已知tan(π⼀4-a)=-1⼀2,则cos2a=

-0.8

解
tana
=tan[π/4-(π/4-a)]
=[tan(π/4)-tan(π/4-a)]/[1+tan(π/4)tan(π/4-a]
=(1+1/2)/(1-1/2)
=3
cos2a
=cos²a-sin²a
=(cos²a-sin²a)/(cos²a+sin²a)
同除以cos²a
=(1-tan²a)/(1+tan²a)
=(1-9)/(1+9)
=-4/5

∵tan(π/4 - a )= tan [ - (a - π/4) ] = - tan (a - π/4) = - 1/2
∴tan ( a - π/4)= 1/2
又∵ tan ( a - π/4 ) = ( tan a - 1 ) / ( 1 + tan a ) = 1/2
∴ tan a =3
∴ a 属于 一三象限
∵cos²a= 1 / （tan²a + 1）
∴ cos²a = 1/10
∴cos2a=2cos²a-1=-4/5

cos2a=sin(π/2-2a)
=2tan(π/4-a)/[1+tan²(π/4-a)]
=2*(-1/2)/[1+(-1/2)²]
=-1/(5/4)=-4/5