Question

已知tan2x=-2 π/2<2x<π求[2cos^2(x/2)-sinx-1]/根号2sin(π/4+x)

Answer 1

tan(2x)=-2,
π/2<2x<π,
[sec(2x)]^2 = [tan(2x)]^2 + 1 = 5, sec(2x) = 1/cos(2x) = -5^(1/2),
cos(2x) = -1/5^(1/2),
sin(2x) = 2/5^(1/2).

2^(1/2)sin(π/4+x) = 2^(1/2)[sin(x)cos(π/4) + cos(x)sin(π/4)]=sin(x)+cos(x).
2[cos(x/2)]^2 - sin(x) - 1 = 2[cos(x/2)]^2 - 1 - sin(x) = cos(x) - sin(x),

{2[cos(x/2)]^2 - sin(x) - 1}/[2^(1/2)sin(π/4+x)]=
[cos(x)-sin(x)]/[sin(x)+cos(x)]= {[cos(x)]^2 - [sin(x)]^2}/[sin(x)+cos(x)]^2
=cos(2x)/[1+2sin(x)cos(x)]
=cos(2x)/[1+sin(2x)]
=[-1/5^(1/2)]/[1+2/5^(1/2)]
=-1/[2+5^(1/2)]
=-[5^(1/2)-2]/[5-2^2]
=2-5^(1/2)

Answer 2

由于cosx=2cos^(x/2)-1
原式＝(cosx-sinx)/根号2(sinπ/4cosx+cosπ/4sinx)
=(cosx-sinx)/根号2((根号2/2)cosx+(根号2/2)sinx)
=(cosx-sinx)/(cosx+sinx)
=(1-sinx/cosx)/(1+sinx/cox)
=(1-tanx)/(1+tanx)

tan2x=2tanx/(1-tan^2x)=-2
2tanx=-2+2tan^2x
tan^2x-tanx-1=0