tanx=-2√2,求(2cos^x/2-sinx-1)/√2sin(л/4+x)
问题描述：
tanx=-2√2,求(2cos^x/2-sinx-1)/√2sin(л/4+x)
答
(2cos^x/2-sinx-1)/√2sin(л/4+x)
=(cosx-sinx)/根号2( 根号2/2cosx+sinx*根号2/2)
=(cosx-sinx)/(cosx+sinx)
=(1-tanx)/(1+tanx),(上下同除以cosx)
=(1+2根号2)/(1-2根号2)
=(1+8+4根号2)/(1-8)
=(9+4根号2)/(-7)