已知函数f(x)=√(1+sinx)+√(1-sinx)(1)求定义域值域（2）求最小正周期和单调递增区间

Question

已知函数f(x)=√(1+sinx)+√(1-sinx)
(1)求定义域值域
（2）求最小正周期和单调递增区间

Answer 1

定义域是R
y>=0
y^2=1+sinx+1-sinx+2√(1-sinx)(1+sinx)
=2+2√[1-(sinx)^2]
=2+2√(cosx)^2
=2+2|cosx|
0<=|cosx|<=1
0<=2|cosx|<=2
2<=2+2|cosx|<=4
2<=y^2<=4
y>=0
√2<=y<=2
值域[√2,2]
1+sinx=sin^2 x/2 + cos^2 x/2 +2sinx/2cosx/2 = (sinx/2+cosx/2)^2
1-sinx=sin^2 x/2 + cos^2 x/2 -2sinx/2cosx/2 = (sinx/2-cosx/2)^2
f(x)=√(1+sinx)+√(1-sinx)
=(sinx/2+cosx/2)+|sinx/2-cosx/2|
x∈[4kπ+π/2,（4k+2)π+π/2]时,f(x)=2sinx/2
x∈[(4k+2)π+π/2,(4k+4)π+π/2]时,f(x)=2cosx/2
所以,最小正周期=4π
单调递增区间：[(4k+2)π+π/2,(4k+4)π+π/2]
递减区间：[4kπ+π/2,（4k+2)π+π/2]