已知函数f(x)=(1+sinx+cosx)sinx/2-cosx/2除以根号下2+2cosx x属于(-派,派) 1求

f(x)={2cos²x/2+2sinx/2cosx/2)sinx/2-cosx/2}/√[2*2cos²x/2] (∴x∈(-π,π）∴x/2∈（-π/2,π/2)∴cosx/2>0)
=cosx/2*(2cosx/2sinx/2+2sin²x/2-1)/2cosx/2
=(sinx+1-cosx-1)/2
=√2/2sin(x-π/4)
(1)∵x∈(-π,π）∴x-π/4∈(-5π/4,3π/4)
∴x-π/4∈(-π/2,π/2)即x∈(-π/4,3π/4)时,f(x)单调递增；x∈(-π,-π/2]∪[3π/4,π）,f(x)单调递减.
（2）f(x/2)=√2/2sin(x/2-π/4)≤-1/2
∴sin(x/2-π/4)≤-√2/2
∵x∈(-π,π）∴x/2-π/4∈(-3π/4,π/4)
∴x/2-π/4∈(-3π/4,-π/4]
∴x∈(-π,0]