cos(3π/2+α)=-3/5,α为第四象限角,求tan(2π-α)
问题描述：
cos(3π/2+α)=-3/5,α为第四象限角,求tan(2π-α)
答
cosα=sqr(1-sinα^2)=sqr[1-(-5/13)^2]=sqr(1-25/169)=sqr144/169=12/13 tanα=sinα/cosα=(-5/13)/(12/13)=-5/12 由sin
答
cos(3π/2+α)=-3/5,α为第四象限角,求tan(2π-α)
cos(3π/2+α)=sinα=-3/5,α为第四象限角,故cosα=√(1-9/25)=4/5
∴tan(2π-α)=-tanα=-sinα/cosα=-(-3/5)/(4/5)=3/4