19题

19题
19题
 

19题
18、
原式===> (sinα*cosα)/(-tanα*cosα)=-√3/3
===> cosα=√3/3
已知α∈(0,π)
所以,sinα=√6/3
则,tanα=sinα/cosα=√2
(1)(cosα-sinα)/(cosα+sinα)
=(1-tanα)/(1+tanα)
=(1-√2)/(1+√2)
=-(1-√2)^2
=-3+2√2
(2)原式=1+(√6/3)*(√3/3)+(√3/3)^2=1+(√2/3)+(1/3)=(4+√2)/3
19、
f(x)=(4cos^4 x-2cos^2 x-1)/[tan(π/4+x)*cos^2 (π/4+x)]
=(4cos^4 x-2cos^2 x-1)/[sin(π/4+x)*cos(π/4+x)]
=2(4cos^4 x-2cos^2 x-1)/[sin(π/2+2x)]
=2(4cos^4 x-2cos^2 x-1)/cos2x
=[8(cos2x+1/2)^2-(cos2x+1)-1]/(cos2x)
=[2(cos2x+1)^2-cos2x-2]/(cos2x)
=[2cos^2 2x+3cos2x]/(cos2x)
=2cos2x+3
(1)f(-17π/12)=2cos(-17π/6)+3
=2cos(-5π/6)+3=2cos(5π/6)+3
=2*(-√3/2)+3
=3-√3
(2)g(x)=(1/2)f(x)+sin2x=cos2x+(3/2)+sin2x=√2sin[2x+(π/4)]+(3/2)
已知x∈[0,π/2],则2x∈[0,π],2x+(π/4)∈[π/4,5π/4]
所以,sin[2x+(π/4)]∈[-√2/2,1]
所以,√2sin[2x+(π/4)]∈[-1,√2]
所以,g(x)最小值为1/2,最大值为√2+(3/2)

看不清啊啊

先解一题,  等等

这个看不清请发一个清楚的

19题 19题 19题. 19题., 19 题... 19题 19题, 19题? 19题, 19题, 19题 19题,&&& 19题, 19题 19题, 19题, 19题 19题,