已知函数f(x)=sin(π/2-x)+sinx
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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
(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
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
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
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