已知f(a)=tan(π-α)cos(2π-α)sin(π/2+α)/cos(-α-π)

解
f(a)=[tan(π-a)cos(2π-a)sin(π/2+a)]/cos(-a-π)
=[-tanacosacosa]/cos(a+π)
=(-cosasina)/(-cosa)
=sina
sina=4/5
∵a是第二象限
∴cosa

f(a)=tan(π-α)cos(2π-α)sin(π/2+α)/cos(-α-π)
=【-tanα】【cosα】【cosα】/【-cosα】
=tana cosa
=sina
sina = 4/5，cosa = -3/5
cos2a = cos²a - sin²a = -7/25
sin2a = 2sinacosa = -24/25
cos(2α+π/4)=cos2a cos(π/4)-sin2asin(π/4) = (-7/25)(√2/2)-(-24/25)(√2/2) = 17√2 / 50