x)-sin²x=1-2sin²x
∴sin²x=(1-cos2x)/2
∵sin2x=2sinxcosx
∴sinxcosx=(sin2x)/2
∴f(x)=(1-cos2x)/2+(sin2x)/2
=1/2+(sin2x)/2-cos(2x)/2
=1/2+√2/2(√2/2sin2x-√2/2cos2x)
=1/2+√2/2(cosπ/4sin2x-sinπ/4cos2x)
=1/2+√2/2sin(2x-π/4)
∴T=2π/2=π
x∈[0,π/2] 2x-π/4∈[-π/4,3π/4] sin(2x-π/4)∈[-√2/2,1]
∴f(x)值域为[0,1/2+√2/2]
(2)f(a)=1/2+√2/2sin(2a-π/4)=5/6
√2/2sin(2a-π/4)=1/3 和差化积
sin2a-cos2a=2/3 cos2a=sin2a-2/3
sin²2a+cos²2a=1
sin²2a+(sin2a-2/3)²=1 设sin2a=m
m²+(m-2/3)²=1
2m²-4m/3-5/9=0 18m²-12m-5=0
求根公式:
m=(12±6√14)/36 [a∈[0,π/2] 2a∈[0,π] sin2a≥0]
∶sin2a=(2+√14)/6
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Answer 2

f(x)=sin²x+sinxcosx=(1-cos2x)/2+1/2sin2x=1/2+1/2(sin2x-cos2x)=1/2+√2/2sin(2x-π/4)
最小正周期T=π
Kπ-π/8<x<kπ+3π/8递增， x∈[0,π/2】 ，值域[0, (1+√2)/2]
2）若f（α)=5/6，求sin2α的值
f(α)=sin²α+sinαcosα=5/6 2sin²α+2sinαcosα=5/3 , sin2α=5/3-2sin²α=5/3+cos2α-1=2/3+cos2α
(sin2α-2/3)²=(cos2α)²=1-(sin2α)²
(sin2α)²-4/3sin2α+4/9=1-(sin2α)²
2(sin2α)²-4/3sin2α-5/9=0 , 2(sin2α-1/3)²=7/9 , sin2α=(2±√14)/6