Question

已知函数y=sinxcosx+sinx+cosx求值域

Answer 1

y=sinxcosx+sinx+cosx
=1/2(2sinxcosx+1-1)+sinx+cosx
=1/2(sinx+cosx)^2-1/2+(sinx+cosx)
=1/2[(sinx+cosx)^2+2(sinx+cosx)+1-2]
=1/2(sinx+cosx+1)^2-1
=1/2{(√2)*[sin(x+π/4)]+1}^2-1
所以值域为[-1,√2+1/2]

Answer 2

解：设t=sinx+cosx=(√2)*[sin(x+π/4)],则t^2=1+2sinxcosx, sinxcosx=(t^2-1)/2,所以有 y=(t^2-1)/2+t=(t^2+2t-1)/2, t属于[负根号2，正根号2]，利用二次函数的性质，当t＝－1，y有最小值-1, 当t=正根号2, y 有最大值:（根号2）+1/2

Answer 3

y=sinxcosx+sinx+cosx
=（sinx+cosx）^2/2+（sinx+cosx）-1/2
=2sin²(x+π/4)+√2sin(x+π/4)-1/2
=2[sin(x+π/4)+√2/4]²-3/4
设sin(x+π/4)=u ,(-1≤u≤1)则，
y=2(u+√2/4)²-3/4
当u=-√2/4时，ymin=-3/4,
当u=1时，ymax=√2+3/2。
所求值域为：【-3/4,√2+3/2】