已知函数f(x)＝1＋sin x×cos x．(1)求函数f(x)的最小正周期和单调递减区间；(2)若tan x＝2,求

1.
f(x)=1+sinxcosx=1+(1/2)sin(2x)
最小正周期Tmin=2π/2=π
2kπ+π/2≤2x≤2kπ+3π/2 (k∈Z)时,f(x)单调递减,此时
kπ+π/4≤x≤kπ+3π/4 (k∈Z)
函数的单调递减区间为[kπ+π/4,kπ+3π/4] (k∈Z)
2.
tanx=sinx/cosx=2
sinx=2cosx
(sinx)^2+(cosx)^2=1
(2cosx)^2+(cosx)^2=1
5(cosx)^2=1
(cosx)^2=1/5
f(x)=1+sinxcosx=1+(2cosx)cosx=1+2(cosx)^2=1+2(1/5)=1+2/5=7/5