化简：[sin(﹣α)cos(π－α)]∕[tan(π＋α)sin﹙3π／2＋α)]

sinα+cosα 化简 化简：[sin(﹣α)cos(π－α)]∕[tan(π＋α)sin﹙3π／2＋α)] 1.证明 cos(α-π/2)=sinα sin（α－π/2)=﹣sinαcos(α-π/2)=sinαsin（α－π/2)=﹣sinα 化简:sin(2π-α)cos(π+α)cos(11π/2-α)/cos(π-α)sin(3π-α)sin(-π-α)sin(9π/2+α) [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简 (1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 （3/2*π 化简sin(-π/2-α)sin(πα)cos(-α-π)/cos(π-α)sin(3π α) 化简sin(π/4+α)cosα-sin(π/4-α)sinα 2sinα*cosα*cos(2π-α)+cos(π+2α)*cosα*tanα化简! 化简：tanα(cosα-sinα)+sinα(sinα+tanα)/1+cosα. 化简sin(α-5π)/cos(3π-α)×cos(π/2-α)/sin(α-3π)×cos(8π-α)/sin(-α-4π) 化简[sin(π/2-α）cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π+α) 化简sin(α-2π)+sin(-a-3π)cos(α-3π)/cos(π-α)-cos(-π-α)cos(α-4π) (sinα+cosα)(sin2α-sinαcosα+cos2α)(sinα＋cosα)(sin的平方α－sinαcosα＋cos的平方α)＝a(1－sinαcosα)为什么? 化简：(sin²αtanα+cos²α/tanα+2sinαcosα)sinαcosα 化简(sin²α/1-cosα)-cosα 化简：cos(π/3+α)+sin(π/6+α) 化简sin(π/4-α)+cos(π/4+α)