已知f(x)=2sin(x+π6)-433tanα•cos2x2,α∈(0,π)且f(π2=3-2).(1)求α;(2)当x∈[π2,π]时,求函数y=f(x+α)的值域.-数学

题目简介

已知f(x)=2sin(x+π6)-433tanα•cos2x2,α∈(0,π)且f(π2=3-2).(1)求α;(2)当x∈[π2,π]时,求函数y=f(x+α)的值域.-数学

题目详情

已知f(x)=2sin(x+
π
6
)-
4
3
3
tanα•cos2
x
2
,α∈(0,π) 且f(
π
2
=
3
-2).
(1)求α;
(2)当x∈[
π
2
,π
]时,求函数y=f(x+α)的值域.
题型:解答题难度:中档来源:不详

答案

(1)因为f(x)=2sin(x+class="stub"π
6
)-
4
3
3
tanα•cos2class="stub"x
2
,∴f(class="stub"π
2
)=2sin(class="stub"π
2
+class="stub"π
6
)-
4
3
3
tanα•cos2class="stub"π
4
=
3
-
4
3
3
tanα•class="stub"1
2
=
3
-2,
所以,tanα=
3
,又 α∈(0,π),故 α=class="stub"π
3

(2)由(1)得,f(x)=2sin(x+class="stub"π
6
)-
4
3
3
tanα•cos2class="stub"x
2
=2sin(x+class="stub"π
6
)-4cos2class="stub"x
2
=
3
sinx+cosx-2(1+cosx)=2(
3
2
sinx-class="stub"1
2
cosx)-2=2sin(x-class="stub"π
6
)-2,
所以,y=f(x+α)=f(x+class="stub"π
3
)=2sin(x+class="stub"π
3
-class="stub"π
6
)-2=2sin(x+class="stub"π
6
)-2.
因为 class="stub"π
2
≤x≤π,所以 class="stub"2π
3
≤x+class="stub"π
6
class="stub"7π
6
,∴-class="stub"1
2
≤sin(x+class="stub"π
6
)≤
3
2
,∴-3≤2sin(x-class="stub"π
6
)-2≤
3
-2,
因此,函数y=f(x+α)的值域为[-3,
3
-2].

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