tan2x=-2√2 且满足 π/4

tan2x=-2√2 且满足 π/4
tan2x=-2√2 且满足 π/4

tan2x=-2√2 且满足 π/4
tan2x=2tanx/(1-tan²x)=-2√2
令a=tanx
2a/(1-a²)=-2√2
a=-√2+√2a²
√2a²-a-√2=0
(√2a+1)(a-√2)=0
a=-1/√2,a=√2
π/4所以tanx>1
即a>1
所以tanx=a=√2
{2cos²(x/2)-sinx-1}/{√2sin[(π/4)+x]
={[2cos²(x/2)-1]-sinx}/√2(sinπ/4cosx+cosπ/4sinx)
=(cosx-sinx)/(cosx+sinx)
上下除以cosx,且sinx/cosa=tanx
所以原式=(1-tanx)/(1+tanx)
=(1-√2)/(1+√2)
=-3+2√2

原式=（cosx-sinx)/(cosx+sinx)
=[1-sin(2x)]/cos2x 分号上下都乘以cosx-sinx
由tan2x=-2√2，设cos2x=k,sin2x=-2√2k,可求的k=1/3
所以，原式=3+2√2