设-π/6≤ⅹ≤π/4,求f(x)=sin(x-π/6)sinx的值域
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解决时间 2021-04-06 12:57
- 提问者网友:我们很暧昧
- 2021-04-05 12:11
设-π/6≤ⅹ≤π/4,求f(x)=sin(x-π/6)sinx的值域
最佳答案
- 五星知识达人网友:你可爱的野爹
- 2021-04-05 12:53
解:
f(x)=sin(x-π/6)sinx
=(sinxcosπ/6-cosxsinπ/6)sinx
=(√3/2)sin²x-(1/2)sinxcosx
=(√3/4)(1-cos2x)-(1/4)sin2x
=-½[½sin2x+(√3/2)cos2x] +√3/4
=-½sin(2x+π/3) +√3/4
-π/6≤x≤π/4,则0≤2x+π/3≤5π/6
0≤sin(2x+π/3)≤1
(√3-2)/4≤-½sin(2x+π/3) +√3/4≤√3/4
(√3-2)/4≤f(x)≤√3/4
函数的值域为[(√3-2)/4,√3/4]
f(x)=sin(x-π/6)sinx
=(sinxcosπ/6-cosxsinπ/6)sinx
=(√3/2)sin²x-(1/2)sinxcosx
=(√3/4)(1-cos2x)-(1/4)sin2x
=-½[½sin2x+(√3/2)cos2x] +√3/4
=-½sin(2x+π/3) +√3/4
-π/6≤x≤π/4,则0≤2x+π/3≤5π/6
0≤sin(2x+π/3)≤1
(√3-2)/4≤-½sin(2x+π/3) +√3/4≤√3/4
(√3-2)/4≤f(x)≤√3/4
函数的值域为[(√3-2)/4,√3/4]
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