题目

已知tan2x=-2,兀/4提问时间：2020-11-10

答案

[(2cos^2(x/2))-sinx-1]/[根号2*sin(兀/4+x)]
={[2*1/2(1+cosx)-sinx-1]}/[根号2*sin(兀/4+x)]
={1+cosx-1-sinx}/[根号2*sin(兀/4+x)]
=[cosx-sinx]/[根号2*sin(兀/4+x)]
=[根号2*sin(x+3/4兀)]/[根号2*sin(兀/4+x)]
=[sin(x+3/4兀)]/[sin(兀/4+x)]
=[sin(x+1/2兀+1/4兀)]/[sin(兀/4+x)]
=[cos(x+1/4兀)]/[sin(兀/4+x)]
=1/[tan(x+1/4兀)]
=[1-tan1/4兀*tanx]/[tan1/4兀+tanx]
=(1-tanx)/(1+tanx)
又tan2x=-2
2tanx/(1+tan^2 x)=-2
tan^2 x-tanx-1=0
又 兀/4