已知sinx+cosx=根号2/2,求sin^4x+cos^4x

显示全部楼层 · 2010-12-19 19:26:17

（1）已知sinx+cosx=根号2/2,求sin^4x+cos^4x
（2）某x在第三象限,化简根号(1+tanx)^2+（1-tanx)^2
（3）化简（2cosθ/根号1-sin^2θ）+根号1-cos^2θ/sinθ

千问 · 2010-12-19 19:26:17

sinx+cosx=√2/2(sinx+cosx)^2=1/2(sinx)^2+2sinxcosx+(cosx)^2=1/21+2sinxcosx=1/22sinxcosx=-1/2sinxcosx=-1/4sin^4x+cos^4x =sin^4x+2(sinxcosx)^2+cos^4x -2(sinxcosx)^2=[sin^2x+cos^2x ]^2-2(sinxcosx)^2=1^2-2*(1/4)^2=1-1/8=7/8√[(1+tanx)^2+（1-tanx)^2]=√[1+tan^2x+2tanx+1+tan^2x-2tanx]=√[2+2tan^2x]=√[2(1+t