已知sin(π-α)=45,α∈(0,π2).(1)求sin2α-cos2α2的值;(2)求函数f(x)=56cosαsin2x-12cos2x的单调递增区间.-数学

题目简介

已知sin(π-α)=45,α∈(0,π2).(1)求sin2α-cos2α2的值;(2)求函数f(x)=56cosαsin2x-12cos2x的单调递增区间.-数学

题目详情

已知sin(π-α)=
4
5
,α∈(0,
π
2
).
(1)求sin2α-cos2
α
2
的值;
(2)求函数f(x)=
5
6
cosαsin2x-
1
2
cos2x的单调递增区间.
题型:解答题难度:中档来源:蓝山县模拟

答案

∵sin(π-α)=class="stub"4
5
,∴sinα=class="stub"4
5

又∵α∈(0,class="stub"π
2
),∴cosα=class="stub"3
5

(1)sin2α-cos2class="stub"α
2

=2sinαcosα-class="stub"1+cosα
2

=2×class="stub"4
5
×class="stub"3
5
-
1+class="stub"3
5
2
=class="stub"4
25

(2)f(x)=class="stub"5
6
×class="stub"3
5
sin2x-class="stub"1
2
cos2x
=
2
2
sin(2x-class="stub"π
4
).
令2kπ-class="stub"π
2
≤2x-class="stub"π
4
≤2kπ+class="stub"π
2
,k∈Z,
得kπ-class="stub"π
8
≤x≤kπ+class="stub"3
8
π,k∈Z.
∴函数f(x)的单调递增区间为[kπ-class="stub"π
8
,kπ+class="stub"3
8
π],k∈Z.

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