已知tan（α+π/4）=-1/7.求（cos2α+1）/【2cos(α-π/4)*sin(α+π/4)-sin2α】的值.
问题描述：
已知tan（α+π/4）=-1/7.求（cos2α+1）/【2cos(α-π/4)*sin(α+π/4)-sin2α】的值.
答
由已知：tan(α+π/4)=(tanα+1)/(1-tanα)=-1/7，可得tanα=-4/3，所以：
(cos2α+1)/[2cos(α-π/4)*sin(α+π/4)-sin2α]
=2cos^2(α)/{2cos(α-π/4)*sin(α+π/4)-sin[(a-π/4)+(a+π/4)]}
=2cos^2(α)/[cos(α-π/4)*sin(α+π/4)-sin(a-π/4)*cos(a+π/4)]
=2cos^2(a)/sin[(a+π/4)-(a-π/4)]
=2cos^2(a)/sinπ/2
=2cos^2(a)
=2/[tan^2(a)+1]
=2/[(-4/3)^2+1]
=18/25
答
tan(α+π/4)=(tanα+tanπ/4)/(1-tanα*tanπ/4)=(tanα+1)/(1-tanα)=-1/7解得tanα=-4/3（cos2α+1）/【2cos(α-π/4)*sin(α+π/4)-sin2α】=2cos²α/[sin2α+sin(π/2)-sin2α]=2/(1+tan²α)=2/[1+(-...