已知函数f(x)=sinx*cosx+sin²x.求,1,f(x)的值域和最小正周期

题目

已知函数f(x)=sinx*cosx+sin²x.求,1,f(x)的值域和最小正周期
2,设a∈（0,π）,且f(a)=1,求a的值.

答案

(1)
f(x) = 1/2 sin 2x + 1/2 (1 - cos 2x)
= 1/2 (sin2x-cos2x) + 1/2
= 1/√2 (1/√2 sin2x - 1/√2cos2x) + 1/2
= 1/√2(sin(2x-π/4) + 1/2
所以值域是 [1/2 - 1/√2,1/2 + 1/√2]
最小正周期是π
(2)
因a∈(0,π) 故 2a-π/4∈(-π/4,7/4 π)
当 sin(2a-π/4) = 1/√2 时,f(a) = 1
所以 2a-π/4 = π/4,3/4 π
即 a = π/4 ,π/2