tan(a+π/4)=k,则cos2a=
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tan(a+π/4)=k,则cos2a=tan(a+π/4)=k,则cos2a=tan(a+π/4)=k,则cos2a=由tan的和角公式：tan(a+π/4)=(tana+tanπ/4)/(1-ta
tan(a+π/4)=k,则cos2a=
tan(a+π/4)=k,则cos2a=
tan(a+π/4)=k,则cos2a=
由 tan 的和角公式：
tan(a+π/4)
=(tana+tanπ/4)/(1-tanatanπ/4)
=(1+tana)/(1-tana)
=k
所以 tana = (k-1)/(k+1).sina/cosa = tana = (k-1)/(k+1).
结合 (sina)^2+(cosa)^2=1 即知 (sina)^2 = (k-1)^2/(2k^2+2).
再由倍角公式即知 cos2a = 1-2(sina)^2 = 2k/(k^2+1).
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tan(a+π/4)=k,则cos2a=?
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