Question

已知函数f(x)=sin(π-x)sin(π/2-x)+cos²x(1)求函数f(x)的最小正周期

（2）当x∈[-π/8,3π/8]时，求函数f(x)的单调区间

详细过程！！

Answer 1

解答：sin(π-x)=sinx sin(π/2-x)=cosx
所以f(x)=sin(π-x)sin(π/2-x)+cos²x=1/2sin2x+(1+cos2x)/2=√2/2sin(2x+π/4)+1/2
所以f(x)最小正周期为T=π
（2）由正弦函数单调性知道：sinx在[2kπ-π/2,2kπ+π/2]单调递增，在[2kπ+π/2,2kπ+π3/2]递减（其中k为整数）
又x∈[-π/8,3π/8] 2x+π/4∈[0,π]
不妨设：2x+π/4∈[2kπ-π/2,2kπ+π/2]
此时函数f(x)的单调递增区间为[-π/8,π/8],同理得到递减区间为[π/8,π3/8]

Answer 2

f(x)=sin(π-x)sin(π/2-x)+cos²x
f(x)=sinxcosx+cos²x
=1/2sin2x+1/2+1/2cos2x
=1/2(sin2x+cos2x)+1/2
=√2/2sin(2x+π/4)+1/2
T=2π/2=π

答：函数f(x)的最小正周期是π 画图出来可知 x∈ [-π/8,π/8]单调递增 [π/8,3π/8]单调递减

追问

=1/2(sin2x+cos2x)+1/2
=√2/2sin(2x+π/4)+1/2
怎么得到的？

追答

你看这个的变换就知道了
sin2x+cos2x
=√2(sin2x*√2/2+cos2x*√2/2) 这步是提取根号2
= √2(sin2xcosπ/4+cos2xsinπ/4)
=√2sin(2x+π/4)

Answer 3

f(x)=sinxcosx+cos^2x
=1/2sin2x+(1+cos2x)/2
=1/2sin2x+1/2cos2x+1/2
=√2/2sin(2x+π/4)+1/2
函数f(x)的最小正周期T=2π/2=π