已知向量a=(sinx,cosx),b=(sinx,sinx)若x∈[-3π/8,π/4]函数f(x)=λa*b的最大值

f(x)=λa*b=λ(sin^2x+sinx*cosx)
=λ[(1-cos2x)/2+sin2x/2]
=(λ/2)[1+sin2x-cos2x]
=(λ/2)[1+√2sin(2x-π/4)]
因为x∈[-3π/8,π/4],所以(2x-π/4)∈[-π,π/4],
令t(x)=1+√2sin(2x-π/4),所以-1