Question

已知f(a)=sin(π-a)cos(2π-a)tan(-π+3π/2)/tan(π/2)sin(-π-a)若a是第三象限角且cos(a-3π/2)=1/5求f

求步骤 答案是 5分之2根6

Answer 1

a是第三象限角
cos(a-3π/2)=1/5
cos(a-3π/2)=cos(-2π+(π/2+a))=cos(π/2+a)=cos(π-（π/2-a))=-cos(π/2-a)=-sina=1/5
sina=-1/5
cosa=-根号(1-sin^2a)=-2根号6 /5
f(a)=sin(π-a)cos(2π-a)tan(-π+3π/2) / {tan(π/2)sin(-π-a)}
=sinacosa tan(-π+π+π/2) / {tan(π/2)[-sin(-a)]}
=sinacosa tan(π/2) / {tan(π/2)sina}
=cosa
=-2根号6 /5

Answer 2

解：f(a)=sin(π-a)cos(2π-a)tan(-π+3π/2)/tan(π/2)sin(-π-a)
=sina cosa tan(π/2)/tan(π/2)sina
=cos a
cos(a-3π/2)=cos (a-3π/2)+2π=cos π/2+a=-sina=1/5
因为a是第三象限角
f（a）=cos a= -√(1-sin^2)=-2√6/5