由1-x≥0,x≥0解得0≤x≤1
设x=sin²a,a∈[0,π/2]
则1-x=cos²a
所以y=√cos²a-√sin²a
=cosa-sina
=√2(√2/2cosa- √2/2 sina)
=√2(sinπ/4cosa-cosπ/4sina)
=√2sin(π/4-a)
因为π/4-a∈[-π/4,π/4]
所以-√2/2≤sin(π/4-a)≤√2/2
所以-1≤√2sin(π/4-a)≤1
即原函数的值域[-1,1]
由1-x≥0,x≥0解得0≤x≤1
设x=sin²a,a∈[0,π/2]
则1-x=cos²a
所以y=√cos²a-√sin²a
=cosa-sina
=√2(√2/2cosa- √2/2 sina)
=√2(sinπ/4cosa-cosπ/4sina)
=√2sin(π/4-a)
因为π/4-a∈[-π/4,π/4]
所以-√2/2≤sin(π/4-a)≤√2/2
所以-1≤√2sin(π/4-a)≤1
即原函数的值域[-1,1]