设函数f(x)=cos(πx4-π3)-cosπx4.(Ⅰ)求f(x)的最小正周期;(Ⅱ)求函数y=f(-2-x)在[0,2]上的值域.-数学

题目简介

设函数f(x)=cos(πx4-π3)-cosπx4.(Ⅰ)求f(x)的最小正周期;(Ⅱ)求函数y=f(-2-x)在[0,2]上的值域.-数学

题目详情

设函数f(x)=cos(
πx
4
-
π
3
)-cos
πx
4

(Ⅰ)求f(x)的最小正周期;
(Ⅱ)求函数y=f(-2-x)在[0,2]上的值域.
题型:解答题难度:中档来源:日照模拟

答案

(Ⅰ)f(x)=cos(class="stub"πx
4
-class="stub"π
3
)-cosclass="stub"πx
4
=cosclass="stub"πx
4
cosclass="stub"π
3
+sinclass="stub"πx
4
sinclass="stub"π
3
-cosclass="stub"πx
4

=class="stub"1
2
cosclass="stub"πx
4
+
3
2
sinclass="stub"πx
4
-cosclass="stub"πx
4
=
3
2
sinclass="stub"πx
4
-class="stub"1
2
cosclass="stub"πx
4

=sin(class="stub"π
4
x-class="stub"π
6

∴f(x)的最小正周期T=class="stub"2π
class="stub"π
4
=8.
(Ⅱ)由(Ⅰ)知  y=f(-2-x)=sin[class="stub"π
4
(-2-x)-class="stub"π
6
]
=sin(-class="stub"π
2
-class="stub"π
4
x-class="stub"π
6
)=-cos(class="stub"π
4
x+class="stub"π
6

∵0≤x≤2,∴class="stub"π
6
class="stub"π
4
x+class="stub"π
6
class="stub"2π
3

∴-class="stub"1
2
≤cos(class="stub"π
4
x+class="stub"π
6
)≤
3
2

∴-
3
2
≤-cos(class="stub"π
4
x+class="stub"π
6
)≤class="stub"1
2

故函数y=f(-2-x)在[0,2]上的值域为[-
3
2
class="stub"1
2
].

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