已知f(x)=cos2x+sinxcosx,g(x)=2sin(x+π4)sin(x-π4).(1)求f(x)的最小正周期及单调增区间;(2)若f(α)+g(α)=56,且α∈[3π8,5π8]求si

题目简介

已知f(x)=cos2x+sinxcosx,g(x)=2sin(x+π4)sin(x-π4).(1)求f(x)的最小正周期及单调增区间;(2)若f(α)+g(α)=56,且α∈[3π8,5π8]求si

题目详情

已知f(x)=cos2x+sinxcosx,g(x)=2sin(x+
π
4
)sin(x-
π
4
)

(1)求f(x)的最小正周期及单调增区间;
(2)若f(α)+g(α)=
5
6
,且α∈[
8
8
]
求sin2α的值.
题型:解答题难度:中档来源:不详

答案

(1)y=cos2x+sinxcosx=class="stub"1+cos2x
2
+class="stub"1
2
sin2x=
2
2
sin(2x+class="stub"π
4
+class="stub"1
2

∴T=class="stub"2π
2
=π,由 2kπ-class="stub"π
2
≤2x+class="stub"π
4
≤class="stub"π
2
+2kπ   k∈Z
,即 kπ-class="stub"3π
8
≤x≤class="stub"π
8
+kπ   k∈Z

所以函数的单调增区间为:[-class="stub"3
8
π+kπ,class="stub"π
8
+kπ] (k∈Z)

(2)g(x)=2sin(x+class="stub"π
4
)sin(x-class="stub"π
4
)
=-sin(2x+class="stub"π
2
)=-cos2x,
因为f(x)+g(x)=class="stub"1+cos2x
2
+class="stub"1
2
sin2x-cos2x=class="stub"1
2
+class="stub"1
2
sin2x-class="stub"1
2
cos2x
=class="stub"1
2
+
2
2
sin(2x-class="stub"π
4

f(α)+g(α)=class="stub"5
6
class="stub"1
2
+
2
2
sin(2α-class="stub"π
4
)=class="stub"5
6

sin(2α-class="stub"π
4
)=
2
3
  α∈[class="stub"3π
8
,class="stub"5π
8
]

2α∈[class="stub"3π
4
,class="stub"5π
4
]
  2α-class="stub"π
4
∈[class="stub"π
2
,π]
cos(2α-class="stub"π
4
)=-
7
3

sin2α=sin(2α-class="stub"π
4
+class="stub"π
4

=sin(2α-class="stub"π
4
)cosclass="stub"π
4
+cos(2α-class="stub"π
4
)sinclass="stub"π
4

=
2
3
× 
2
2
+(-
7
3
)×
2
2
=class="stub"1
3
-
14
6
=
2-
14
6

更多内容推荐