Question

已知f(a)=sin(π-a)cos(2π-a)tan(-a+3π/2)/tan(π/2+a)sin(-π-a)若a是第三象限角且a=-31π/3求f(a)值

解题步骤

Answer 1

解:f(a)=sin(π-a)cos(2π-a)tan(-a+3π/2)/tan(π/2+a)sin(-π-a)
=sinacosatan(π/2-a)/[cot(-a)sina
=cosacota/(-cota)
=-cosa
把a=-31π/3=-5×2π-π/3代入得
f(a)=-cos(-5×2π-π/3)
=-cos(-π/3)
=-cosπ/3
=-1/2

Answer 2

解：原式变换为：f(a)=sinα*cosα*tan[π+(π/2-α)]/(-ctgα*sinα).
f(α)=sinα*cosα(-ctgα)/(-ctgα*sinα].
=cosα
=cos(-31π/3)
=cos(-10π-π/3).
=cos[-(10π+π/3)
=cos(10π+π/3).
= cosπ/3
∴f(a)=1/2.