Question

已知f(x）=sin(π-x)sin(π/2-x)+cos^2x.求函数的最小正周期；当x∈{-π/8,3π/8}时，求函数f(x）的单调区

间

Answer 1

f(x）=sin(π-x)sin(π/2-x)+cos^2x
=sinxcosx+(1+cos2x )/2
=sin2x/2+cos2x/2 +1/2
=√2/2sin(2x +π/4) +1/2
T=2π/w=2π/2=π
2kπ-π/2<=2x+π/4<=2kπ+π/2
2kπ-3π/4<=2x<=2kπ+π/4
kπ-3π/8<=x<=kπ+π/8
所以。函数f(x)的单调区间为x∈[kπ-3π/8,kπ+π/8]

Answer 2

思路：利用诱导公式、二倍角公式和和差化积公式，将函数变成只含同一个三角函数的形式
f(x）=sin(π-x)sin(π/2-x)+cos^2x
=sinxcosx+(1+cos2x )/2
=(sin2x +cos2x)/2 +1/2
=(√2/2)sin(2x +π/4) +1/2
从而可知其最小正周期=2π/2 =π
自己再写出单调区间即可