Question

y=sin^2x+sin (π/2-x)+3sin^2(3π/2-x)若tanx=0.5.求y值;若x∈【0,π/2】求y值域

Answer 1

y=1-cos^2x+cosx+3cos^2x=2cos^2x+cosx+1

tanx=sinx/cosx=1/2
cosx=2sinx
(sinx)^2+(cosx)^2=5/4(cosx)^2=1
(cosx)^2=4/5
cosx=(+/-)2/5根号5
故y=2*4/5+1(+/-)2/5根号5=(13(+/-)2根号5)/5
y=2(cosx+1/4)^2+7/8
而0<=cosx<=1
故当cosx=0时有最小值是y=1,当cosx=1时有最大值是4
即值域是[1,4]

Answer 2

解：y=sin^2x+2sinxcosx+3cos^2x
=sin^2x+cosx+3cos^2x
=2cos^2x+cosx+1
=2(cosx+1/4)^2+7/8
[1]当tanx=0.5时，x为第一、三象限角
所以cos^2x=1/sec^2x=1/(1+tan^2x)=4/5
Cosx=±√(4/5)
∴y=13/5±(2√5)/5
[2]当x∈∈【0,π/2】时，0≤cosx≤1
∴1/4≤cosx+1/4≤5/4
1≤2(cosx+1/4)^2+7/8≤4
∴y= sin^2x+2sinxcosx+3cos^2x=2(cosx+1/4)^2+7/8的值域为：[1，4]