已知函数f(x)=sinxcosx+sin2x.(1)求f(x)的值域和最小正周期;(2)设α∈(0,π),且f(α)=1,求α的值.-数学

题目简介

已知函数f(x)=sinxcosx+sin2x.(1)求f(x)的值域和最小正周期;(2)设α∈(0,π),且f(α)=1,求α的值.-数学

题目详情

已知函数f(x)=sinxcosx+sin2x.
(1)求f(x)的值域和最小正周期;
(2)设α∈(0,π),且f(α)=1,求α的值.
题型:解答题难度:中档来源:西城区二模

答案

(1)f(x)=sinx•cosx+sin2x=class="stub"1
2
sin2x+class="stub"1-cos2x
2

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

因为-1≤sin(2x-class="stub"π
4
)≤1

所以
1-
2
2
2
2
sin(2x-class="stub"π
4
)+class="stub"1
2
1+
2
2

即函数f(x)的值域为[
1-
2
2
1+
2
2
]

函数f(x)的最小正周期为T=class="stub"2π
2

(2)由(Ⅰ)得f(α)=
2
2
sin(2α-class="stub"π
4
)+class="stub"1
2
=1

所以sin(2α-class="stub"π
4
)=
2
2

因为0<α<π,所以-class="stub"π
4
<2α-class="stub"π
4
<class="stub"7π
4

所以2α-class="stub"π
4
=class="stub"π
4
 或2α-class="stub"π
4
=class="stub"3π
4

所以α=class="stub"π
4
 或α=class="stub"π
2

更多内容推荐