tan(7π+α)=tan(6π+（π+α）)=tan(π+α)=tanα
sin(α-3π)+cos（π-α）/sin(-α)-cos（π+α）
=sin(-(3π-a))-cosα/(-sinα)+cosα
=-sin(3π-α)-cosα/(-sinα)+cosα
=sinα-cosα/-（sinα-cosα）
=-1
若tan(7π+α)=a,则sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)=?
4个回答
相关问题
-
若tan(7π+α)=α,求sin(α+3π)+cos(π-α)/sin(-α)-cos(π+α)的值
-
设tan（π+α）=2，则sin(α−π)+cos(π−α)sin(π+α)−cos(π+α)=（ ）
-
tanα=m,则sin(α+3π)+cos(π+α)/sin(－α)-cos(π+α)等于?
-
..sin^2（π+α）*cos（π+α）*cos（-α-2π）/tan（π+α）*sin^3（π/2+α）*sin（-
-
化简：（Ⅰ）sin(α−2π)cos(α+π)tan(α−99π)cos(π−α)sin(3π−α)sin(−α−π)；
-
tan(π+α)=3 则 sin(π+α)cos(π-α)=?
-
已知sin(7π-α)-3cos(3π/2+α)=2,则(sin(π-α)+cos(π+α))/(sinα+cos(-α
-
tan(3π-α)/sin(π-α)sin(3/2π-α)+sin(2π-α)cos(α-7/2π)/sin(3/2π+
-
化简：sin(3π+α)•cos(π−α)•tan(3π2+α)cos(π3)•sin(π2−α)•cos(−α)=__
-
化简：sin²（α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)