已知tan2α=-2根号2且tanα>1,求[2cos平方α/2-sinα-1]/[根号2sin(π/4+α）]的值

tan2a=2tana/[1-tan^2a]=-2√2 , tanα>1,
∴ tana= √2
[2cos^2α/2-sinα-1]/[√2 *sin(π/4+α）]
=[(1+cosa)-sinα-1]/[√2 *sin(π/4+α）]
=[cosa-sinα]/[√2 *sin(π/4+α）]
=-√2[ -√2/2*cosa+√2/2*sinα]/[√2 *sin(π/4+α）]
=-[ sin(a-π/4)]/[sin(π/4+α）]
=[ cos(a+π/4)]/[sin(π/4+α）]
=tan(π/4+α)
=[tan(π/4)+tanα]/[1-tan(π/4)*tanα]
=[1+√2 ]/[1-√2 ]
=-3-2√2

解析：∵tan2a=2tana/[1-(tana)^2]=-2√2,
∴√2(tana)^2-tana-√2=0,
∴tana=√2,tana=-√2/2(舍去）
[2(cosa/2)^2-sina-1]/[√2sin(π/4+α）]
=(cosa-sina)/(sina+cosa)
=（1-tana)/(1+tana)
=(1-√2)/(1+√2)
=(1-√2)^2/(1-2)
=2√2-3