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_数学解答

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

人气：287 ℃　时间：2020-05-23 19:57:15

解答

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