化简cos(3/2π+π)sin(-π+α)

2个回答

化简：sin(3π+α)•cos(π−α)•tan(3π2+α)cos(π3)•sin(π2−α)•cos(−α)=__
化简[sin(π/2-α）cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π
化简：tan(π+α)cos(2π+α)sin(α−3π2)cos(−α−3π)sin(−3π−α)=______．
化简：（Ⅰ）sin(α−2π)cos(α+π)tan(α−99π)cos(π−α)sin(3π−α)sin(−α−π)；
化简[sin(π/2+α)*cos(π/2-α)]/cos(π+α)+[sin(π-α)*cos(π/2+α)]/sin
（1）化简：sin(2π−α)cos(π+α)cos(3π2+α)cos(13π2−α)cos(π−α)sin(3π−α
化简（1）tan(π−α)cos(2π−α)sin(−α+3π2)cos(−α−π)sin(−π−α)．
化简①sin（π+α）×cos（3π/α）+sin（π/2+α）×cos（π+α）
化简：（1）cos(α+π)sin(−α)cos(−3π−α)sin(−α−4π)（2）cos(α−π2)sin(5π2
化简sin(2π−α)cos(π+α)cos(π−α)sin(3π−α)sin(−α−π)=−1sinα−1sinα．