Question

已知函数f(x)=cos²x+sinxcosx(x∈R)(1)求f(3π/8)的值(2)求f(x)的单调区间

第一问不用了 第二问求过程

Answer 1

f(x)=cos²x+sinxcosx
=(cos2x+1)/2+sinxcosx
=1/2cos2x+1/2+1/2sin2x
=√2/2sin(2x+π/4)+1/2
f(3π/8)=√2/2*sinπ+1/2=1/2
先求增区间
令2kπ-π/2<=2x+π/4<=π/2+2kπ,k∈Z
2kπ-3π/4<=2x<=π/4+2kπ,k∈Z
kπ-3π/8<=x<=π/8+kπ,k∈Z
增区间[kπ-3π/8,π/8+kπ],k∈Z
再求减区间
令π/2+2kπ<=2x+π/4<=3π/2+2kπ,k∈Z
π/4+2kπ<=2x<=5π/4+2kπ,k∈Z
π/8+kπ<=x<=5π/8+kπ,k∈Z
减区间是[π/8+kπ,5π/8+kπ]k∈Z
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