化简sin(-α)cos(-α-π)tan(2π+α)过程详细点关键这个我没学过

sinα+cosα 化简 若sinα-cosα=-1/5,α∈(3/2π,7/4π),求sinα与cosα 若sin（α+β）=1/2,sin(α-β)=1/3,求log√5（ta若sinα-cosα=-1/5,α∈(3/2π,7/4π),求sinα与cosα 若sin（α+β）=1/2,sin(α-β)=1/3,求log√5（tan*cotβ）若sinα-2cosα/3sinα+5 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) 已知sin[(π/4)-x)]=5/13,0<x<π/4.求cos2x/(cosπ/4+已知sin[(π/4)-x)]=5/13,0<x<π/4.求cos2x/[cos(π/4)+x] 已知0<β<π/2<α<π,cos[α-(β/2)]=-1/9,sin[(α/2)-β]=2/3 求cos(α+β) 已知α,β∈(0,π)且tan(α-β)=1/2,ta 化简sin(α-2π)+sin(-a-3π)cos(α-3π)/cos(π-α)-cos(-π-α)cos(α-4π) 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) 化简：2sin（π+α）cos（π-α）