已知cos(π+α)=1/2,计算sin(2π-α

已知cos(π+α)=1/2,计算sin(2π-α) sin[(2n+1)π+α]+sin[α-(2n+1)π]/sin

2个回答

化简：（1）sin[α+(2n+1)π]•2sin[α−(2n+1)π]sin(α−2nπ)cos(2nπ−α)(n∈Z
已知1/2sin(α-π)=cos(α-2π),求sin（π-α）+5cos（2π-α)/3sin（3π/2+α-sin
化简sin(2π−α)cos(π+α)cos(π−α)sin(3π−α)sin(−α−π)=−1sinα−1sinα．
已知：sin(π+α）=-1/2 计算：sin(π/2+α） cos(α-3π/2) tan(π/2-α）
已知cos(π+α)=-1/2,且α是第四象限角,计算｛sin[α+（2n+1）π]+sin[α-（2n+1)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(π/2+α)·cos(π/2-α)/cos(π+α)+sin（π-α）·cos（π/2+α）/sin（π+α）=
已知f（α）=sin(α−3π)•cos(π+α)cos(2π−α)•sin(−π−α)•sin(3π2−α)
已知f(α)＝sin(α−3π)•cos(2π−α)•sin(−α+3π2)cos(−π−α)•sin(−π−α)