Question

已知函数f(x)=cosx(sinx+cosx)-1/2

已知函数f(x)=cosx(sinx+cosx)-1/2
若0＜α＜π/2，且sinα=（根号2）/2 求f(α）的值
求函数f(x)的最小正周期及单调递增区间

Answer 1

f(x)=sinxcosx+cos²x-1/2
=1/2*sin2x+1/2*(1+cos2x)-1/2
=1/2*sin2x+1/2*cos2x
=√2/2*(√2/2*sin2x+√2/2*cos2x)
=√2/2*sin(2x+π/4)
(1)∵0<α<π/2，sinα=√2/2，∴α=π/4
∴f(α)=√2/2*sin(3π/4)=1/2
(2)最小正周期T=2π/2=π
令-π/2+2kπ≤2x+π/4≤π/2+2kπ，得：-3π/8+kπ≤x≤π/8+kπ
∴单调递增区间为[-3π/8+kπ，π/8+kπ] (k∈Z)

望采纳

Answer 2

0＜α＜π/2,sinα=（根号2）/2

α=π/4
f(α)=cosπ/4(sinπ/4+cosπ/4)-1/2
=√2/2(√2/2+√2/2)-1/2
=1/2
f(x)=cosx(sinx+cosx)-1/2

=sinxcosx+cos^2x-1/2
=1/2sin2x+(cos2x+1)/2-1/2
=1/2(sin2x+cos2x)
=√2/2sin(2x+π/4)
T=2π/2=π
-π/2+2kπ<2x+π/4<π/2+2kπ
-3π/4+2kπ<2x<π/4+2kπ
单增区间：-3π/8+kπ<x<π/8+kπ

Answer 3

f(x)=cos²x+sinxcosx-1/2
=(1+cos2x)/2+(sin2x)/2-1/2
=1/2(sin2x+cos2x)
=√2/2sin(2x+π/4)
最大值2x+π/4=π/2 x=π/8+2kπ 最小值x=5π/8+2kπ