Question

已知函数y=cos平方x+2sinxcosx+3sin平方x,求该最小正周期和最小值并写出该函数的单调递增区间(有过程）

Answer 1

y=cos²x+2sinxcosx+3sin²x

=1+sin(2x)+2sin²x

=sin(2x)+cos(2x)

=√2sin(2x+π/4)

最小正周期Tmin=2π/2=π

当sin(2x+π/4)=-1时,有ymin=-√2

2kπ-π/2≤2x+π/4≤2kπ+π/2 (k∈Z)时,即kπ-3π/8≤x≤kπ+π/8 (k∈Z)时,函数单调递增.

单调递增区间为[kπ-3π/8,kπ+π/8] (k∈Z)