作业帮已知函数f(x)=（4cos^4x-2cos2x-1）/[tan(π/4+x)·sin^2(π/4-x)]（1）求f(-17π/12)的值（2）当x∈[0,π/2]时,求g(x)=1/2f(x)+sin2x的最大值和最小值
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![已知函数f(x)=（4cos^4x-2cos2x-1）/[tan(π/4+x)·sin^2(π/4-x)]（1）求f(-17π/12)的值（2）当x∈[0,π/2]时,求g(x)=1/2f(x)+sin2x的最大值和最小值](/uploads/image/z/4111731-27-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%EF%BC%884cos%5E4x-2cos2x-1%EF%BC%89%2F%5Btan%28%CF%80%2F4%2Bx%29%C2%B7sin%5E2%28%CF%80%2F4-x%29%5D%EF%BC%881%EF%BC%89%E6%B1%82f%28-17%CF%80%2F12%29%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E5%BD%93x%E2%88%88%5B0%2C%CF%80%2F2%5D%E6%97%B6%2C%E6%B1%82g%28x%29%3D1%2F2f%28x%29%2Bsin2x%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC)
已知函数f(x)=（4cos^4x-2cos2x-1）/[tan(π/4+x)·sin^2(π/4-x)]（1）求f(-17π/12)的值（2）当x∈[0,π/2]时,求g(x)=1/2f(x)+sin2x的最大值和最小值
已知函数f(x)=（4cos^4x-2cos2x-1）/[tan(π/4+x)·sin^2(π/4-x)]
（1）求f(-17π/12)的值
（2）当x∈[0,π/2]时,求g(x)=1/2f(x)+sin2x的最大值和最小值
已知函数f(x)=（4cos^4x-2cos2x-1）/[tan(π/4+x)·sin^2(π/4-x)]（1）求f(-17π/12)的值（2）当x∈[0,π/2]时,求g(x)=1/2f(x)+sin2x的最大值和最小值
4(cosx)^4-2cos2x-1=[2(cosx)^2]^2-1-2cos2x=[(2(cosx)^2-1]*[2(cosx)^2+1]-2cos2x
=cos2x*[2(cosx)^2+1]-2cos2x=cos2x*[2(cos)^2-1]=(cos2x)^2
tan(π/4+x)·[sin(π/4-x)]^2=(1+tanx)/(1-tanx)*[√2/2*cosx- √2/2*sinx]^2
=(cosx+sinx)/(cosx-sinx)*1/2*(cosx-sinx)^2
=1/2*(cosx+sinx)*(cosx-sinx)=1/2*[(cosx)^2-(sinx)^2]
=1/2*cos2x
f(x)=(cosx)^2/[1/2*cos2x]=2cos2x
(1),f(-17π/12)=2cos(-17π/6)=2cos[-2π-5π/6]=2cos(-5π/6)=2cos(-π+π/6)
=-2cosπ/6=-2*√3/2=-√3
(2),g(x)=1/2f(x)+sin2x=cos2x+sin2x=√2(√2/2*cos2x+√2/2sin2x)
=√2*(sinπ/4*cos2x+cosπ/4*sin2x)=√2*sin(2x+π/4)
x∈[0,π/2],x+π/4∈[π/4,3π/4]
sin(2x+π/4)∈[√2/2,1]
g(x)最大值:√2 ,最小值:1.