cos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)²]=-4/5
sin(2x)=-cos(2x+π/2)=-cos[2(x+π/4)]=1-2cos²(x+π/4)=1-2•(3/5)²=7/25
[sin(2x)+2sin²x]/(1-tanx)
=2(sinxcosx+sin²x)/(1-sinx/cosx)
=2(cosx+sinx)/(1/sinx-1/cosx)
=2(cosx+sinx)sinxcosx/(cosx-sinx)
=cos(x-π/4)sin(2x)/cos(x+π/4)
=-4/5•7/25/(3/5)
=-28/75不對吧 是cos²x 不是sin²x而且是cos(π/4+x）不是cos(x-π/4)你再看看 別玩復制粘貼 OK？cos(π/4+x)=cosπ/4cosx-sinπ/4sinx=√2/2(cosx-sinx)=3/5cosx-sinx=3√2/5 …………（1）（cosx-sinx）^2=(cosx)^2+(sinx)^2-2sinxcosx=1-2sinxcosx=18/252sinxcosx=1-18/25=7/25所以sin2x=2sinxcosx=7/25由（1）得sinx=cosx-3√2/5兩邊平方sin²x=1-cos²x=cos²x-6√2/5*cosx+18/252cos²x-6√2/5*cosx-7/25=050cos²x-30√2cosx-7=0cosx=(3√2±4√2)/10cosx=-√2/10或7√2/1017π/12