Question

已知tan(a+π/4)=-1/7，a属于（π/2，π),求tana+(cos2a+1)/根号2cos(a-π/4)-sin2a

Answer 1

tan（α+π／4）=-1／7，α∈（π／2，π）
tan（α+π／4）= (tanα + tanπ/4)/(1-tanαtanπ/4) = (tanα + 1)/(1-tanα) = -1/7
7tanα + 7= - 1+tanα
tanα = -4/3
α∈（π／2，π）
cosα = -1/根号(1+tan^2α)=-1/根号(1+16/9)=-3/5
sinα=|tanα|/根号(1+tan^2α)=(4/3)/(5/3)=4/5
cos2α=2cos^2α-1=2*9/25-1= -7/25
√2cos（α－π／4）=根号2(cosαcosπ/4+sinαsinπ/4)=根号2*根号2/2(-3/4+4/5)=1/5
sin2α=2sinαcosα=2*4/5*（-3/5）=-24/25

tanα+（cos2a+1）／[√2cos（α－π／4）－sin2a]
=[(-4/3)+(-7/25+1)] / [1/5-(-24/25)]
=(-46/75)/(29/25)
=-46/87希望对您有所帮助。望采纳哦~