已知函数f(x)=sin(x+7π4)+cos(x-3π4),x∈R.(1)求f(x)的最小正周期和最小值;(2)已知cos(β-α)=45,cos(β+α)=-45,0<α<β≤π2,求f(β)的值

题目简介

已知函数f(x)=sin(x+7π4)+cos(x-3π4),x∈R.(1)求f(x)的最小正周期和最小值;(2)已知cos(β-α)=45,cos(β+α)=-45,0<α<β≤π2,求f(β)的值

题目详情

已知函数f(x)=sin(x+
4
)+cos(x-
4
)
,x∈R.
(1)求f(x)的最小正周期和最小值;
(2)已知cos(β-α)=
4
5
cos(β+α)=-
4
5
0<α<β≤
π
2
,求f(β)的值.
题型:解答题难度:中档来源:不详

答案

(1)∵f(x)=sin(x+class="stub"7π
4
)+cos(x-class="stub"3π
4
)

=sinxcosclass="stub"7π
4
+sinclass="stub"7π
4
cosx
+cosxcosclass="stub"3π
4
+sinxsinclass="stub"3π
4

=
2
2
sinx
-
2
2
cosx
-
2
2
cosx+
2
2
sinx

f(x)=
2
sinx-
2
cosx=2sin(x-class="stub"π
4
)

∴T=2π,f(x)max=2
(2)∵cos(β-α)=cosαcosβ+sinαsinβ=class="stub"4
5
,cos(β+α)=cosαcosβ-sinαsinβ=-class="stub"4
5

∴cosαcosβ=0
0<α<β≤class="stub"π
2
⇒cosβ=0⇒β=class="stub"π
2

f(β)=
2

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