cos(π/2+α)=-sinα转换过程

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sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin（π-α）·cos（π/2+α）/sin（π+α）=
已知tanα=√2/2,cos2(π-α)+sin(π+α)cos(π-α)+2sin2(α-π)要详细过程~
化简[sin(π/2+α)*cos(π/2-α)]/cos(π+α)+[sin(π-α)*cos(π/2+α)]/sin
化简[sin(π/2-α）cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2
..sin^2（π+α）*cos（π+α）*cos（-α-2π）/tan（π+α）*sin^3（π/2+α）*sin（-
若sin(α-π）＝2cos（2π-α),求[sin(π－α）＋5cos（2π－α）]／[3cos（π－α）－sin（－
求值【sin(2π-α)sin(π+α)cos(-π-α)】/【sin(3π-α)-cos(π-α)】
已知f（α）=sin(π−α)cos(2π−α)cos(−α+32π)cos(π2−α)sin(−π−α)