已知α是第三象限角,且f(α)=sin(π-α)cos(2π-α)cos(-α+3π/2)/sin(-α-π)sin(-π-α)化简若cos(α-3π/2)=1/5,求f(α),若α=-1860°,求f(α)

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/04 21:52:49
已知α是第三象限角,且f(α)=sin(π-α)cos(2π-α)cos(-α+3π/2)/sin(-α-π)sin(-π-α)化简若cos(α-3π/2)=1/5,求f(α),若α=-1860°,求f(α)
x){}Km|6c5k|q˙^,dǔ4s5m347F&6>ߠoRo3!L{ER{:`t<ؔ UhkhafphMR>] b }ԱR׀yg.HHAՂ5)+h%B"[( o@A{ Pgkc&T P!ЛpBl:TCT4`!<0361!cB47/.H̳

已知α是第三象限角,且f(α)=sin(π-α)cos(2π-α)cos(-α+3π/2)/sin(-α-π)sin(-π-α)化简若cos(α-3π/2)=1/5,求f(α),若α=-1860°,求f(α)
已知α是第三象限角,且f(α)=sin(π-α)cos(2π-α)cos(-α+3π/2)/sin(-α-π)sin(-π-α)化简
若cos(α-3π/2)=1/5,求f(α),若α=-1860°,求f(α)

已知α是第三象限角,且f(α)=sin(π-α)cos(2π-α)cos(-α+3π/2)/sin(-α-π)sin(-π-α)化简若cos(α-3π/2)=1/5,求f(α),若α=-1860°,求f(α)
∵f(α)=sin(π-α)cos(2π-α)cos(-α+3π/2)/sin(-α-π)sin(-π-α)
=sin(α)cos(α)sin(-α)/sin² (α)
=-cosa
又α是第三象限角且cos(α-3π/2)=-sina=1/5
∴sina=-1/5 ,cosa=(-2√ 6)/5
(1)f(α)=-cosa=(2√ 6)/5
(2)f(-1860°)=-cos1860°=-cos[1440°+420°]=-cos[360°+60°]=-cos[60°]=-1/2