cos(π-α)sin(-π-α) / cos(11/2π-β)tan(3π+β)
=cos(π-α)sin[-(π+α)] / cos[4π+(3/2π-β)]tan[2π+(π+β)]
=-cosαsinα / (-sinβ)tanβ
=cosαsinαcosβ / (sinβ)^2
化简：cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)
3个回答
相关问题
-
化简:(1)cos(π/4+α)-cos(π/4-α) (2)cos(α+β)cosβ+sin(α+β)sinβ 求详解
-
化简：sin2(α+π)•cos(π+α)tan(π+α)•cos3(−α−π)•tan(−α−2π)．
-
已知α∈(―π﹨2,π﹨2),β∈(0,π),且sinα=√2cosβ,tanα=√3/tanβ,求α,β
-
化简：sin²（α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)
-
化简：sin(3π+α)•cos(π−α)•tan(3π2+α)cos(π3)•sin(π2−α)•cos(−α)=__
-
化简：（Ⅰ）sin(α−2π)cos(α+π)tan(α−99π)cos(π−α)sin(3π−α)sin(−α−π)；
-
化简：tan(π+α)cos(2π+α)sin(α−3π2)cos(−α−3π)sin(−3π−α)=______．
-
化简（1）tan(π−α)cos(2π−α)sin(−α+3π2)cos(−α−π)sin(−π−α)．
-
sin｛π－α｝cos｛2π－α｝tan｛－α＋π｝/sin｛π＋α｝化简
-
化简cos(π-α)sin(α+2π)/cos^2(-α-π)tan(2π-α