Question

已知函数f(x)=sin(π/2-x)+sinx

已知函数f(x)=sin(π/2-x)+sinx
(1)求函数y=f(x)的单调递增区间
(2)若f(a-π/4)=根号2/3，求f(2a+π/4)的值

Answer 1

解：∵f(x)=sin[(π/2)-x]+sinx
=√2[(√2/2)cosx+(√2/2)sinx]
=√2[sin(π/4)cosx+cos(π/4)sinx]
=√2sin[x+(π/4)]
又∵y=sinx在[-π/2，π/2]上单调递增，
即：-π/2≤x≤π/2
∴-π/2≤x+(π/4)≤π/2
整理得：-3π/4≤x≤π/4
∴f(x)在2kπ-（3π/4)≤x≤2kπ+(π/4）（k∈Z）上单调递增；
同理，∵sinx在[π/2,3π/2]上单调递减；
∴π/2≤x+(π/4)≤3π/2
整理得：π/4≤x≤5π/4
∴f(x)在2kπ+(π/4)≤x≤2kπ+(5π/4)(k∈Z）上单调递减；
∵f(a-π/4)=√2/3
∴f(a-π/4)=√2sin[(a-π/4)+(π/4)
=√2sina
即√2sina=√2/3
∴sina=1/3
sina^2=1/9
cosa^2=1-(1/3)^2
=8/9
f(2a+π/4)=√2sin[(2a+π/4)+(π/4)]
=√2sin(2a+π/2]
=-√2cos2a
=-√2(cosa^2-sina^2)
=-√2[(8/9)-(1/9)]
=-7√2/9

Answer 2

f(x)=cosx＋sinx
f(x)=√2sin(x＋π/4)
(1)递增区间：2kπ－π/2≤x＋π/4≤2kπ＋π/2
得：2kπ－3/4π≤x≤2kπ＋π/4
递增区间是：[2kπ－3π/4，2kπ＋π/4]，其中k∈Z

(2)f(a－π/4)=√2sina=√2/3
则：sina=1/3
f(2a＋π/4)=√2sin(2a＋π/2)=√2cos2a=√2[1－2sin²a]=(7/9)√2

Answer 3

(1)f(x)=cosx+sinx=根号2乘以sin（x+π/4）
由-π/2+2Kπ<=x+π/4<=π/2+2Kπ解得单调区间[-π3/4+2Kπ,π/4+2Kπ]
(2)f(x)=cosx+sinx=根号2乘以sin（x+π/4）
f(a-π/4)=根号2乘以sin（a）=根号2/3,所以sina=1/3
f(2a+π/4)=根号2乘以sin（2a+π/2）=sin2a=2sinacosa