1.证明 cos(α-π/2)=sinα sin(α-π/2)=﹣sinα
1个回答
cos(α-π/2)=cosαcosπ/2+sinαsinπ/2 =sinα (cosπ/2=0,sinπ/2=1)
sin(α-π/2)=sinαcosπ/2-sinπ/2cosα=-cosα
相关问题
化简sin(2π−α)cos(π+α)cos(π−α)sin(3π−α)sin(−α−π)=−1sinα−1sinα.
sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin(π-α)·cos(π/2+α)/sin(π+α)=
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
化简[sin(π/2+α)*cos(π/2-α)]/cos(π+α)+[sin(π-α)*cos(π/2+α)]/sin
..sin^2(π+α)*cos(π+α)*cos(-α-2π)/tan(π+α)*sin^3(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
化简(1)sin(α+π)cos(-α)sin(-α-π) (2)sin³(-α)cos(2π+α)tan(-
求值【sin(2π-α)sin(π+α)cos(-π-α)】/【sin(3π-α)-cos(π-α)】
已知tanα=2,sinα+cosα<0,求[sin(2π-α)*sin(π+α)*cos(-π+α)]/[sin(3π