已知tan（α+π/4）=-1/7.求（cos2α+1）/【2cos(α-π/4)*sin(α+π/4)-sin2α】的值.
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已知tan（α+π/4）=-1/7.求（cos2α+1）/【2cos(α-π/4)*sin(α+π/4)-sin2α】的值.
已知tan（α+π/4）=-1/7.求（cos2α+1）/【2cos(α-π/4)*sin(α+π/4)-sin2α】的值.
已知tan（α+π/4）=-1/7.求（cos2α+1）/【2cos(α-π/4)*sin(α+π/4)-sin2α】的值.
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+(-4/3)²]
=18/25
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)-si...
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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
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