Question

已知函数f(x)=sin(5π/4-x)-cos(π/4+x) (1)求f(x)的单调递增区间

已知函数f(x)=sin(5π/4-x)-cos(π/4+x) (1)求f(x)的单调递增区间 (2)已知cos（α-β）=3/5, cos（α+β）=-3/5, 0< α< β<=π/2, 求f(β)

Answer 1

解：
（1）
f(x)=sin(5π/4-x)-cos(π/4+x)
=sin(π+π/4-x)-cos(π/4+x)
=-sin(π/4-x)-cos(π/4+x)
=sin(-π/4+x)-cos(π/4+x)
=sin(-π/2+π/4+x)-cos(π/4+x)
=-2cos(π/4+x)
∵f(x)递增
∴cos(π/4+x)递减
∴2kπ-π/4<=x<=2kπ+3π/4
（2）
0＜α＜b≤π/2
-π/2≤-b<0
-π/2≤a-b<0
cos(a-b)=3/5
sin(a-b)=-√(1-cos^2(a-b))=-4/5
0<a+b≤π
cos(a+b)=-3/5
sin(a+b)=√(1-cos^2(a+b))=4/5
cos2b=cos(a+b-(a-b))
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=-3/5*3/5 +4/5*(-4/5)
=-9/25-16/25
=-1
=2cos^2b-1
cosb=0
b=π/2
f(b)=-2cos(π/4+x)=√2

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不明白请及时追问，满意敬请采纳，O(∩_∩)O谢谢~~

追问

=sin(-π/4+x)-cos(π/4+x)
=sin(-π/2+π/4+x)-cos(π/4+x) 这个用诱导公式的话 -π/2前面是-1 ，应该变为cos啊

追答

你说的对啊
是变成cos
=sin(-π/4+x)-cos(π/4+x)
=sin(-π/2+π/4+x)-cos(π/4+x)
=-sin[π/2-(π/4+x)]-cos(π/4+x)
=-cos(π/4+x)-cos(π/4+x)
=-2cos(π/4+x)

追问

那和前面那一步sin 就连不上啊 为啥sin(-π/4+x)要 变成这样sin(-π/2+π/4+x)

追答

连的上啊

-π/4+x
=-π/2+π/4+x

-π/2+π/4还是等于-π/4

追问

噢~~~~ 没反应过来 懂了懂了 谢谢哈

追答

懂了就采纳吧

Answer 2

(1)∵f(x)=sin(5π/4-x)-cos(π/4+x)=sin[π+(π/4-x)]-cos[π/2+(-π/4+x)]=2sin(x-π/4)
∴2kπ-π/2<x-π/4<2kπ+π/2，即2kπ-π/4<x<2kπ+3π/4
(2)∵cos(α-β)+cos(α+β)=2cosαcosβ=0，又∵0<α<β≤π/2 ∴β=π/2
∵f(x)=2sin(x-π/4) ∴f(β)=2sin(π/2-π/4)=√2

追问

我算的咋是f(x）= -2sin(x-π/4)