Question

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

Answer 1

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.