tan（π/4+α）=k 求cos2α

cos2α
=sin(Pi/2+2α)
=sin( 2(Pi/4+α) )
=2tan(π/4+α))/(1+tan^2(π/4+α))
=2k/(1+k^2)

tanπ/4=1
所以tan（π/4+α）=(1+tanα)/(1-tanα)=k
tanα=(k-1)/(k+1)

因为sin²α+cos²α=1
所以cos2α=cos²α-sin²α=(cos²α-sin²α)/(cos²α+sin²α)
上下除cos²α
则sin²α/cos²α=tan²α
所以cos2α=(1-tan²α)(1+tan²α)
=[1-(k-1)²/(k+1)²]/[1+(k-1)²/(k+1)²]
=[(k+1)²-(k-1)²]/[(k+1)²+(k-1)²]
=2k/(k²+1)