作业帮若tan(7π+α)=a,则sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)=?
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若tan(7π+α)=a,则sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)=?
若tan(7π+α)=a,则sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)=?
若tan(7π+α)=a,则sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)=?
tan(7π+α)=tan(6π+（π+α）)=tan(π+α)=tanα
sin(α-3π)+cos（π-α）/sin(-α)-cos（π+α）
=sin(-(3π-a))-cosα/(-sinα)+cosα
=-sin(3π-α)-cosα/(-sinα)+cosα
=sinα-cosα/-（sinα-cosα）
=-1
tan(7π+α)=α,
tanα=α,
sin(α+3π)+cos(π-α)/sin(-α)-cos(π+α)
=(-sinα-cosα)/(-sinα+cosα), (分子分母同除以cosα)
=(-tanα-1)/(-tanα+1)
=(-α-1)/(-α+1)
=(α+1)/(α-1)
tan(7π+α)=tan(α)=a;cos²(α)=1/(1+a²);sin²(α)=a²/(1+a²);cos(α)=±√1/(1+a²);
sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)=-sin(α)+cos(α)/sinα)+cos(α)=1/a±(1-a)1/(1+a²);
郭敦顒回答：
∵tan(7π+α)=a
∴tan(7π+α)= tan(π+α)= tanα=a
sin(α-3π)= sin(π+α)=－sinα
cos(π-α)/sin(-α)＝－cosα／（－sinα）＝cosα／sinα＝ctgα
cos(π+α)＝－cosα
∴sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)
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郭敦顒回答：
∵tan(7π+α)=a
∴tan(7π+α)= tan(π+α)= tanα=a
sin(α-3π)= sin(π+α)=－sinα
cos(π-α)/sin(-α)＝－cosα／（－sinα）＝cosα／sinα＝ctgα
cos(π+α)＝－cosα
∴sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)
＝－sinα＋ctgα－cosα
＝－sinα＋cosα／sinα－cosα
＝cosα（1／sinα－1）－sinα
tanα＝sinα／cosα，sinα＝tanαcosα＝acosα
∴cosα（1／sinα－1）－sinα
＝cosα（1／acosα－1）－acosα
＝1／a- cosα- acosα
∴sin(α-3π)+cos(π-α)/sin(-α)-cos(π+α)
=1／a- cosα- acosα
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