已知函数f(x)=tan(2x+π4)(I)求该函数的定义域,周期及单调区间;(II)若f(θ)=17,求2cos2θ2-sinθ-12sin(θ+π4)的值.-数学

题目简介

已知函数f(x)=tan(2x+π4)(I)求该函数的定义域,周期及单调区间;(II)若f(θ)=17,求2cos2θ2-sinθ-12sin(θ+π4)的值.-数学

题目详情

已知函数f(x)=tan(2x+
π
4

(I)求该函数的定义域,周期及单调区间;
(II)若f(θ)=
1
7
,求
2cos2
θ
2
-sinθ-1
2
sin(θ+
π
4
)
的值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由题意得,T=class="stub"π
2

2x+class="stub"π
4
class="stub"π
2
+kπ
(k∈Z)得,x≠class="stub"kπ
2
+class="stub"π
8

-class="stub"π
2
+kπ<2x+class="stub"π
4
<class="stub"π
2
+kπ
(k∈Z)得,class="stub"kπ
2
-class="stub"3π
8
<x<class="stub"kπ
2
+class="stub"π
8

综上得,函数的周期是class="stub"π
2
,定义域是{x|x≠class="stub"kπ
2
+class="stub"π
8
,k∈Z},
单调增区间是(class="stub"kπ
2
-class="stub"3π
8
class="stub"kπ
2
+class="stub"π
8
)(k∈Z).
(Ⅱ)式子
2cos2class="stub"θ
2
-sinθ-1
2
sin(θ+class="stub"π
4
)
=class="stub"cosθ-sinθ
sinθ+cosθ
=class="stub"1-tanθ
tanθ+1
①,
∵f(θ)=class="stub"1
7
,∴tan(2θ+class="stub"π
4
)=class="stub"1
7

则tan2θ=tan[(2θ+class="stub"π
4
)-class="stub"π
4
]=
class="stub"1
7
-1
1+class="stub"1
7
=-class="stub"3
4

由tan2θ=class="stub"2tanθ
1-tan2θ
=-class="stub"3
4
得,tanθ=3或-class="stub"1
3

把tanθ=3代入上式①得,
2cos2class="stub"θ
2
-sinθ-1
2
sin(θ+class="stub"π
4
)
=-class="stub"1
2

把tanθ=-class="stub"1
3
代入上式①得,
2cos2class="stub"θ
2
-sinθ-1
2
sin(θ+class="stub"π
4
)
=2.

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