Question

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

化简f（a）
若f(a)=4/5,且α是第二象限角。求cos(2α+π/4)的值
急急急，在线等，求详解啊！！谢谢！！！

Answer 1

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<0
∵sin²a+cos²a=1
∴cosa=-3/5
∴sin2a=2sinacosa=-24/25
cos2a=2cos²a-1=-7/25
∴cos(2a+π/4)
=cos2acosπ/4-sin2asinπ/4
=√2/2(-7/25+24/25)
=17√2/50

Answer 2

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