已知函数f(1+cotx)sinx^2-2sin(x+π/4)sin(x-π/4)1、若tana=2,求f(a)2、若x∈[π/12,π/2],求f(x)的取值范围

问题描述:

已知函数f(1+cotx)sinx^2-2sin(x+π/4)sin(x-π/4)
1、若tana=2,求f(a)
2、若x∈[π/12,π/2],求f(x)的取值范围

f(x)=(1+cotx)sinx^2-2sin(x+π/4)sin(x-π/4)
=(1+cosx/sinx)*sinx ^2-(sinx ^2-cosx ^2)
=cosx ^2+sinxcosx

1、若tana=2,1/(cosa ^2)=seca ^2=1+2*2=5
f(a)=cosa ^2+sinacosa=cosa ^2(1+tana)=1/5*(1+2)=3/5
2、f(x)=cosx ^2+sinxcosx
=cosx(cosx+sinx)
=√2sin(x+∏/4)cosx
=√2/2 〔sin(x+∏/4+x)+sin(x+∏/4-x)〕
=√2/2 sin(2x+∏/4)+1/2
x∈[π/12,π/2],则5π/12≤2x+π/4≤5π/4
当x=π/2,f(x)min=√2/2 sin(5∏/4)+1/2=√2/2 *(-√2/2)+1/2=0
当2x+π/4=π/2,即x=π/8,时f(x)max=√2/2 +1/2
所以f(x)的取值范围是〔0,√2/2 +1/2〕

(1+cotx)sin^2x=sin^2x+sinxcosx
2sin(x+π/4)sin(x-π/4)=根号2(sinx+cosx)*根号2/2(sinx-cosx)=sin^2x-cos^2x
f(x)=(1+cotx)sinx^2-2sin(x+π/4)sin(x-π/4)=sinxcosx+cos^2x
f(x)=1/2sin2x+1/2(2cos^2x-1)+1/2
f(x)=1/2[sin2x+cos2x]+1/2
f(x)=根号2/2sin(2x+π/4)+1/2
1.tana=2=> sina/cosa=2 sin^2a+cos^2a=1=>cos^2a=1/5
f(a)=sinacosa+cos^2a=3cos^2a=3/5
2.x∈[π/12,π/2],2x属于[π/6,π]
2x+π/4属于[5/12π,5/4π]
sin(2x+π/4)属于[根号2/2,1]
f(x)取值范围为[1,(根号2+1)/2]