作业帮已知函数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
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![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](/uploads/image/z/3742391-47-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%281%2Bcotx%29sinx%5E2-2sin%28x%2B%CF%80%2F4%29sin%28x-%CF%80%2F4%291%E3%80%81%E8%8B%A5tana%3D2%2C%E6%B1%82f%28a%292%E3%80%81%E8%8B%A5x%E2%88%88%5B%CF%80%2F12%2C%CF%80%2F2%5D%2C%E6%B1%82f%28x%29%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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
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
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
(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]
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=co...
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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〕
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