已知f(x)=sin2wx+32sin2wx-12(x∈R,w>0),若f(x)的最小正周期为2π.(1)求f(x)的表达式和f(x)的单调递增区间;(2)求f(x)在区间[-π6,5π6]上的最大值

题目简介

已知f(x)=sin2wx+32sin2wx-12(x∈R,w>0),若f(x)的最小正周期为2π.(1)求f(x)的表达式和f(x)的单调递增区间;(2)求f(x)在区间[-π6,5π6]上的最大值

题目详情

已知f(x)=sin2wx+
3
2
sin2wx-
1
2
(x∈R,w>0),若f(x)的最小正周期为2π.
(1)求f(x)的表达式和f(x)的单调递增区间;
(2)求f(x)在区间[-
π
6
6
]上的最大值和最小值.
题型:解答题难度:中档来源:不详

答案

(1)由已知f(x)=sin2wx+
3
2
sin2wx-class="stub"1
2

=class="stub"1
2
(1-cos2wx)+
3
2
sin2wx-class="stub"1
2

=
3
2
sin2wx-class="stub"1
2
cos2wx
=sin(2wx-class="stub"π
6
).
又由f(x)的周期为2π,则2π=class="stub"2π
2w
⇒2w=1⇒w=class="stub"1
2

⇒f(x)=sin(x-class="stub"π
6
),
2kπ-class="stub"π
2
≤x-class="stub"π
6
≤2kπ+class="stub"π
2
(k∈Z)⇒2kπ-class="stub"π
3
≤x≤2kπ+class="stub"2π
3
(k∈Z),
即f(x)的单调递增区间为
[2kπ-class="stub"π
3
,2kπ+class="stub"2π
3
](k∈Z).
(2)由x∈[-class="stub"π
6
class="stub"5π
6
]⇒-class="stub"π
6
≤x≤class="stub"5π
6

⇒-class="stub"π
6
-class="stub"π
6
≤x-class="stub"π
6
class="stub"5π
6
-class="stub"π
6
⇒-class="stub"π
3
≤x-class="stub"π
6
class="stub"2π
3

⇒sin(-class="stub"π
3
)≤sin(x-class="stub"π
6
)≤sinclass="stub"π
2
.∴-
3
2
≤sin(x-class="stub"π
6
)≤1.
故f(x)在区间[-class="stub"π
6
class="stub"5π
6
]的最大值和最小值分别为1和-
3
2

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